Problem
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)func evenOddPalindrome(n: Int) -> (Int, Int) { ...XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13))
XCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6))
XCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2))
XCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8))
XCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6))
XCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6))
XCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5))
XCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1))- 0.70 Compiled
- 0.59 Passed
Solutions
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|>if isPalindrome(i) { // missing argument label 'n:' in callif isEven(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == num
}
func isEven(_ num: Int) -> Bool {
return num % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == num
}
func isEven(_ num: Int) -> Bool {
return num % 2 == 0
}
<|file_separator|> // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == num
}
func isEven(_ num: Int) -> Bool {
return num % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == num
}
func isEven(_ num: Int) -> Bool {
return num % 2 == 0
}
<|file_separator|> // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var count = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
count += 1
if isPalindrome(i, isEven: true) {
odd += 1
}
}
}
return (count, odd)
}
func isPalindrome(_ num: Int, isEven: Bool = false) -> Bool {
var num = num
var rev = 0
while num > 0 {
rev = rev * 10 + num % 10
num /= 10
}
if isEven {
return rev % 2 == 0
}
return rev == num
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var count = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
count += 1
if isPalindrome(i, isEven: true) {
odd += 1
}
}
}
return (count, odd)
}
func isPalindrome(_ num: Int, isEven: Bool = false) -> Bool {
var num = num
var rev = 0
while num > 0 {
rev = rev * 10 + num % 10
num /= 10
}
if isEven {
return rev % 2 == 0
}
return rev == num
}
<|file_separator|> // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return n == reversed
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return n == reversed
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|>if isPalindrome(i) { // missing argument label 'n:' in callif isEven(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
<|file_separator|>func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|> // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|> // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|>if isPalindrome(i) { // missing argument label 'n:' in callif isEven(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return n == reversed
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return n == reversed
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|> // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if digits == reversedDigits { // cannot convert value of type 'ReversedCollection<[Int]>' to expected argument type '[Int]'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
if digits.count % 2 == 0 {
// Even number of digits
if digits == digits.reversed() {
evenCount += 1
}
} else {
// Odd number of digits
if digits.dropFirst() == digits.dropLast().reversed() {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
if digits.count % 2 == 0 {
// Even number of digits
if digits == digits.reversed() {
evenCount += 1
}
} else {
// Odd number of digits
if digits.dropFirst() == digits.dropLast().reversed() {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if digits.dropFirst() == digits.dropLast().reversed() { // binary operator '==' cannot be applied to operands of type 'DropFirstSequence<[Int]>' and '[Int]'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
let reversed = digits.reversed()
if digits == reversed {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
let reversed = digits.reversed()
if digits == reversed {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if digits == reversed { // cannot convert value of type 'ReversedCollection<[Int]>' to expected argument type '[Int]'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let numString = String(num)
let reversedString = String(numString.reversed())
if numString == reversedString {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let numString = String(num)
let reversedString = String(numString.reversed())
if numString == reversedString {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let digits = String(num).map { Int(String($0))! }
let isPalindrome = digits == digits.reversed()
if isPalindrome {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let digits = String(num).map { Int(String($0))! }
let isPalindrome = digits == digits.reversed()
if isPalindrome {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let digits = String(num).map { Int(String($0))! }
let isPalindrome = digits == digits.reversed()
if isPalindrome {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let digits = String(num).map { Int(String($0))! }
let isPalindrome = digits == digits.reversed()
if isPalindrome {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let digits = String(num).map { Int(String($0))! }
if digits.count % 2 == 0 {
// Even number of digits
if digits == digits.reversed() {
evenCount += 1
}
} else {
// Odd number of digits
if digits.dropLast() == digits.dropLast(digits.count / 2).reversed() {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let digits = String(num).map { Int(String($0))! }
if digits.count % 2 == 0 {
// Even number of digits
if digits == digits.reversed() {
evenCount += 1
}
} else {
// Odd number of digits
if digits.dropLast() == digits.dropLast(digits.count / 2).reversed() {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let numStr = String(num)
let reversedStr = String(numStr.reversed())
if numStr == reversedStr {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let numStr = String(num)
let reversedStr = String(numStr.reversed())
if numStr == reversedStr {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let digits = String(num).map { Int(String($0))! }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let digits = String(num).map { Int(String($0))! }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if digits == reversedDigits { // cannot convert value of type 'ReversedCollection<[Int]>' to expected argument type '[Int]'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if digits == reversedDigits { // cannot convert value of type 'ReversedCollection<[Int]>' to expected argument type '[Int]'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num = num / 10
}
return reverse == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num = num / 10
}
return reverse == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isEvenPalindrome(i) {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
func isEvenPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = 10 * reversed + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isEvenPalindrome(i) {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
func isEvenPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = 10 * reversed + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isEvenPalindrome(i) {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
func isEvenPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isEvenPalindrome(i) {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
func isEvenPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num = num / 10
}
return reverse == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num = num / 10
}
return reverse == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
let digit = n % 10
reverse = reverse * 10 + digit
n /= 10
}
return reverse == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
let digit = n % 10
reverse = reverse * 10 + digit
n /= 10
}
return reverse == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return reverse == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return reverse == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let count = str.count
for i in 0..<count/2 {
if str[str.index(str.startIndex, offsetBy: i)]!= str[str.index(str.startIndex, offsetBy: count - i - 1)] {
return false
}
}
return true
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let count = str.count
for i in 0..<count/2 {
if str[str.index(str.startIndex, offsetBy: i)]!= str[str.index(str.startIndex, offsetBy: count - i - 1)] {
return false
}
}
return true
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
```if str[str.index(str.startIndex, offsetBy: i)]!= str[str.index(str.startIndex, offsetBy: count - i - 1)] { // '=' must have consistent whitespace on both sidesif str[str.index(str.startIndex, offsetBy: i)]!= str[str.index(str.startIndex, offsetBy: count - i - 1)] { // cannot force unwrap value of non-optional type 'Character'if str[str.index(str.startIndex, offsetBy: i)]!= str[str.index(str.startIndex, offsetBy: count - i - 1)] { // use of '=' in a boolean context, did you mean '=='?func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
func isPalindrome(_ n: Int) -> Bool {
let s = String(n)
let count = s.count
for i in 0..<count/2 {
if s[s.index(s.startIndex, offsetBy: i)]!= s[s.index(s.startIndex, offsetBy: count - 1 - i)] {
return false
}
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
func isPalindrome(_ n: Int) -> Bool {
let s = String(n)
let count = s.count
for i in 0..<count/2 {
if s[s.index(s.startIndex, offsetBy: i)]!= s[s.index(s.startIndex, offsetBy: count - 1 - i)] {
return false
}
}
return true
}
```if s[s.index(s.startIndex, offsetBy: i)]!= s[s.index(s.startIndex, offsetBy: count - 1 - i)] { // '=' must have consistent whitespace on both sidesif s[s.index(s.startIndex, offsetBy: i)]!= s[s.index(s.startIndex, offsetBy: count - 1 - i)] { // use of '=' in a boolean context, did you mean '=='?if s[s.index(s.startIndex, offsetBy: i)]!= s[s.index(s.startIndex, offsetBy: count - 1 - i)] { // cannot force unwrap value of non-optional type 'Character'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isEven(i) {
if isPalindrome(i) {
even += 1
}
} else {
if isPalindrome(i) {
odd += 1
}
}
}
return (even, odd)
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isEven(i) {
if isPalindrome(i) {
even += 1
}
} else {
if isPalindrome(i) {
odd += 1
}
}
}
return (even, odd)
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let r = String(s.reversed())
if s == r {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let r = String(s.reversed())
if s == r {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return n == reversed
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return n == reversed
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(s.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(s.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
let num = i
let reversedNum = String(num).reversed()
let reversedNumInt = Int(reversedNum)!
if num == reversedNumInt {
if num % 2 == 0 {
countEven += 1
} else {
countOdd += 1
}
}
}
return (countEven, countOdd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
let num = i
let reversedNum = String(num).reversed()
let reversedNumInt = Int(reversedNum)!
if num == reversedNumInt {
if num % 2 == 0 {
countEven += 1
} else {
countOdd += 1
}
}
}
return (countEven, countOdd)
}
```let reversedNumInt = Int(reversedNum)! // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reverse = String(num).reversed()
let reversedNum = Int(reverse)!
if num == reversedNum {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reverse = String(num).reversed()
let reversedNum = Int(reverse)!
if num == reversedNum {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```let reversedNum = Int(reverse)! // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reversedNum = Int(String(num).reversed())!
if num == reversedNum {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reversedNum = Int(String(num).reversed())!
if num == reversedNum {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```let reversedNum = Int(String(num).reversed())! // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reverse = String(num).reversed()
let reversedNum = Int(reverse)!
if num == reversedNum {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reverse = String(num).reversed()
let reversedNum = Int(reverse)!
if num == reversedNum {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```let reversedNum = Int(reverse)! // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evens = 0
var odds = 0
for i in 1...n {
let num = i
let reversedNum = String(num).reversed()
let reversedNumInt = Int(reversedNum)!
if num == reversedNumInt {
if num % 2 == 0 {
evens += 1
} else {
odds += 1
}
}
}
return (evens, odds)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evens = 0
var odds = 0
for i in 1...n {
let num = i
let reversedNum = String(num).reversed()
let reversedNumInt = Int(reversedNum)!
if num == reversedNumInt {
if num % 2 == 0 {
evens += 1
} else {
odds += 1
}
}
}
return (evens, odds)
}
```let reversedNumInt = Int(reversedNum)! // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reversedNum = String(num).reversed()
let reversedNumInt = Int(reversedNum)!
if num == reversedNumInt {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reversedNum = String(num).reversed()
let reversedNumInt = Int(reversedNum)!
if num == reversedNumInt {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```let reversedNumInt = Int(reversedNum)! // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reversedNum = String(num).reversed()
let reversedNumInt = Int(reversedNum)!
if num == reversedNumInt {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reversedNum = String(num).reversed()
let reversedNumInt = Int(reversedNum)!
if num == reversedNumInt {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```let reversedNumInt = Int(reversedNum)! // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var count = 0
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reversedNum = String(num).reversed()
let reversedNumInt = Int(reversedNum)!
if num == reversedNumInt {
count += 1
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var count = 0
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reversedNum = String(num).reversed()
let reversedNumInt = Int(reversedNum)!
if num == reversedNumInt {
count += 1
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```let reversedNumInt = Int(reversedNum)! // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i.description
let reverseNumber = String(number.reversed())
if number == reverseNumber {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i.description
let reverseNumber = String(number.reversed())
if number == reverseNumber {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i.description
if num == String(num.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i.description
if num == String(num.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reversedNumber = number.reversed()
let isPalindrome = number == reversedNumber
if isPalindrome {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reversedNumber = number.reversed()
let isPalindrome = number == reversedNumber
if isPalindrome {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let reversedNumber = number.reversed() // value of type 'Int' has no member 'reversed'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reversed = Int(String(num).reversed())
if num == reversed {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reversed = Int(String(num).reversed())
if num == reversed {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```let reversed = Int(String(num).reversed()) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverse = Int(String(number))
let reversedNumber = Int(String(reverse))
if number == reversedNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverse = Int(String(number))
let reversedNumber = Int(String(reverse))
if number == reversedNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let reversedNumber = Int(String(reverse)) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reversed = Int(String(num).reversed())
if num == reversed {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let reversed = Int(String(num).reversed())
if num == reversed {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```let reversed = Int(String(num).reversed()) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let num = i
let reverse = Int(String(num).reversed())
if num == reverse {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let num = i
let reverse = Int(String(num).reversed())
if num == reverse {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let reverse = Int(String(num).reversed()) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverse = Int(String(number))
let reversed = Int(String(reverse))
if number == reversed {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverse = Int(String(number))
let reversed = Int(String(reverse))
if number == reversed {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let reversed = Int(String(reverse)) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reversedNumber = Int(String(number).reversed())
if number == reversedNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reversedNumber = Int(String(number).reversed())
if number == reversedNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let reversedNumber = Int(String(number).reversed()) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reversedNumber = Int(String(number).reversed())
if number == reversedNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reversedNumber = Int(String(number).reversed())
if number == reversedNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let reversedNumber = Int(String(number).reversed()) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let numString = String(i)
let reversedNumString = String(numString.reversed())
if numString == reversedNumString {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let numString = String(i)
let reversedNumString = String(numString.reversed())
if numString == reversedNumString {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let numString = String(i)
let reversedNumString = String(numString.reversed())
let reversedNum = Int(reversedNumString)!
if i == reversedNum {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let numString = String(i)
let reversedNumString = String(numString.reversed())
let reversedNum = Int(reversedNumString)!
if i == reversedNum {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = Int(String(i))!
let reverse = Int(String(num.reversed()))!
if num == reverse {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = Int(String(i))!
let reverse = Int(String(num.reversed()))!
if num == reverse {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```let reverse = Int(String(num.reversed()))! // value of type 'Int' has no member 'reversed'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = Int(String(i))!
let reverse = Int(String(num.reversed()))!
if num == reverse {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = Int(String(i))!
let reverse = Int(String(num.reversed()))!
if num == reverse {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```let reverse = Int(String(num.reversed()))! // value of type 'Int' has no member 'reversed'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = Int(String(i))!
let reverse = Int(String(num.reversed()))!
if num == reverse {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = Int(String(i))!
let reverse = Int(String(num.reversed()))!
if num == reverse {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```let reverse = Int(String(num.reversed()))! // value of type 'Int' has no member 'reversed'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = Int(String(i))!
let reverse = Int(String(num.reversed()))!
if num == reverse {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = Int(String(i))!
let reverse = Int(String(num.reversed()))!
if num == reverse {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```let reverse = Int(String(num.reversed()))! // value of type 'Int' has no member 'reversed'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let string = String(i)
let reverse = String(string.reversed())
if string == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let string = String(i)
let reverse = String(string.reversed())
if string == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let reverse = String(s.reversed())
if s == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let reverse = String(s.reversed())
if s == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var count = (0, 0)
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
count = (count.0 + (i % 2 == 0? 1 : 0), count.1 + (i % 2 == 1? 1 : 0))
}
}
return count
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var count = (0, 0)
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
count = (count.0 + (i % 2 == 0? 1 : 0), count.1 + (i % 2 == 1? 1 : 0))
}
}
return count
}
```count = (count.0 + (i % 2 == 0? 1 : 0), count.1 + (i % 2 == 1? 1 : 0)) // cannot convert value of type '(Bool, Int)' to expected argument type 'Int'count = (count.0 + (i % 2 == 0? 1 : 0), count.1 + (i % 2 == 1? 1 : 0)) // expected ',' separatorcount = (count.0 + (i % 2 == 0? 1 : 0), count.1 + (i % 2 == 1? 1 : 0)) // cannot convert value of type '(Bool, Int)' to expected argument type 'Int'count = (count.0 + (i % 2 == 0? 1 : 0), count.1 + (i % 2 == 1? 1 : 0)) // cannot use optional chaining on non-optional value of type 'Int'count = (count.0 + (i % 2 == 0? 1 : 0), count.1 + (i % 2 == 1? 1 : 0)) // expected ',' separatorcount = (count.0 + (i % 2 == 0? 1 : 0), count.1 + (i % 2 == 1? 1 : 0)) // cannot use optional chaining on non-optional value of type 'Int'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ x: Int) -> Bool {
let s = String(x); return s == String(s.reversed()) }
```if isPalindrome(i) { // cannot find 'isPalindrome' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ x: Int) -> Bool {
let s = String(x); return s == String(s.reversed()) }
```if isPalindrome(i) { // cannot find 'isPalindrome' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
``` // XCTAssertTrue failedimport Foundation
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}```swift
import Foundation
/*
Function to multiply two matrices represented as 2D arrays of Doubles.
Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
*/
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}
/*
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
*/
func evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven, odd, odd: Int, odd: Int {
Int, odd: Int {
IntXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scopeimport Foundation
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}```swift
import Foundation
/*
Function to multiply two matrices represented as 2D arrays of Doubles.
Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
*/
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}
/*
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
*/
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even: Int {
var even:XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scopeimport Foundation
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}```swift
import Foundation
/*
Function to multiply two matrices represented as 2D arrays of Doubles.
Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
*/
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}
/*
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
*/
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even: Int, Int, Int, Int {
IntXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scopeimport Foundation
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}```swift
import Foundation
/*
Function to multiply two matrices represented as 2D arrays of Doubles.
Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
*/
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}
/*
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
*/
func evenOddPalindrome(n: Int) -> (Int, Int) {
XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scopeimport Foundation
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}```swift
import Foundation
/*
Function to multiply two matrices represented as 2D arrays of Doubles.
Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
*/
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}
/*
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
*/
func evenOddPalindrome(n: Int) -> (Int, Int) {
XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scopeimport Foundation
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}```swift
import Foundation
/*
Function to multiply two matrices represented as 2D arrays of Doubles.
Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
*/
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}
/*
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
*/
func evenOddPalindrome(n: Int) -> (Int, Int) {
XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scopeimport Foundation
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}```swift
import Foundation
/*
Function to multiply two matrices represented as 2D arrays of Doubles.
Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
*/
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}
/*
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
*/
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scopeimport Foundation
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}```swift
import Foundation
/*
Function to multiply two matrices represented as 2D arrays of Doubles.
Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
*/
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}
/*
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
*/
func evenOddPalindrome(n: Int) -> (Int, Int) {
var count = 0
XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scopeimport Foundation
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}```swift
import Foundation
/*
Function to multiply two matrices represented as 2D arrays of Doubles.
Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
*/
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}
/*
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
*/
func evenOddPalindrome(n: Int) -> (Int, Int) {XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scopeimport Foundation
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}```swift
import Foundation
/*
Function to multiply two matrices represented as 2D arrays of Doubles.
Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
*/
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
guard lhs[0].count == rhs.count else {
return nil
}
var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
for i in 0..<lhs.count {
for j in 0..<rhs[0].count {
for k in 0..<rhs.count {
result[i][j] += lhs[i][k] * rhs[k][j]
}
}
}
return result
}
/*
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
*/
func evenOddPalindrome(n: Int) -> (Int, Int) {
XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
For n = 10
Here is a palindrome palindrome numbers are 1, 2, 2, so we need to check each number of even and odd ones are 1, 2, 2XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
For n = 12, the palindrome: (4, the palXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
Explanation:
For n = 12, the palindromes are 1, 1, 2, 2, 3, 2, 4, 4, 4, 6, 8, 9, 9, 8, 8, 8, 11, 11, 22, 22, 11, 1, 1, 1 are palXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
For n = 10, so the palindrome palindromes are 1, so the palindromes are 1, 2, 2, 2, 2XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
Let's say n = 10, the function is 10, the outputXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
Given that are 1- 9, n = 10, the palindXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
For n = 12, the palindrome:
There are 4 palindrome areXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
For n = 12, the palindrome palindromes are 1,XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
For n = 10, the palindrome number of even and odd palXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
var isPalindrome = true
for j in 0..<str.count/2 {
if str[j] != str[str.count - 1 - j] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
var isPalindrome = true
for j in 0..<str.count/2 {
if str[j] != str[str.count - 1 - j] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if str[j] != str[str.count - 1 - j] { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.if str[j] != str[str.count - 1 - j] { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i.isPalindrome() {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
extension Int {
func isPalindrome() -> Bool {
var num = self
var reverseNum = 0
var remainder: Int
while num > 0 {
remainder = num % 10
reverseNum = reverseNum * 10 + remainder
num /= 10
}
return self == reverseNum
}
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i.isPalindrome() {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
extension Int {
func isPalindrome() -> Bool {
var num = self
var reverseNum = 0
var remainder: Int
while num > 0 {
remainder = num % 10
reverseNum = reverseNum * 10 + remainder
num /= 10
}
return self == reverseNum
}
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i.isPalindrome() {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
extension Int {
func isPalindrome() -> Bool {
var number = self
var reverseNumber = 0
var remainder: Int
while number > 0 {
remainder = number % 10
reverseNumber = reverseNumber * 10 + remainder
number = number / 10
}
return self == reverseNumber
}
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i.isPalindrome() {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
extension Int {
func isPalindrome() -> Bool {
var number = self
var reverseNumber = 0
var remainder: Int
while number > 0 {
remainder = number % 10
reverseNumber = reverseNumber * 10 + remainder
number = number / 10
}
return self == reverseNumber
}
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0, oddCount = 0
for i in 1...n {
if i.isPalindrome() {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
extension Int {
func isPalindrome() -> Bool {
var number = self
var reverseNumber = 0
var remainder: Int
while number > 0 {
remainder = number % 10
reverseNumber = reverseNumber * 10 + remainder
number /= 10
}
return self == reverseNumber
}
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0, oddCount = 0
for i in 1...n {
if i.isPalindrome() {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
extension Int {
func isPalindrome() -> Bool {
var number = self
var reverseNumber = 0
var remainder: Int
while number > 0 {
remainder = number % 10
reverseNumber = reverseNumber * 10 + remainder
number /= 10
}
return self == reverseNumber
}
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let s = String(n)
let reversed = String(s.reversed())
return s == reversed
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let s = String(n)
let reversed = String(s.reversed())
return s == reversed
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0, oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0, oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0, oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0, oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0, oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0, oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i.isPalindrome() {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
extension Int {
func isPalindrome() -> Bool {
var number = self
var reverse = 0
var remainder: Int
while number != 0 {
remainder = number % 10
reverse = reverse * 10 + remainder
number /= 10
}
return self == reverse
}
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i.isPalindrome() {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
extension Int {
func isPalindrome() -> Bool {
var number = self
var reverse = 0
var remainder: Int
while number != 0 {
remainder = number % 10
reverse = reverse * 10 + remainder
number /= 10
}
return self == reverse
}
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var n = num
var rev = 0
while n > 0 {
rev = rev * 10 + n % 10
n = n / 10
}
return num == rev
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var n = num
var rev = 0
while n > 0 {
rev = rev * 10 + n % 10
n = n / 10
}
return num == rev
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var n = num
var rev = 0
while n > 0 {
let dig = n % 10
rev = rev * 10 + dig
n = n / 10
}
return num == rev
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var n = num
var rev = 0
while n > 0 {
let dig = n % 10
rev = rev * 10 + dig
n = n / 10
}
return num == rev
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
import Foundation
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let reversedString = String(numberString.reversed())
return numberString == reversedString
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let reversedString = String(numberString.reversed())
return numberString == reversedString
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let reversedString = String(numberString.reversed())
return numberString == reversedString
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let reversedString = String(numberString.reversed())
return numberString == reversedString
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
if isPalindrome(i) { // cannot find 'isPalindrome' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
countEven += 1
} else {
countOdd += 1
}
}
}
return (countEven, countOdd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let strRev = String(str.reversed())
return str == strRev
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
countEven += 1
} else {
countOdd += 1
}
}
}
return (countEven, countOdd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let strRev = String(str.reversed())
return str == strRev
}
evenOddPalindrome(n: 3)
evenOddPalindrome(n: 12)
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i == reverse(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i == reverse(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
if i == reverse(i) { // cannot find 'reverse' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let rev = String(str.reversed())
return str == rev
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let rev = String(str.reversed())
return str == rev
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let rev = String(str.reversed())
return str == rev
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let rev = String(str.reversed())
return str == rev
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let s = String(n)
let r = String(s.reversed())
return s == r
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let s = String(n)
let r = String(s.reversed())
return s == r
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversed = String(str.reversed())
return str == reversed
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversed = String(str.reversed())
return str == reversed
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var count = 0
for i in 1...n {
if isPalindrome(i) {
count += 1
}
}
return (count / 2, count - count / 2)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var count = 0
for i in 1...n {
if isPalindrome(i) {
count += 1
}
}
return (count / 2, count - count / 2)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```
// XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let rev = String(str.reversed())
return str == rev
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let rev = String(str.reversed())
return str == rev
}
```
```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var count = 0
var count2 = 0
for i in 1...n {
if i == 1 {
count += 1
count2 += 1
} else if i == 2 {
count += 1
count2 += 1
} else if i == 3 {
count += 1
count2 += 1
} else if i == 4 {
count += 1
count2 += 1
} else if i == 5 {
count += 1
count2 += 1
} else if i == 6 {
count += 1
count2 += 1
} else if i == 7 {
count += 1
count2 += 1
} else if i == 8 {
count += 1
count2 += 1
} else if i == 9 {
count += 1
count2 += 1
} else if i == 11 {
count += 1
count2 += 1
} else if i == 101 {
count += 1
count2 += 1
} else if i == 111 {
count += 1
count2 += 1
} else if i == 121 {
count += 1
count2 += 1
} else if i == 131 {
count += 1
count2 += 1
} else if i == 141 {
count += 1
count2 += 1
} else if i == 151 {
count += 1
count2 += 1
} else if i == 161 {
count += 1
count2 += 1
} else if i == 171 {
count += 1
count2 += 1
} else if i == 181 {
count += 1
count2 += 1
} else if i == 191 {
count += 1
count2 += 1
} else if i == 1001 {
count += 1
count2 += 1
} else if i == 1101 {
count += 1
count2 += 1
} else if i == 1111 {
count += 1
count2 += 1
} else if i == 1121 {
count += 1
count2 += 1
} else if i == 1131 {
count += 1
count2 += 1
} else if i == 1141 {
count += 1
count2 += 1
} else if i == 1151 {
count += 1
count2 += 1
} else if i == 1161 {
count += 1
count2 += 1
} else if i == 1171 {
count += 1
count2 += 1
} else if i == 1181 {
count += 1
count2 += 1
} else if i == 1191 {
count += 1
count2 += 1
} else if i == 10001 {
count += 1
count2 += 1
} else if i == 10101 {
count += 1
count2 += 1
} else if i == 10201 {
count += 1
count2 += 1
} else if i == 10301 {
count += 1
count2 += 1
} else if i == 10401 {
count += 1
count2 += 1
} else if i == 10501 {
count += 1
count2 += 1
} else if i == 10601 {
count += 1
count2 += 1
} else if i == 10701 {
count += 1
count2 += 1
} else if i == 10801 {
count +=XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if isPalindrome(i) { // cannot find 'isPalindrome' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
// TODO: Implement me!
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
// TODO: Implement me!
}
```} // missing return in global function expected to return '(Int, Int)'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var reversed = 0
var original = n
while n > 0 {
reversed = reversed * 10 + n % 10
n = n / 10
}
return reversed == original
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var reversed = 0
var original = n
while n > 0 {
reversed = reversed * 10 + n % 10
n = n / 10
}
return reversed == original
}
```if isPalindrome(i) { // missing argument label 'n:' in calln = n / 10 // cannot assign to value: 'n' is a 'let' constantfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
var isPalindrome = true
for j in 0..<s.characters.count/2 {
if s[j]!= s[s.characters.count-j-1] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
var isPalindrome = true
for j in 0..<s.characters.count/2 {
if s[j]!= s[s.characters.count-j-1] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```for j in 0..<s.characters.count/2 { // 'characters' is unavailable: Please use String directlyif s[j]!= s[s.characters.count-j-1] { // '=' must have consistent whitespace on both sidesif s[j]!= s[s.characters.count-j-1] { // cannot force unwrap value of non-optional type 'Character'if s[j]!= s[s.characters.count-j-1] { // use of '=' in a boolean context, did you mean '=='?func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == str.reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == str.reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
### Problemif str == str.reversed() { // binary operator '==' cannot be applied to operands of type 'String' and '[String.Element]' (aka 'Array<Character>')func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversed = String(str.reversed())
return str == reversed
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversed = String(str.reversed())
return str == reversed
}
```
### Problemfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var rev = 0
while n > 0 {
rev = rev * 10 + n % 10
n /= 10
}
return n == rev
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var rev = 0
while n > 0 {
rev = rev * 10 + n % 10
n /= 10
}
return n == rev
}
```
### Problem // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var rev = 0
while n > 0 {
rev = rev * 10 + n % 10
n /= 10
}
return n == rev
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var rev = 0
while n > 0 {
rev = rev * 10 + n % 10
n /= 10
}
return n == rev
}
```
### Problem // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var rev = 0
while n > 0 {
rev = rev * 10 + n % 10
n /= 10
}
return n == rev
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var rev = 0
while n > 0 {
rev = rev * 10 + n % 10
n /= 10
}
return n == rev
}
```
### Problem // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
```
### Problemfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return n == reversed
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return n == reversed
}
```
### Problem // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
```
### Problemfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let rev = String(str.reversed())
return str == rev
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let rev = String(str.reversed())
return str == rev
}
```
### Problemfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var rev = 0
while n > 0 {
rev = rev * 10 + n % 10
n /= 10
}
return n == rev
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var rev = 0
while n > 0 {
rev = rev * 10 + n % 10
n /= 10
}
return n == rev
}
```
### Problem // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == str.reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == str.reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
if str == str.reversed() { // binary operator '==' cannot be applied to operands of type 'String' and '[String.Element]' (aka 'Array<Character>')func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i.isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
extension Int {
var isPalindrome: Bool {
let str = String(self)
return str == String(str.reversed())
}
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i.isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
extension Int {
var isPalindrome: Bool {
let str = String(self)
return str == String(str.reversed())
}
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let rev = String(str.reversed())
if str == rev {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let rev = String(str.reversed())
if str == rev {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
### Testfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let numString = String(num)
let numReversed = String(numString.reversed())
return numString == numReversed
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let numString = String(num)
let numReversed = String(numString.reversed())
return numString == numReversed
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reverse = 0
while num != 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return reverse == num
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reverse = 0
while num != 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return reverse == num
}
```
// XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let numStr = String(num)
let numStrReversed = String(numStr.reversed())
return numStr == numStrReversed
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let numStr = String(num)
let numStrReversed = String(numStr.reversed())
return numStr == numStrReversed
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let reversedNumberString = String(numberString.reversed())
return numberString == reversedNumberString
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let reversedNumberString = String(numberString.reversed())
return numberString == reversedNumberString
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let numberString = String(number)
let numberStringReversed = String(numberString.reversed())
return numberString == numberStringReversed
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let numberString = String(number)
let numberStringReversed = String(numberString.reversed())
return numberString == numberStringReversed
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let numString = String(num)
let numStringReversed = String(numString.reversed())
return numString == numStringReversed
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let numString = String(num)
let numStringReversed = String(numString.reversed())
return numString == numStringReversed
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let reversed = String(String(i).reversed())
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let reversed = String(String(i).reversed())
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```
This code defines a function `evenOddPalindrome` that takes a positive integer `n` and returns a tuple containing the number of even and odd integer palindromes that fall within the range `(1, n)`, inclusive.
The function iterates over each number from 1 to `n`, checks if it is a palindrome, and then determines if it is even or odd. It then increments the respective count for even or odd palindromes.
The function returns a tuple containing the counts of even and odd palindromes.func evenOddPalindrome(n: Int) -> (Int, Int) {
// Initialize counters for even and odd palindromes
var evenPalindromes = 0
var oddPalindromes = 0
// Iterate over the range from 1 to n (inclusive)
for i in 1...n {
// Convert the number to a string to check if it's a palindrome
let str = String(i)
if str == String(str.reversed()) {
// If the number is a palindrome, check if it's even or odd
if i % 2 == 0 {
evenPalindromes += 1
} else {
oddPalindromes += 1
}
}
}
// Return the count of even and odd palindromes as a tuple
return (evenPalindromes, oddPalindromes)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
// Initialize counters for even and odd palindromes
var evenPalindromes = 0
var oddPalindromes = 0
// Iterate over the range from 1 to n (inclusive)
for i in 1...n {
// Convert the number to a string to check if it's a palindrome
let str = String(i)
if str == String(str.reversed()) {
// If the number is a palindrome, check if it's even or odd
if i % 2 == 0 {
evenPalindromes += 1
} else {
oddPalindromes += 1
}
}
}
// Return the count of even and odd palindromes as a tuple
return (evenPalindromes, oddPalindromes)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```
This Swift code defines a function `evenOddPalindrome` that takes a positive integer `n` and returns a tuple containing the number of even palindromes and odd palindromes within the range of 1 to `n`, inclusive.
The function iterates over the range of numbers from 1 to `n`, converts each number to a string to check if it's a palindrome, and increments the appropriate counter if it is.
The function then returns the counts of even and odd palindromes as a tuple.
The time complexity of this solution is O(n), where n is the number of digits in `n`, as it iterates over each number in the range. The space complexity is O(1), as the memory usage does not depend on the size of `n`.func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let reversed = String(String(i).reversed())
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let reversed = String(String(i).reversed())
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```
This function iterates through all numbers from 1 to n and checks if they are palindromes. If they are, it increments the count of even or odd palindromes accordingly.func evenOddPalindrome(n: Int) -> (Int, Int) {
// Initialize counters for even and odd palindromes
var evenPalindromes = 0
var oddPalindromes = 0
// Iterate over the range from 1 to n (inclusive)
for i in 1...n {
// Convert the number to a string to check if it's a palindrome
let str = String(i)
if str == String(str.reversed()) {
// Check if the number of digits is odd or even
if i % 2 == 0 {
evenPalindromes += 1
} else {
oddPalindromes += 1
}
}
}
return (evenPalindromes, oddPalindromes)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
// Initialize counters for even and odd palindromes
var evenPalindromes = 0
var oddPalindromes = 0
// Iterate over the range from 1 to n (inclusive)
for i in 1...n {
// Convert the number to a string to check if it's a palindrome
let str = String(i)
if str == String(str.reversed()) {
// Check if the number of digits is odd or even
if i % 2 == 0 {
evenPalindromes += 1
} else {
oddPalindromes += 1
}
}
}
return (evenPalindromes, oddPalindromes)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```
This function uses a simple loop to iterate over the range of numbers and checks if each number is a palindrome. If it is, it increments the appropriate counter (even or odd). The function returns a tuple with the counts of even and odd palindromes.func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```
This code defines a function `evenOddPalindrome` that takes a positive integer `n` and returns a tuple containing the number of even and odd integer palindromes that fall within the range `(1, n)`, inclusive.
The function iterates over each number from 1 to `n`, checks if it is a palindrome, and then checks if it is even or odd to increment the respective counters.
The function then returns the tuple containing the counts of even and odd palindromes.func evenOddPalindrome(n: Int) -> (Int, Int) {
// Initialize counters for even and odd palindromes
var evenCount = 0
var oddCount = 0
// Iterate over the range from 1 to n (inclusive)
for i in 1...n {
// Convert the number to a string to check if it's a palindrome
let str = String(i)
if str == String(str.reversed()) {
// If the number is a palindrome, increment the appropriate counter
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
// Return the tuple with the counts of even and odd palindromes
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
// Initialize counters for even and odd palindromes
var evenCount = 0
var oddCount = 0
// Iterate over the range from 1 to n (inclusive)
for i in 1...n {
// Convert the number to a string to check if it's a palindrome
let str = String(i)
if str == String(str.reversed()) {
// If the number is a palindrome, increment the appropriate counter
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
// Return the tuple with the counts of even and odd palindromes
return (evenCount, oddCount)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```
This Swift code defines a function `evenOddPalindrome` that takes a positive integer `n` and returns a tuple with the number of even and odd integer palindromes that fall within the range `(1, n)`, inclusive.
The function iterates over the range from `1` to `n` and checks if each number is a palindrome. If it is, it increments the appropriate counter (even or odd).
The function then returns the tuple with the counts of even and odd palindromes.
The time complexity of this solution is O(n), where n is the number of iterations required to check all numbers up to `n`. The space complexity is O(1) as no additional space is required that scales with input size.func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|>```if isPalindrome(i) { // missing argument label 'n:' in callif isEven(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|>if isPalindrome(i) { // missing argument label 'n:' in callif isEven(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return n == reversed
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return n == reversed
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|>if isPalindrome(i) { // missing argument label 'n:' in callif isEven(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
reverse = reverse * 10 + n % 10
n = n / 10
}
return reverse == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
reverse = reverse * 10 + n % 10
n = n / 10
}
return reverse == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|>if isPalindrome(i) { // missing argument label 'n:' in callif isEven(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
}
if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
func isOdd(n: Int) -> Bool {
return n % 2 == 1
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
}
if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
func isOdd(n: Int) -> Bool {
return n % 2 == 1
}
<|file_separator|>```if isPalindrome(i) && isEven(i) { // missing argument label 'n:' in callif isPalindrome(i) && isEven(i) { // missing argument label 'n:' in callif isPalindrome(i) && isOdd(i) { // missing argument label 'n:' in callif isPalindrome(i) && isOdd(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
evenCount += 1
} else if isPalindrome(i) && isOdd(i) {
oddCount += 1
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
func isOdd(n: Int) -> Bool {
return n % 2 == 1
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
evenCount += 1
} else if isPalindrome(i) && isOdd(i) {
oddCount += 1
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
func isOdd(n: Int) -> Bool {
return n % 2 == 1
}
<|file_separator|>if isPalindrome(i) && isEven(i) { // missing argument label 'n:' in callif isPalindrome(i) && isEven(i) { // missing argument label 'n:' in call} else if isPalindrome(i) && isOdd(i) { // missing argument label 'n:' in call} else if isPalindrome(i) && isOdd(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
}
if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return reverse == num
}
func isEven(_ num: Int) -> Bool {
return num % 2 == 0
}
func isOdd(_ num: Int) -> Bool {
return num % 2 == 1
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
}
if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return reverse == num
}
func isEven(_ num: Int) -> Bool {
return num % 2 == 0
}
func isOdd(_ num: Int) -> Bool {
return num % 2 == 1
}
<|file_separator|> // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var rev = 0
while num > 0 {
rev = rev * 10 + num % 10
num /= 10
}
return n == rev
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var rev = 0
while num > 0 {
rev = rev * 10 + num % 10
num /= 10
}
return n == rev
}
<|file_separator|>func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
reverse = reverse * 10 + n % 10
n /= 10
}
return reverse == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
reverse = reverse * 10 + n % 10
n /= 10
}
return reverse == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|> // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
} else if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == num
}
func isEven(_ num: Int) -> Bool {
return num % 2 == 0
}
func isOdd(_ num: Int) -> Bool {
return num % 2 == 1
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
} else if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == num
}
func isEven(_ num: Int) -> Bool {
return num % 2 == 0
}
func isOdd(_ num: Int) -> Bool {
return num % 2 == 1
}
<|file_separator|>``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
} else if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
func isOdd(n: Int) -> Bool {
return n % 2 == 1
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
} else if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
func isOdd(n: Int) -> Bool {
return n % 2 == 1
}
<|file_separator|>if isPalindrome(i) && isEven(i) { // missing argument label 'n:' in callif isPalindrome(i) && isEven(i) { // missing argument label 'n:' in call} else if isPalindrome(i) && isOdd(i) { // missing argument label 'n:' in call} else if isPalindrome(i) && isOdd(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|>if isPalindrome(i) { // missing argument label 'n:' in callif isEven(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let s = String(i)
if s == String(s.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func maxSubArray(nums: [Int]) -> Int {
var maxSum = Int.min
var sum = 0
for i in nums {
sum += i
if sum > maxSum {
maxSum = sum
}
if sum < 0 {
sum = 0
}
}
return maxSum
}
func maxProduct(nums: [Int]) -> Int {
var maxProduct = Int.min
var minProduct = Int.min
var currentProduct = 1
for i in nums {
currentProduct *= i
maxProduct = max(maxProduct, currentProduct)
if currentProduct == 0 {
currentProduct = 1
}
}
currentProduct = 1
for i in nums.reversed() {
currentProduct *= i
minProduct = min(minProduct, currentProduct)
if currentProduct == 0 {
currentProduct = 1
}
}
return max(maxProduct, minProduct)
}
func lengthOfLongestSubstring(s: String) -> Int {
var maxLen = 0
var start = 0
var end = 0
var charSet = Set<Character>()
while end < s.count {
if charSet.contains(s[end]) {
while s[start] != s[end] {
charSet.remove(s[start])
start += 1
}
start += 1
} else {
charSet.insert(s[end])
end += 1
maxLen = max(maxLen, end - start)
}
}
return maxLen
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let s = String(i)
if s == String(s.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Given an integer array nums, find the contiguous subarray (containing at least one number)
// which has the largest sum and return its sum.
// Example 1:
// Explanation:
// [−2,1,−3,4,−1,2,1,−5,4]
// contiguous subarray [4,−1,2,1] has the largest sum = 6.
// Example 2:
// Explanation:
// [1]
// contiguous subarray [1] has the largest sum = 1.
func maxSubArray(nums: [Int]) -> Int {
var maxSum = Int.min
var sum = 0
for i in nums {
sum += i
if sum > maxSum {
maxSum = sum
}
if sum < 0 {
sum = 0
}
}
return maxSum
}
// Given an integer array nums, find the contiguous subarray (containing at least one number)
// which has the largest product and return its product.
// Example 1:
// Explanation:
// [2,3,-2,4]
// contiguous subarray [2,3] has the largest product = 6.
// Example 2:
// Explanation:
// [-2,0,-1]
// contiguous subarray [-1] has the largest product = -1.
func maxProduct(nums: [Int]) -> Int {
var maxProduct = Int.min
var minProduct = Int.min
var currentProduct = 1
for i in nums {
currentProduct *= i
maxProduct = max(maxProduct, currentProduct)
if currentProduct == 0 {
currentProduct = 1
}
}
currentProduct = 1
for i in nums.reversed() {
currentProduct *= i
minProduct = min(minProduct, currentProduct)
if currentProduct == 0 {
currentProduct = 1
}
}
return max(maxProduct, minProduct)
}
// Given a string s, find the length of the longest substring without repeating characters.
// Example 1:
// Explanation:
// The answer is "abc", with the length of 3.
// Example 2:
// Explanation:
// The answer is "bb", with the length of 2.
func lengthOfLongestSubstring(s: String) -> Int {
var maxLen = 0
var start = 0
var end = 0
var charSet = Set<Character>()
while end < s.count {
if charSet.contains(s[end]) {
while s[start] != s[end] {
charSet.remove(s[start])
start += 1
}
start += 1
} else {
charSet.insert(s[end])
end += 1
maxLen = max(maxLen, end - start)
}
}
return maxLen
}
// Given an array of integers, find the first missing positive integer in linear time
// and constant space. In other words, find the lowest positive integer that does not
// exist in the array. The array can contain duplicates and negative numbers as well.
// Example 1:
// Explanation:
// The smallest positive integer is 1.
// Example 2:
// Explanation:
// The smallest positive integer is 1.
func firstMissingPositive(nums: [Int]) -> Int {
var nums = nums
for i in 0..<nums.count {
while nums[i] > 0 && nums[i] <= nums.count && nums[nums[i] - 1] != nums[i] {
nums.swapAt(i, nums[i] - 1)
}
}
for i in 0..<nums.count {
if nums[i] != i + 1 {
return i + 1
}
}
return nums.count + 1if charSet.contains(s[end]) { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.while s[start] != s[end] { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.while s[start] != s[end] { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.charSet.remove(s[start]) // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.charSet.insert(s[end]) // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return n == reversed
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return n == reversed
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|>if isPalindrome(i) { // missing argument label 'n:' in callif isEven(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
countEven += 1
} else if isPalindrome(i) && isOdd(i) {
countOdd += 1
}
}
return (countEven, countOdd)
}
func isPalindrome(num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
let digit = num % 10
reversed = reversed * 10 + digit
num /= 10
}
return reversed == num
}
func isEven(num: Int) -> Bool {
return num % 2 == 0
}
func isOdd(num: Int) -> Bool {
return num % 2 != 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
countEven += 1
} else if isPalindrome(i) && isOdd(i) {
countOdd += 1
}
}
return (countEven, countOdd)
}
func isPalindrome(num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
let digit = num % 10
reversed = reversed * 10 + digit
num /= 10
}
return reversed == num
}
func isEven(num: Int) -> Bool {
return num % 2 == 0
}
func isOdd(num: Int) -> Bool {
return num % 2 != 0
}
print(evenOddPalindrome(n: 3))
print(evenOddPalindrome(n: 12))
<|file_separator|>if isPalindrome(i) && isEven(i) { // missing argument label 'num:' in callif isPalindrome(i) && isEven(i) { // missing argument label 'num:' in call} else if isPalindrome(i) && isOdd(i) { // missing argument label 'num:' in call} else if isPalindrome(i) && isOdd(i) { // missing argument label 'num:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
} else if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == num
}
func isEven(_ num: Int) -> Bool {
return num % 2 == 0
}
func isOdd(_ num: Int) -> Bool {
return !isEven(num)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
} else if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == num
}
func isEven(_ num: Int) -> Bool {
return num % 2 == 0
}
func isOdd(_ num: Int) -> Bool {
return !isEven(num)
}
<|file_separator|>``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return n == reversed
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return n == reversed
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
<|file_separator|>func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
} else if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
func isOdd(n: Int) -> Bool {
return n % 2 == 1
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
} else if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
func isEven(n: Int) -> Bool {
return n % 2 == 0
}
func isOdd(n: Int) -> Bool {
return n % 2 == 1
}
<|file_separator|>```if isPalindrome(i) && isEven(i) { // missing argument label 'n:' in callif isPalindrome(i) && isEven(i) { // missing argument label 'n:' in call} else if isPalindrome(i) && isOdd(i) { // missing argument label 'n:' in call} else if isPalindrome(i) && isOdd(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
} else if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let s = String(n)
let l = s.count
for i in 0..<l/2 {
if s[i] != s[l-i-1] {
return false
}
}
return true
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
func isOdd(_ n: Int) -> Bool {
return n % 2 == 1
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) && isEven(i) {
even += 1
} else if isPalindrome(i) && isOdd(i) {
odd += 1
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let s = String(n)
let l = s.count
for i in 0..<l/2 {
if s[i] != s[l-i-1] {
return false
}
}
return true
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
func isOdd(_ n: Int) -> Bool {
return n % 2 == 1
}
<|file_separator|>```if s[i] != s[l-i-1] { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.if s[i] != s[l-i-1] { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
<|file_separator|>```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(n: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
let str = String(n)
var i = 0
var j = str.count - 1
while i < j {
if str[i] != str[j] {
return false
}
i += 1
j -= 1
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(n: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
let str = String(n)
var i = 0
var j = str.count - 1
while i < j {
if str[i] != str[j] {
return false
}
i += 1
j -= 1
}
return true
}
```if str[i] != str[j] { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.if str[i] != str[j] { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if isPalindrome(i) { // cannot find 'isPalindrome' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i == reverse(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func reverse(_ n: Int) -> Int {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i == reverse(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func reverse(_ n: Int) -> Int {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
var left = 0
var right = str.count - 1
while left < right {
if str[str.index(str.startIndex, offsetBy: left)] != str[str.index(str.startIndex, offsetBy: right)] {
return false
}
left += 1
right -= 1
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
var left = 0
var right = str.count - 1
while left < right {
if str[str.index(str.startIndex, offsetBy: left)] != str[str.index(str.startIndex, offsetBy: right)] {
return false
}
left += 1
right -= 1
}
return true
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func countPalindromes(n: Int) -> Int {
var count = 0
for i in 1...n {
if isPalindrome(i) {
count += 1
}
}
return count
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Given a positive integer n, return the number of palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func countPalindromes(n: Int) -> Int {
var count = 0
for i in 1...n {
if isPalindrome(i) {
count += 1
}
}
return count
}
// Given a positive integer n, return the number of palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func countPalindromes(n: Int) -> Int {
var count = 0
for i in 1...n {
if isPalindrome(i) {
count += 1
}
}
return count
}
// Given a positive integer n, return the number of palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func countPalindromes(n: Int) -> Int {
var count = 0
for i in 1...n {
if isPalindrome(i) {
count += 1
}
}
return count
}
// Given a positive integer n, return the number of palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func countPalindromes(n: Int) -> Int {
var count = 0
for i in 1...n {
if isPalindrome(i) {
count += 1
}
}
return count
}
// Given a positive integer n, return the number of palindif isPalindrome(i) { // cannot find 'isPalindrome' in scopeif isPalindrome(i) { // cannot find 'isPalindrome' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = "\(n)"
let reversed = String(str.reversed())
return str == reversed
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = "\(n)"
let reversed = String(str.reversed())
return str == reversed
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
return (evenCount, oddCount)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ x: Int) -> Bool {
let str = String(x)
for i in 0..<str.count / 2 {
if str[str.index(str.startIndex, offsetBy: i)] != str[str.index(str.endIndex, offsetBy: -i - 1)] {
return false
}
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Given an integer, return true if the number is a palindrome.
// Example 1:
// Explanation:
// 121 is a palindrome.
// Example 2:
// Explanation:
// 123 is not a palindrome.
func isPalindrome(_ x: Int) -> Bool {
let str = String(x)
for i in 0..<str.count / 2 {
if str[str.index(str.startIndex, offsetBy: i)] != str[str.index(str.endIndex, offsetBy: -i - 1)] {
return false
}
}
return true
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(num: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(num: Int) -> Bool {
let str = String(num)
var leftIndex = 0
var rightIndex = str.count - 1
while leftIndex < rightIndex {
if str[str.index(str.startIndex, offsetBy: leftIndex)] != str[str.index(str.startIndex, offsetBy: rightIndex)] {
return false
}
leftIndex += 1
rightIndex -= 1
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(num: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Given an integer, return true if it's a palindrome
func isPalindrome(num: Int) -> Bool {
let str = String(num)
var leftIndex = 0
var rightIndex = str.count - 1
while leftIndex < rightIndex {
if str[str.index(str.startIndex, offsetBy: leftIndex)] != str[str.index(str.startIndex, offsetBy: rightIndex)] {
return false
}
leftIndex += 1
rightIndex -= 1
}
return true
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversed = String(str.reversed())
return str == reversed
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversed = String(str.reversed())
return str == reversed
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if i.isEven() {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if i.isEven() {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if i.isEven() {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if i.isEven() {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if i.isEven() {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2for i in 1 .. n { // cannot find operator '..' in scope; did you mean '...'?func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(num: Int) -> Bool {
var reversed = 0
var original = num
while num > 0 {
reversed = reversed * 10 + num % 10
num = num / 10
}
return reversed == original
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(num: Int) -> Bool {
var reversed = 0
var original = num
while num > 0 {
reversed = reversed * 10 + num % 10
num = num / 10
}
return reversed == original
}
```for i in 1 .. n { // cannot find operator '..' in scope; did you mean '...'?num = num / 10 // cannot assign to value: 'num' is a 'let' constantfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
var reversed = 0
var original = n
while n > 0 {
reversed = reversed * 10 + n % 10
n = n / 10
}
return original == reversed
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
var reversed = 0
var original = n
while n > 0 {
reversed = reversed * 10 + n % 10
n = n / 10
}
return original == reversed
}
```for i in 1 .. n { // cannot find operator '..' in scope; did you mean '...'?n = n / 10 // cannot assign to value: 'n' is a 'let' constantfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
var s = i.String()
var reverse = s.Reversed()
if s == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func firstUniqChar(s: String) -> Int {
var m = make(map[String]Int)
for i in 0 .. s.Count() - 1 {
var c = s[i]
var key = c.String()
if _, ok := m[key]; ok {
m[key] = m[key] + 1
} else {
m[key] = 1
}
}
for i in 0 .. s.Count() - 1 {
var c = s[i]
var key = c.String()
if m[key] == 1 {
return i
}
}
return -1
}
func singleNumber(nums: [Int]) -> Int {
var m = make(map[Int]Int)
for i in 0 .. nums.Count() - 1 {
var key = nums[i]
if _, ok := m[key]; ok {
m[key] = m[key] + 1
} else {
m[key] = 1
}
}
for i in 0 .. nums.Count() - 1 {
var key = nums[i]
if m[key] == 1 {
return key
}
}
return 0
}
func containsDuplicate(nums: [Int]) -> Bool {
var m = make(map[Int]Int)
for i in 0 .. nums.Count() - 1 {
var key = nums[i]
if _, ok := m[key]; ok {
return true
} else {
m[key] = 1
}
}
return false
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
var s = i.String()
var reverse = s.Reversed()
if s == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Given a string s, return the first non-repeating character in it and return its index. If it doesn't exist, return -1.
// Example 1:
// Input: s = "leetcode"
// Output: 0
// Example 2:
// Input: s = "loveleetcode"
// Output: 2
// Example 3:
// Input: s = "aabb"
// Output: -1
// Constraints:
// 1 <= s.length <= 3000
// s consists of only lowercase English letters.
func firstUniqChar(s: String) -> Int {
var m = make(map[String]Int)
for i in 0 .. s.Count() - 1 {
var c = s[i]
var key = c.String()
if _, ok := m[key]; ok {
m[key] = m[key] + 1
} else {
m[key] = 1
}
}
for i in 0 .. s.Count() - 1 {
var c = s[i]
var key = c.String()
if m[key] == 1 {
return i
}
}
return -1
}
// Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.
// You must implement a solution with a linear runtime complexity and use only constant extra space.
// Example 1:
// Input: nums = [2,2,1]
// Output: 1
// Example 2:
// Input: nums = [4,1,2,1,2]
// Output: 4
// Example 3:
// Input: nums = [1]
// Output: 1
// Constraints:
// 1 <= nums.length <= 3 * 10^4
// -3 * 10^4 <= nums[i] <= 3 * 10^4
// Each element in the array appears twice except for one element which appears only once.
func singleNumber(nums: [Int]) -> Int {
var m = make(map[Int]Int)
for i in 0 .. nums.Count() - 1 {
var key = nums[i]
if _, ok := m[key]; ok {
m[key] = m[key] + 1
} else {
m[key] = 1
}
}
for i in 0 .. nums.Count() - 1 {
var key = nums[i]
if m[key] == 1 {
return key
}
}
return 0
}
// Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.
// Example 1:
// Input: nums = [1,2,3,1]
// Output: true
// Example 2:
// Input: nums = [1,2,3,4]
// Output: false
// Example 3:
// Input: nums = [1,1,1,3,3,4,3,2,4,2]
// Output: true
// Constraints:
// 1 <= nums.length <= 10^5
// -10^9 <= nums[i] <= 10^9
func containsDuplicate(nums: [Int]) -> Bool {
var m = make(map[Int]Int)
for i in 0 .. nums.Count() - 1 {
var key = nums[i]
if _, ok := m[key]; ok {
return true
} else {
m[key] = 1
}
}
return false
}
// Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
// You may assume that each input would have exactly one solution, and you may not use the same element twice.
//for i in 1 .. n { // cannot find operator '..' in scope; did you mean '...'?var m = make(map[String]Int) // cannot find 'make' in scopevar m = make(map[String]Int) // cannot find 'map' in scopevar m = make(map[String]Int) // expected ',' separatorfor i in 0 .. s.Count() - 1 { // cannot find operator '..' in scope; did you mean '...'?for i in 0 .. s.Count() - 1 { // value of type 'String' has no member 'Count'if _, ok := m[key]; ok { // expected '{' after 'if' conditionfor i in 0 .. s.Count() - 1 { // value of type 'String' has no member 'Count'for i in 0 .. s.Count() - 1 { // cannot find operator '..' in scope; did you mean '...'?var m = make(map[Int]Int) // cannot find 'make' in scopevar m = make(map[Int]Int) // expected ',' separatorvar m = make(map[Int]Int) // cannot find 'map' in scopefor i in 0 .. nums.Count() - 1 { // cannot find operator '..' in scope; did you mean '...'?for i in 0 .. nums.Count() - 1 { // value of type '[Int]' has no member 'Count'if _, ok := m[key]; ok { // expected '{' after 'if' conditionfor i in 0 .. nums.Count() - 1 { // value of type '[Int]' has no member 'Count'for i in 0 .. nums.Count() - 1 { // cannot find operator '..' in scope; did you mean '...'?var m = make(map[Int]Int) // cannot find 'make' in scopevar m = make(map[Int]Int) // cannot find 'map' in scopevar m = make(map[Int]Int) // expected ',' separatorfor i in 0 .. nums.Count() - 1 { // cannot find operator '..' in scope; did you mean '...'?for i in 0 .. nums.Count() - 1 { // value of type '[Int]' has no member 'Count'if _, ok := m[key]; ok { // expected '{' after 'if' conditionfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
var num = i
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num = num / 10
}
if i == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
var num = i
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num = num / 10
}
if i == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
var num = i
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num = num / 10
}
if i == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```for i in 1 .. n { // cannot find operator '..' in scope; did you mean '...'?func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if i.isEven() {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if i.isEven() {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if i.isEven() {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
```for i in 1 .. n { // cannot find operator '..' in scope; did you mean '...'?func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var str = n.toString()
var len = str.count()
for i in 0 .. len / 2 {
if str[i] != str[len - i - 1] {
return false
}
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var str = n.toString()
var len = str.count()
for i in 0 .. len / 2 {
if str[i] != str[len - i - 1] {
return false
}
}
return true
}
```for i in 1 .. n { // cannot find operator '..' in scope; did you mean '...'?var str = n.toString() // value of type 'Int' has no member 'toString'for i in 0 .. len / 2 { // cannot find operator '..' in scope; did you mean '...'?func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
var reversed = 0
var original = n
while n > 0 {
reversed = reversed * 10 + n % 10
n = n / 10
}
return reversed == original
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
var reversed = 0
var original = n
while n > 0 {
reversed = reversed * 10 + n % 10
n = n / 10
}
return reversed == original
}
```for i in 1 .. n { // cannot find operator '..' in scope; did you mean '...'?n = n / 10 // cannot assign to value: 'n' is a 'let' constantfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palfor i in 1 .. n { // cannot find operator '..' in scope; did you mean '...'?func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1 .. n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
for i in 1 .. n { // cannot find operator '..' in scope; did you mean '...'?func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var number = i
var reversedNumber = 0
while number > 0 {
let digit = number % 10
reversedNumber = reversedNumber * 10 + digit
number /= 10
}
if i == reversedNumber {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var number = i
var reversedNumber = 0
while number > 0 {
let digit = number % 10
reversedNumber = reversedNumber * 10 + digit
number /= 10
}
if i == reversedNumber {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = "\(number)"
let reversedNumberString = numberString.reversed()
return numberString == "\(reversedNumberString)"
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = "\(number)"
let reversedNumberString = numberString.reversed()
return numberString == "\(reversedNumberString)"
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
if digits == digits.reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
if digits == digits.reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ i: Int) -> Bool {
var digits = [Int]()
var num = i
while num > 0 {
digits.append(num % 10)
num /= 10
}
for i in 0..<digits.count / 2 {
if digits[i] != digits[digits.count - i - 1] {
return false
}
}
return true
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ i: Int) -> Bool {
var digits = [Int]()
var num = i
while num > 0 {
digits.append(num % 10)
num /= 10
}
for i in 0..<digits.count / 2 {
if digits[i] != digits[digits.count - i - 1] {
return false
}
}
return true
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var number = i
var reversedNumber = 0
while number > 0 {
let digit = number % 10
reversedNumber = reversedNumber * 10 + digit
number /= 10
}
if i == reversedNumber {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var number = i
var reversedNumber = 0
while number > 0 {
let digit = number % 10
reversedNumber = reversedNumber * 10 + digit
number /= 10
}
if i == reversedNumber {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
var isPalindrome = true
for j in 0..<digits.count / 2 {
if digits[j] != digits[digits.count - j - 1] {
isPalindrome = false
break
}
}
if isPalindrome {
if digits.count % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
var isPalindrome = true
for j in 0..<digits.count / 2 {
if digits[j] != digits[digits.count - j - 1] {
isPalindrome = false
break
}
}
if isPalindrome {
if digits.count % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var number = i
var reversedNumber = 0
while number > 0 {
let digit = number % 10
reversedNumber = reversedNumber * 10 + digit
number /= 10
}
if reversedNumber == i {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var number = i
var reversedNumber = 0
while number > 0 {
let digit = number % 10
reversedNumber = reversedNumber * 10 + digit
number /= 10
}
if reversedNumber == i {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let digits = String(num).map { Int(String($0))! }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
let digits = String(num).map { Int(String($0))! }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if digits == reversedDigits { // cannot convert value of type 'ReversedCollection<[Int]>' to expected argument type '[Int]'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for number in 1...n {
let stringNumber = String(number)
let reversedNumber = String(stringNumber.reversed())
if stringNumber == reversedNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for number in 1...n {
let stringNumber = String(number)
let reversedNumber = String(stringNumber.reversed())
if stringNumber == reversedNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
if isPalindrome(num) {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let numStr = String(num)
let reversedStr = String(numStr.reversed())
return numStr == reversedStr
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for num in 1...n {
if isPalindrome(num) {
if num % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let numStr = String(num)
let reversedStr = String(numStr.reversed())
return numStr == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { String($0) }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { String($0) }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if digits == reversedDigits { // cannot convert value of type 'ReversedCollection<[String]>' to expected argument type '[String]'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { String($0) }
let reversed = digits.reversed()
if digits == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { String($0) }
let reversed = digits.reversed()
if digits == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if digits == reversed { // cannot convert value of type 'ReversedCollection<[String]>' to expected argument type '[String]'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { String($0) }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { String($0) }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if digits == reversedDigits { // cannot convert value of type 'ReversedCollection<[String]>' to expected argument type '[String]'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { String($0) }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { String($0) }
let reversedDigits = digits.reversed()
if digits == reversedDigits {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if digits == reversedDigits { // cannot convert value of type 'ReversedCollection<[String]>' to expected argument type '[String]'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { String($0) }
let reversed = digits.reversed()
let palindrome = Int(digits + reversed) ?? 0
if palindrome % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { String($0) }
let reversed = digits.reversed()
let palindrome = Int(digits + reversed) ?? 0
if palindrome % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
return (evenCount, oddCount)
}
```let palindrome = Int(digits + reversed) ?? 0 // cannot convert value of type '[String]' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
var isPalindrome = true
for j in 0..<digits.count / 2 {
if digits[j] != digits[digits.count - 1 - j] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
var isPalindrome = true
for j in 0..<digits.count / 2 {
if digits[j] != digits[digits.count - 1 - j] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
var reversed = digits
reversed.reverse()
if digits == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
var reversed = digits
reversed.reverse()
if digits == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
var reversed = digits
reversed.reverse()
if digits == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
var reversed = digits
reversed.reverse()
if digits == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
var reversed = digits
reversed.reverse()
if digits == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
var reversed = digits
reversed.reverse()
if digits == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
var isPalindrome = true
for j in 0..<digits.count / 2 {
if digits[j] != digits[digits.count - j - 1] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<end_of_turn>
<start_of_turn>model
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let digits = String(i).map { Int(String($0))! }
var isPalindrome = true
for j in 0..<digits.count / 2 {
if digits[j] != digits[digits.count - j - 1] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
```if isPalindrome(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
```if isPalindrome(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
```if isPalindrome(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
```if isPalindrome(i) { // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i.isPalindrome {
if i.isEven {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
extension Int {
var isPalindrome: Bool {
return String(self).reversed() == String(self)
}
var isEven: Bool {
return self % 2 == 0
}
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i.isPalindrome {
if i.isEven {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
extension Int {
var isPalindrome: Bool {
return String(self).reversed() == String(self)
}
var isEven: Bool {
return self % 2 == 0
}
}
```return String(self).reversed() == String(self) // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let numString = String(num)
return numString == String(numString.reversed())
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let numString = String(num)
return numString == String(numString.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6) [/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let str = String(num)
if str == String(num.reversed()) {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let str = String(num)
if str == String(num.reversed()) {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if str == String(num.reversed()) { // value of type 'Int' has no member 'reversed'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var numCopy = num
var reversedNum = 0
while numCopy > 0 {
let remainder = numCopy % 10
reversedNum = reversedNum * 10 + remainder
numCopy /= 10
}
return num == reversedNum
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var numCopy = num
var reversedNum = 0
while numCopy > 0 {
let remainder = numCopy % 10
reversedNum = reversedNum * 10 + remainder
numCopy /= 10
}
return num == reversedNum
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let str = String(num)
if str == String(num.reversed()) && num % 2 == 0 {
even += 1
} else if str == String(num.reversed()) && num % 2 == 1 {
odd += 1
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = i
let str = String(num)
if str == String(num.reversed()) && num % 2 == 0 {
even += 1
} else if str == String(num.reversed()) && num % 2 == 1 {
odd += 1
}
}
return (even, odd)
}
```if str == String(num.reversed()) && num % 2 == 0 { // value of type 'Int' has no member 'reversed'} else if str == String(num.reversed()) && num % 2 == 1 { // value of type 'Int' has no member 'reversed'func evenOddPalindrome(n: Int) -> (Int, Int) {
var count = 0
var even = 0
var odd = 0
for i in 1...n {
let number = i
let reversedNumber = Int(String(number).reversed())!
if number == reversedNumber {
count += 1
}
if number % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var count = 0
var even = 0
var odd = 0
for i in 1...n {
let number = i
let reversedNumber = Int(String(number).reversed())!
if number == reversedNumber {
count += 1
}
if number % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
```let reversedNumber = Int(String(number).reversed())! // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
var number = number
var reversedNumber = 0
while number > 0 {
let remainder = number % 10
reversedNumber = reversedNumber * 10 + remainder
number /= 10
}
return number == reversedNumber
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
var number = number
var reversedNumber = 0
while number > 0 {
let remainder = number % 10
reversedNumber = reversedNumber * 10 + remainder
number /= 10
}
return number == reversedNumber
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let string = String(number)
return string == String(string.reversed())
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let string = String(number)
return string == String(string.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
var number = number
var reversedNumber = 0
while number > 0 {
let remainder = number % 10
reversedNumber = reversedNumber * 10 + remainder
number /= 10
}
return number == reversedNumber
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
var number = number
var reversedNumber = 0
while number > 0 {
let remainder = number % 10
reversedNumber = reversedNumber * 10 + remainder
number /= 10
}
return number == reversedNumber
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
return numberString == String(numberString.reversed())
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
return numberString == String(numberString.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let string = String(number)
return string == String(string.reversed())
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let string = String(number)
return string == String(string.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let string = String(number)
return string == String(string.reversed())
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let string = String(number)
return string == String(string.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverseNumber = Int(String(number).reversed())
if number == reverseNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverseNumber = Int(String(number).reversed())
if number == reverseNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let reverseNumber = Int(String(number).reversed()) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverseNumber = Int(String(number).reversed())
if number == reverseNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverseNumber = Int(String(number).reversed())
if number == reverseNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let reverseNumber = Int(String(number).reversed()) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverseNumber = Int(String(number).reversed())
if number == reverseNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverseNumber = Int(String(number).reversed())
if number == reverseNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let reverseNumber = Int(String(number).reversed()) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverseNumber = Int(String(number).reversed())
if number == reverseNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverseNumber = Int(String(number).reversed())
if number == reverseNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let reverseNumber = Int(String(number).reversed()) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverseNumber = Int(String(number).reversed())
if number == reverseNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverseNumber = Int(String(number).reversed())
if number == reverseNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let reverseNumber = Int(String(number).reversed()) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let numString = String(i)
let reversedNumString = String(numString.reversed())
let reversedNum = Int(reversedNumString)!
if i == reversedNum {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let numString = String(i)
let reversedNumString = String(numString.reversed())
let reversedNum = Int(reversedNumString)!
if i == reversedNum {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let numString = String(i)
let reversedNumString = String(numString.reversed())
if numString == reversedNumString {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let numString = String(i)
let reversedNumString = String(numString.reversed())
if numString == reversedNumString {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverse = Int(String(number).reversed())
if number == reverse {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reverse = Int(String(number).reversed())
if number == reverse {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let reverse = Int(String(number).reversed()) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reversedNumber = Int(String(number).reversed())
if number == reversedNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let number = i
let reversedNumber = Int(String(number).reversed())
if number == reversedNumber {
if number % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let reversedNumber = Int(String(number).reversed()) // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let numString = String(i)
let reversedNumString = String(numString.reversed())
let reversedNum = Int(reversedNumString)!
if i == reversedNum {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let numString = String(i)
let reversedNumString = String(numString.reversed())
let reversedNum = Int(reversedNumString)!
if i == reversedNum {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let reversed = String(i).reversed()
if i == Int(String(reversed)) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let reversed = String(i).reversed()
if i == Int(String(reversed)) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i.isPalindrome() {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
extension Int {
func isPalindrome() -> Bool {
let originalNumber = self
var reversedNumber = 0
var number = self
while number > 0 {
let remainder = number % 10
reversedNumber = (reversedNumber * 10) + remainder
number = number / 10
}
return originalNumber == reversedNumber
}
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if i.isPalindrome() {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
extension Int {
func isPalindrome() -> Bool {
let originalNumber = self
var reversedNumber = 0
var number = self
while number > 0 {
let remainder = number % 10
reversedNumber = (reversedNumber * 10) + remainder
number = number / 10
}
return originalNumber == reversedNumber
}
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let reversed = String(i).reversed()
if i == Int(String(reversed)) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let reversed = String(i).reversed()
if i == Int(String(reversed)) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
var reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
var reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```var reversedStr = String(str.reversed()) // variable 'reversedStr' was never mutated; consider changing to 'let' constantfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
var isPalindrome = true
for j in 0..<str.count/2 {
if str[j] != str[str.count-j-1] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
var isPalindrome = true
for j in 0..<str.count/2 {
if str[j] != str[str.count-j-1] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if str[j] != str[str.count-j-1] { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.if str[j] != str[str.count-j-1] { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num = num / 10
}
return reverse == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num = num / 10
}
return reverse == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return reverse == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return reverse == n
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
return str == String(str.reversed())
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let string = String(n)
let reversed = String(string.reversed())
return string == reversed
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let string = String(n)
let reversed = String(string.reversed())
return string == reversed
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversed = String(str.reversed())
return str == reversed
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversed = String(str.reversed())
return str == reversed
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let string = String(n)
let reversed = String(string.reversed())
return string == reversed
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let string = String(n)
let reversed = String(string.reversed())
return string == reversed
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversed = String(str.reversed())
return str == reversed
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversed = String(str.reversed())
return str == reversed
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == n
}
func isEven(_ n: Int) -> Bool {
return n % 2 == 0
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversed = String(str.reversed())
return str == reversed
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
let str = String(n)
let reversed = String(str.reversed())
return str == reversed
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(s.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(s.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(s.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(s.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == num
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == num
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return n == reversed
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return n == reversed
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(s.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(s.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == num
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == num
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
reverse = reverse * 10 + n % 10
n /= 10
}
return reverse == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
reverse = reverse * 10 + n % 10
n /= 10
}
return reverse == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
reverse = reverse * 10 + n % 10
n /= 10
}
return reverse == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
reverse = reverse * 10 + n % 10
n /= 10
}
return reverse == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return reversed == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
reverse = reverse * 10 + n % 10
n /= 10
}
return reverse == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
reverse = reverse * 10 + n % 10
n /= 10
}
return reverse == n
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var odd = 0
var even = 0
for i in 1...n {
let s = String(i)
if s == s.reverse() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var odd = 0
var even = 0
for i in 1...n {
let s = String(i)
if s == s.reverse() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if s == s.reverse() { // value of type 'String' has no member 'reverse'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if isPalindrome(i) { // cannot find 'isPalindrome' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
var isPalindrome = true
for j in 0..<s.characters.count/2 {
if s[j]!= s[s.characters.count-j-1] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
var isPalindrome = true
for j in 0..<s.characters.count/2 {
if s[j]!= s[s.characters.count-j-1] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```for j in 0..<s.characters.count/2 { // 'characters' is unavailable: Please use String directlyif s[j]!= s[s.characters.count-j-1] { // use of '=' in a boolean context, did you mean '=='?if s[j]!= s[s.characters.count-j-1] { // cannot force unwrap value of non-optional type 'Character'if s[j]!= s[s.characters.count-j-1] { // '=' must have consistent whitespace on both sidesfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
// TODO: Implement me
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
// TODO: Implement me
}
```} // missing return in global function expected to return '(Int, Int)'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == s.reverse() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == s.reverse() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if s == s.reverse() { // value of type 'String' has no member 'reverse'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func oddNumbers(n: Int) -> Int {
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2!= 0 {
odd += 1
}
}
}
return odd
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Given a positive integer n, return the number of odd integers that fall
// within the range(1, n), inclusive.
// Example 1:
// Explanation:
// The odd integers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.
// Example 2:
// Explanation:
// The odd integers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59.
// Note:
// 1. 1 <= n <= 10^6
// 2. returned integer has the number of odd integers that fall within the range(1, n), inclusive.
// oddNumbers(n: 3) == 1
// oddNumbers(n: 12) == 6
func oddNumbers(n: Int) -> Int {
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2!= 0 {
odd += 1
}
}
}
return odd
}
// Given a positive integer n, return the number of odd integers that fall
// within the range(1, n), inclusive.
// Example 1:
// Explanation:
// The odd integers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.
// Example 2:
// Explanation:
// The odd integers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59.
// Note:
// 1. 1 <= n <= 10^6
// 2. returned integer has the number of odd integers that fall within the range(1, n), inclusive.
// oddNumbers(n: 3) == 1
// oddNumbers(n: 12) == 6
func oddNumbers(n: Int) -> Int {
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2!= 0 {
odd += 1
}
}
}
return odd
}
// Given a positive integer n, return the number of odd integers that fall
// within the range(1, n), inclusive.
// Example 1:
// Explanation:
// The odd integers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.
// Example 2:
// Explanation:
// The odd integers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59.
// Note:
// 1. 1 <= n <= 10^6
// 2. returned integer has the number of odd integers that fall within the range(1, n), inclusive.
// oddNumbers(n: 3) == 1
// oddNumbers(n: 12) == 6
func oddNumbers(n: Int) -> Int {
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2!= 0 {
odd += 1
}
}
}
return odd
}
// Given a positive integer n, return the number of odd integers that fall
// within the range(1, n), inclusive.
// Example 1:
// Explanation:
// The odd integers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.
// Example 2:
// Explanation:
// The odd integers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59.
if isPalindrome(i) { // cannot find 'isPalindrome' in scopeif isPalindrome(i) { // cannot find 'isPalindrome' in scopeif i % 2!= 0 { // expression is not assignable: binary operator returns immutable valueif i % 2!= 0 { // '=' must have consistent whitespace on both sidesif i % 2!= 0 { // use of '=' in a boolean context, did you mean '=='?if i % 2!= 0 { // cannot force unwrap value of non-optional type 'Int'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == s.reverse() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == s.reverse() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if s == s.reverse() { // value of type 'String' has no member 'reverse'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
var isPalindrome = true
for j in 0..<s.characters.count/2 {
if s[j]!= s[s.characters.count-j-1] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
var isPalindrome = true
for j in 0..<s.characters.count/2 {
if s[j]!= s[s.characters.count-j-1] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```for j in 0..<s.characters.count/2 { // 'characters' is unavailable: Please use String directlyif s[j]!= s[s.characters.count-j-1] { // use of '=' in a boolean context, did you mean '=='?if s[j]!= s[s.characters.count-j-1] { // cannot force unwrap value of non-optional type 'Character'if s[j]!= s[s.characters.count-j-1] { // '=' must have consistent whitespace on both sidesfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)<|im_end|>
<|im_start|>assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in stride(from: 1, through: n, by: 1) {
let reversed = String(i).reversed()
if String(i) == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i) == reversed { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let string = String(number)
let reversedString = String(string.reversed())
return string == reversedString
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let string = String(number)
let reversedString = String(string.reversed())
return string == reversedString
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let string = String(number)
return string == String(string.reversed())
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let string = String(number)
return string == String(string.reversed())
}
// Example usage:
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let string = String(number)
let reversedString = String(string.reversed())
return string == reversedString
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let string = String(number)
let reversedString = String(string.reversed())
return string == reversedString
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i).reversed() == String(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i).reversed() == String(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i).reversed() == String(i) { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let reverse = String(s.reversed())
if s == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let reverse = String(s.reversed())
if s == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var odd = 0
var even = 0
for i in 1...n {
let s = String(i)
let r = String(s.reversed())
if s == r {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var odd = 0
var even = 0
for i in 1...n {
let s = String(i)
let r = String(s.reversed())
if s == r {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let r = String(s.reversed())
if s == r {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let r = String(s.reversed())
if s == r {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var odd = 0
var even = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var odd = 0
var even = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let r = String(s.reversed())
if s == r {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let r = String(s.reversed())
if s == r {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
let reverse = String(str.reversed())
if str == reverse {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let string = String(i)
let isPalindrome = string == String(string.reversed())
if isPalindrome {
if string.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let string = String(i)
let isPalindrome = string == String(string.reversed())
if isPalindrome {
if string.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
let str = String(i)
let isPalindrome = str == String(str.reversed())
if isPalindrome {
if str.count % 2 == 0 {
countEven += 1
} else {
countOdd += 1
}
}
}
return (countEven, countOdd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
let str = String(i)
let isPalindrome = str == String(str.reversed())
if isPalindrome {
if str.count % 2 == 0 {
countEven += 1
} else {
countOdd += 1
}
}
}
return (countEven, countOdd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ x: Int) -> Bool {
let str = String(x)
let reversed = String(str.reversed())
return str == reversed
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ x: Int) -> Bool {
let str = String(x)
let reversed = String(str.reversed())
return str == reversed
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ x: Int) -> Bool {
let str = String(x)
let length = str.count
for i in 0..<length/2 {
if str[str.index(str.startIndex, offsetBy: i)]!= str[str.index(str.endIndex, offsetBy: -i - 1)] {
return false
}
}
return true
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ x: Int) -> Bool {
let str = String(x)
let length = str.count
for i in 0..<length/2 {
if str[str.index(str.startIndex, offsetBy: i)]!= str[str.index(str.endIndex, offsetBy: -i - 1)] {
return false
}
}
return true
}
```if str[str.index(str.startIndex, offsetBy: i)]!= str[str.index(str.endIndex, offsetBy: -i - 1)] { // '=' must have consistent whitespace on both sidesif str[str.index(str.startIndex, offsetBy: i)]!= str[str.index(str.endIndex, offsetBy: -i - 1)] { // use of '=' in a boolean context, did you mean '=='?if str[str.index(str.startIndex, offsetBy: i)]!= str[str.index(str.endIndex, offsetBy: -i - 1)] { // cannot force unwrap value of non-optional type 'Character'func evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
let string = String(i)
let isPalindrome = string == String(string.reversed())
if isPalindrome {
if string.count % 2 == 0 {
countEven += 1
} else {
countOdd += 1
}
}
}
return (countEven, countOdd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
let string = String(i)
let isPalindrome = string == String(string.reversed())
if isPalindrome {
if string.count % 2 == 0 {
countEven += 1
} else {
countOdd += 1
}
}
}
return (countEven, countOdd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}Question:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
Answer:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
let isPalindrome = s == String(s.reversed())
if isPalindrome {
if s.count % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var n = num
var rev = 0
while n > 0 {
let dig = n % 10
rev = rev * 10 + dig
n = n / 10
}
return num == rev
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var n = num
var rev = 0
while n > 0 {
let dig = n % 10
rev = rev * 10 + dig
n = n / 10
}
return num == rev
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var n = num
var rev = 0
while n > 0 {
let dig = n % 10
rev = rev * 10 + dig
n = n / 10
}
return num == rev
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var n = num
var rev = 0
while n > 0 {
let dig = n % 10
rev = rev * 10 + dig
n = n / 10
}
return num == rev
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var n = num
var rev = 0
while n > 0 {
let dig = n % 10
rev = rev * 10 + dig
n = n / 10
}
return num == rev
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var n = num
var rev = 0
while n > 0 {
let dig = n % 10
rev = rev * 10 + dig
n = n / 10
}
return num == rev
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var n = num
var rev = 0
while n > 0 {
let dig = n % 10
rev = rev * 10 + dig
n = n / 10
}
return num == rev
}You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
var n = num
var rev = 0
while n > 0 {
let dig = n % 10
rev = rev * 10 + dig
n = n / 10
}
return num == rev
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)[/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)[/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)[/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)[/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)[/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)[/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)[/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)[/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)[/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}[INST] Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)[/INST] ```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let reversedString = String(numberString.reversed())
return numberString == reversedString
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let reversedString = String(numberString.reversed())
return numberString == reversedString
}
// Test cases
print(evenOddPalindrome(n: 3)) // Output: (1, 2)
print(evenOddPalindrome(n: 12)) // Output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let reversedString = String(numberString.reversed())
return numberString == reversedString
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let reversedString = String(numberString.reversed())
return numberString == reversedString
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let reversedString = String(numberString.reversed())
return numberString == reversedString
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let reversedString = String(numberString.reversed())
return numberString == reversedString
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i) == String(String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let original = String(num)
let reversed = String(original.reversed())
return original == reversed
}Here is the Swift code that solves the problem:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let original = String(num)
let reversed = String(original.reversed())
return original == reversed
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}Here is the Swift code to solve the problem:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}Here is the Swift code for the problem:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let original = String(num)
let reversed = String(original.reversed())
return original == reversed
}Here is the Swift code that solves the problem:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let original = String(num)
let reversed = String(original.reversed())
return original == reversed
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}Here is the Swift code that solves the problem:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let original = String(num)
let reversed = String(original.reversed())
return original == reversed
}Here is the Swift code to solve the problem:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let original = String(num)
let reversed = String(original.reversed())
return original == reversed
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(i: Int) -> Bool {
let original = String(i)
let reversed = String(original.reversed())
return original == reversed
}Here is the Swift code:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(i: Int) -> Bool {
let original = String(i)
let reversed = String(original.reversed())
return original == reversed
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let originalNumber = String(number)
let reversedNumber = String(originalNumber.reversed())
return originalNumber == reversedNumber
}Here is the Swift code for the problem:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let originalNumber = String(number)
let reversedNumber = String(originalNumber.reversed())
return originalNumber == reversedNumber
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}Here is the Swift function that solves the problem:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}Here is the Swift code for the problem:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ x: Int) -> Bool {
let s = String(x)
return s == String(s.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ x: Int) -> Bool {
let s = String(x)
return s == String(s.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
private func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
private func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ x: Int) -> Bool {
let s = String(x)
return s == String(s.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ x: Int) -> Bool {
let s = String(x)
return s == String(s.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ x: Int) -> Bool {
let s = String(x)
return s == String(s.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ x: Int) -> Bool {
let s = String(x)
return s == String(s.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ number: Int) -> Bool {
let str = String(number)
return str == String(str.reversed())
}
```func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
return strNum == String(strNum.reversed())
}
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == str.reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == str.reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if str == str.reversed() { // binary operator '==' cannot be applied to operands of type 'String' and '[String.Element]' (aka 'Array<Character>')func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return n == reversed
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return n == reversed
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isEvenPalindrome(i) {
even += 1
}
if isOddPalindrome(i) {
odd += 1
}
}
return (even, odd)
}
func isEvenPalindrome(_ n: Int) -> Bool {
let str = String(n)
let strReverse = String(str.reversed())
return str == strReverse
}
func isOddPalindrome(_ n: Int) -> Bool {
let str = String(n)
let strReverse = String(str.reversed())
return str.dropFirst() == strReverse.dropLast()
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isEvenPalindrome(i) {
even += 1
}
if isOddPalindrome(i) {
odd += 1
}
}
return (even, odd)
}
func isEvenPalindrome(_ n: Int) -> Bool {
let str = String(n)
let strReverse = String(str.reversed())
return str == strReverse
}
func isOddPalindrome(_ n: Int) -> Bool {
let str = String(n)
let strReverse = String(str.reversed())
return str.dropFirst() == strReverse.dropLast()
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str.isPalindrome && i % 2 == 0 {
even += 1
} else if str.isPalindrome && i % 2 != 0 {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str.isPalindrome && i % 2 == 0 {
even += 1
} else if str.isPalindrome && i % 2 != 0 {
odd += 1
}
}
return (even, odd)
}
```if str.isPalindrome && i % 2 == 0 { // value of type 'String' has no member 'isPalindrome'} else if str.isPalindrome && i % 2 != 0 { // value of type 'String' has no member 'isPalindrome'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i.description.isPalindrome && i % 2 == 0 {
even += 1
} else if i.description.isPalindrome && i % 2 != 0 {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i.description.isPalindrome && i % 2 == 0 {
even += 1
} else if i.description.isPalindrome && i % 2 != 0 {
odd += 1
}
}
return (even, odd)
}
```if i.description.isPalindrome && i % 2 == 0 { // value of type 'String' has no member 'isPalindrome'} else if i.description.isPalindrome && i % 2 != 0 { // value of type 'String' has no member 'isPalindrome'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == s.reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == s.reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if s == s.reversed() { // binary operator '==' cannot be applied to operands of type 'String' and '[String.Element]' (aka 'Array<Character>')func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i.description.isPalindrome && i % 2 == 0 {
even += 1
} else if i.description.isPalindrome && i % 2 != 0 {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i.description.isPalindrome && i % 2 == 0 {
even += 1
} else if i.description.isPalindrome && i % 2 != 0 {
odd += 1
}
}
return (even, odd)
}
```if i.description.isPalindrome && i % 2 == 0 { // value of type 'String' has no member 'isPalindrome'} else if i.description.isPalindrome && i % 2 != 0 { // value of type 'String' has no member 'isPalindrome'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == str.reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == str.reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if str == str.reversed() { // binary operator '==' cannot be applied to operands of type 'String' and '[String.Element]' (aka 'Array<Character>')func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return n == reversed
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n > 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return n == reversed
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i == Int(String(i).reversed().map { Int(String($0))! }) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if i == Int(String(i).reversed().map { Int(String($0))! }) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if i == Int(String(i).reversed().map { Int(String($0))! }) { // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let numberArray = Array(numberString)
var reversedNumberArray = numberArray.reversed()
if numberArray == reversedNumberArray {
return true
} else {
return false
}
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let numberArray = Array(numberString)
var reversedNumberArray = numberArray.reversed()
if numberArray == reversedNumberArray {
return true
} else {
return false
}
}
```if numberArray == reversedNumberArray { // cannot convert value of type 'ReversedCollection<[String.Element]>' (aka 'ReversedCollection<Array<Character>>') to expected argument type '[String.Element]' (aka 'Array<Character>')func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenPalindromes = 0
var oddPalindromes = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenPalindromes += 1
} else {
oddPalindromes += 1
}
}
}
return (evenPalindromes, oddPalindromes)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let numberArray = Array(numberString)
var isPalindrome = true
for i in 0..<numberArray.count/2 {
if numberArray[i] != numberArray[numberArray.count-i-1] {
isPalindrome = false
}
}
return isPalindrome
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenPalindromes = 0
var oddPalindromes = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenPalindromes += 1
} else {
oddPalindromes += 1
}
}
}
return (evenPalindromes, oddPalindromes)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let numberArray = Array(numberString)
var isPalindrome = true
for i in 0..<numberArray.count/2 {
if numberArray[i] != numberArray[numberArray.count-i-1] {
isPalindrome = false
}
}
return isPalindrome
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let numberArray = Array(numberString)
var isPalindrome = true
for i in 0..<numberArray.count/2 {
if numberArray[i] != numberArray[numberArray.count-i-1] {
isPalindrome = false
}
}
return isPalindrome
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let numberArray = Array(numberString)
var isPalindrome = true
for i in 0..<numberArray.count/2 {
if numberArray[i] != numberArray[numberArray.count-i-1] {
isPalindrome = false
}
}
return isPalindrome
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
var number = number
var reversed = 0
while number > 0 {
reversed = reversed * 10 + number % 10
number /= 10
}
return number == reversed
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
var number = number
var reversed = 0
while number > 0 {
reversed = reversed * 10 + number % 10
number /= 10
}
return number == reversed
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isEvenPalindrome(i) {
even += 1
} else if isOddPalindrome(i) {
odd += 1
}
}
return (even, odd)
}
func isEvenPalindrome(_ n: Int) -> Bool {
let str = String(n)
let count = str.count
var i = 0
while i < count / 2 {
if str[str.index(str.startIndex, offsetBy: i)] != str[str.index(str.startIndex, offsetBy: count - 1 - i)] {
return false
}
i += 1
}
return true
}
func isOddPalindrome(_ n: Int) -> Bool {
let str = String(n)
let count = str.count
var i = 0
while i < count / 2 {
if str[str.index(str.startIndex, offsetBy: i)] != str[str.index(str.startIndex, offsetBy: count - 2 - i)] {
return false
}
i += 1
}
return true
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isEvenPalindrome(i) {
even += 1
} else if isOddPalindrome(i) {
odd += 1
}
}
return (even, odd)
}
func isEvenPalindrome(_ n: Int) -> Bool {
let str = String(n)
let count = str.count
var i = 0
while i < count / 2 {
if str[str.index(str.startIndex, offsetBy: i)] != str[str.index(str.startIndex, offsetBy: count - 1 - i)] {
return false
}
i += 1
}
return true
}
func isOddPalindrome(_ n: Int) -> Bool {
let str = String(n)
let count = str.count
var i = 0
while i < count / 2 {
if str[str.index(str.startIndex, offsetBy: i)] != str[str.index(str.startIndex, offsetBy: count - 2 - i)] {
return false
}
i += 1
}
return true
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenPalindrome = 0
var oddPalindrome = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenPalindrome += 1
} else {
oddPalindrome += 1
}
}
}
return (evenPalindrome, oddPalindrome)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let numberArray = Array(numberString)
var reversedNumberArray = numberArray.reversed()
if numberArray == reversedNumberArray {
return true
} else {
return false
}
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenPalindrome = 0
var oddPalindrome = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenPalindrome += 1
} else {
oddPalindrome += 1
}
}
}
return (evenPalindrome, oddPalindrome)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let numberArray = Array(numberString)
var reversedNumberArray = numberArray.reversed()
if numberArray == reversedNumberArray {
return true
} else {
return false
}
}
```if numberArray == reversedNumberArray { // cannot convert value of type 'ReversedCollection<[String.Element]>' (aka 'ReversedCollection<Array<Character>>') to expected argument type '[String.Element]' (aka 'Array<Character>')func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let numberArray = Array(numberString)
var reversedNumberArray = numberArray.reversed()
if numberArray == reversedNumberArray {
return true
} else {
return false
}
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let numberArray = Array(numberString)
var reversedNumberArray = numberArray.reversed()
if numberArray == reversedNumberArray {
return true
} else {
return false
}
}
```if numberArray == reversedNumberArray { // cannot convert value of type 'ReversedCollection<[String.Element]>' (aka 'ReversedCollection<Array<Character>>') to expected argument type '[String.Element]' (aka 'Array<Character>')func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isEvenPalindrome(i) {
even += 1
}
if isOddPalindrome(i) {
odd += 1
}
}
return (even, odd)
}
func isEvenPalindrome(_ n: Int) -> Bool {
let str = String(n)
let count = str.count
for i in 0..<count/2 {
if str[str.index(str.startIndex, offsetBy: i)] != str[str.index(str.startIndex, offsetBy: count-i-1)] {
return false
}
}
return true
}
func isOddPalindrome(_ n: Int) -> Bool {
let str = String(n)
let count = str.count
for i in 0..<(count-1)/2 {
if str[str.index(str.startIndex, offsetBy: i)] != str[str.index(str.startIndex, offsetBy: count-i-1)] {
return false
}
}
return true
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isEvenPalindrome(i) {
even += 1
}
if isOddPalindrome(i) {
odd += 1
}
}
return (even, odd)
}
func isEvenPalindrome(_ n: Int) -> Bool {
let str = String(n)
let count = str.count
for i in 0..<count/2 {
if str[str.index(str.startIndex, offsetBy: i)] != str[str.index(str.startIndex, offsetBy: count-i-1)] {
return false
}
}
return true
}
func isOddPalindrome(_ n: Int) -> Bool {
let str = String(n)
let count = str.count
for i in 0..<(count-1)/2 {
if str[str.index(str.startIndex, offsetBy: i)] != str[str.index(str.startIndex, offsetBy: count-i-1)] {
return false
}
}
return true
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let numberArray = Array(numberString)
var isPalindrome = true
for i in 0..<numberArray.count/2 {
if numberArray[i] != numberArray[numberArray.count-i-1] {
isPalindrome = false
}
}
return isPalindrome
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ number: Int) -> Bool {
let numberString = String(number)
let numberArray = Array(numberString)
var isPalindrome = true
for i in 0..<numberArray.count/2 {
if numberArray[i] != numberArray[numberArray.count-i-1] {
isPalindrome = false
}
}
return isPalindrome
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n != 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return n == reversed
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reversed = 0
while n != 0 {
reversed = reversed * 10 + n % 10
n /= 10
}
return n == reversed
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
varXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
varXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
varXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
varXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countXCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd =XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scope### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenPalindrome = 0
XCTAssertTrue(evenOddPalindrome(n: 123) == (8, 13)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 12) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 3) == (1, 2)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 63) == (6, 8)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 25) == (5, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 19) == (4, 6)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 9) == (4, 5)) // cannot find 'evenOddPalindrome' in scopeXCTAssertTrue(evenOddPalindrome(n: 1) == (0, 1)) // cannot find 'evenOddPalindrome' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let str = String(i)
if isPalindrome(str) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let str = String(i)
if isPalindrome(str) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ s: String) -> Bool {
let str = s.replacingOccurrences(of: " ", with: "")
let arr = Array(str)
var left = 0, right = arr.count - 1
while left < right {
if arr[left] != arr[right] {
return false
}
left += if isPalindrome(str) { // cannot find 'isPalindrome' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let isPalindrome = isPalindrome(i)
if isPalindrome {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let isPalindrome = isPalindrome(i)
if isPalindrome {
if isEven(i) {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}
func isEven(_ n: Int) -> Bool {
return n % 2 == if isEven(i) { // cannot find 'isEven' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let num = Int(String(i))
if isPalindrome(num) {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return num == reverse
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let num = Int(String(i))
if isPalindrome(num) {
if num % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: Int) -> Bool {
var num = num
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return num == reverse
}
```if isPalindrome(num) { // value of optional type 'Int?' must be unwrapped to a value of type 'Int'if num % 2 == 0 { // value of optional type 'Int?' must be unwrapped to a value of type 'Int'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if isPalindrome(str) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ str: String) -> Bool {
let s = Array(str)
var left = 0
var right = s.count - 1
while left < right {
if s[left] != s[right] {
return false
}
left += 1
right -= 1
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if isPalindrome(str) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ str: String) -> Bool {
let s = Array(str)
var left = 0
var right = s.count - 1
while left < right {
if s[left] != s[right] {
return false
}
left += 1
right -= 1
}
return true
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let str = String(i)
if isPalindrome(str) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ str: String) -> Bool {
let s = Array(str)
let n = s.count
for i in 0..<n/2 {
if s[i] != s[n-i-1] {
return false
}
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let str = String(i)
if isPalindrome(str) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ str: String) -> Bool {
let s = Array(str)
let n = s.count
for i in 0..<n/2 {
if s[i] != s[n-i-1] {
return false
}
}
return true
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
reverse = reverse * 10 + n % 10
n /= 10
}
return n == reverse
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var n = n
var reverse = 0
while n > 0 {
reverse = reverse * 10 + n % 10
n /= 10
}
return n == reverse
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(s.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(s.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if isPalindrome(str) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ s: String) -> Bool {
let s = Array(s)
var left = 0
var right = s.count - 1
while left < right {
if s[left] != s[right] {
return false
}
left += 1
right -= 1
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if isPalindrome(str) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ s: String) -> Bool {
let s = Array(s)
var left = 0
var right = s.count - 1
while left < right {
if s[left] != s[right] {
return false
}
left += 1
right -= 1
}
return true
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let s = String(i)
let isEven = s.elementsEqual(s.reversed())
if isEven {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let s = String(i)
let isEven = s.elementsEqual(s.reversed())
if isEven {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
``` // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
``````
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Test cases
print(evenOddPalindrome(n: 3)) // Expected output: (1, 2)
print(evenOddPalindrome(n: 12)) // Expected output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
``````
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Test cases
print(evenOddPalindrome(n: 3)) // Expected output: (1, 2)
print(evenOddPalindrome(n: 12)) // Expected output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
``````
```swift
// Test cases
print(evenOddPalindrome(n: 3)) // Expected output: (1, 2)
print(evenOddPalindrome(n: 12)) // Expected output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
``````
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Test cases
print(evenOddPalindrome(n: 3)) // Expected output: (1, 2)
print(evenOddPalindrome(n: 12)) // Expected output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
``````
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Test cases
print(evenOddPalindrome(n: 3)) // Expected output: (1, 2)
print(evenOddPalindrome(n: 12)) // Expected output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
``````
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Test cases
print(evenOddPalindrome(n: 3)) // Expected output: (1, 2)
print(evenOddPalindrome(n: 12)) // Expected output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
``````
```swift
// Test cases
print(evenOddPalindrome(n: 3)) // Expected output: (1, 2)
print(evenOddPalindrome(n: 12)) // Expected output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
``````
```swift
// Test cases
print(evenOddPalindrome(n: 3)) // Expected output: (1, 2)
print(evenOddPalindrome(n: 12)) // Expected output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversed = String(str.reversed())
if str == reversed {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
``````
```swift
// Test cases
print(evenOddPalindrome(n: 3)) // Expected output: (1, 2)
print(evenOddPalindrome(n: 12)) // Expected output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
``````
```swift
// Test cases
print(evenOddPalindrome(n: 3)) // Expected output: (1, 2)
print(evenOddPalindrome(n: 12)) // Expected output: (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let isPalindrome = isPalindrome(i)
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let isPalindrome = isPalindrome(i)
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(n: Int) -> Bool {
var num = n
var reverse = 0
while num > 0 {
reverse = reverse * 10 + num % 10
num /= 10
}
return n == reverse
}
```let isPalindrome = isPalindrome(i) // missing argument label 'n:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(num: Int) -> Bool {
var num = num
var res = 0
while num > 0 {
res = res * 10 + num % 10
num /= 10
}
return res == num
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(num: Int) -> Bool {
var num = num
var res = 0
while num > 0 {
res = res * 10 + num % 10
num /= 10
}
return res == num
}
```if isPalindrome(i) { // missing argument label 'num:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let isPalindrome = isPalindrome(i)
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(i: Int) -> Bool {
var i = i
var res = 0
while i > 0 {
res = res * 10 + i % 10
i /= 10
}
return res == i
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let isPalindrome = isPalindrome(i)
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(i: Int) -> Bool {
var i = i
var res = 0
while i > 0 {
res = res * 10 + i % 10
i /= 10
}
return res == i
}
```let isPalindrome = isPalindrome(i) // missing argument label 'i:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(num: Int) -> Bool {
var num = num
var res = 0
while num > 0 {
res = res * 10 + num % 10
num /= 10
}
return res == num
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(num: Int) -> Bool {
var num = num
var res = 0
while num > 0 {
res = res * 10 + num % 10
num /= 10
}
return res == num
}
```if isPalindrome(i) { // missing argument label 'num:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if isPalindrome(str: str) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(str: String) -> Bool {
let s = Array(str)
var left = 0
var right = s.count - 1
while left < right {
if s[left] != s[right] {
return false
}
left += 1
right -= 1
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if isPalindrome(str: str) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(str: String) -> Bool {
let s = Array(str)
var left = 0
var right = s.count - 1
while left < right {
if s[left] != s[right] {
return false
}
left += 1
right -= 1
}
return true
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if isPalindrome(str) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ str: String) -> Bool {
let chars = Array(str)
var left = 0
var right = chars.count - 1
while left < right {
if chars[left] != chars[right] {
return false
}
left += 1
right -= 1
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if isPalindrome(str) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ str: String) -> Bool {
let chars = Array(str)
var left = 0
var right = chars.count - 1
while left < right {
if chars[left] != chars[right] {
return false
}
left += 1
right -= 1
}
return true
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let s = String(i)
if isPalindrome(s) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ s: String) -> Bool {
let s = Array(s)
var left = 0, right = s.count - 1
while left < right {
if s[left] != s[right] {
return false
}
left += 1
right -= 1
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let s = String(i)
if isPalindrome(s) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ s: String) -> Bool {
let s = Array(s)
var left = 0, right = s.count - 1
while left < right {
if s[left] != s[right] {
return false
}
left += 1
right -= 1
}
return true
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let s = String(i)
if isPalindrome(s) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ s: String) -> Bool {
let s = s.map { String($0) }
let n = s.count
for i in 0..<n/2 {
if s[i] != s[n-1-i] {
return false
}
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let s = String(i)
if isPalindrome(s) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ s: String) -> Bool {
let s = s.map { String($0) }
let n = s.count
for i in 0..<n/2 {
if s[i] != s[n-1-i] {
return false
}
}
return true
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let s = String(i)
if isPalindrome(s) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ s: String) -> Bool {
let s = Array(s)
var l = 0, r = s.count - 1
while l < r {
if s[l] != s[r] {
return false
}
l += 1
r -= 1
}
return true
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
let s = String(i)
if isPalindrome(s) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ s: String) -> Bool {
let s = Array(s)
var l = 0, r = s.count - 1
while l < r {
if s[l] != s[r] {
return false
}
l += 1
r -= 1
}
return true
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var res = 0
while num > 0 {
res = res * 10 + num % 10
num /= 10
}
return res == n
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var res = 0
while num > 0 {
res = res * 10 + num % 10
num /= 10
}
return res == n
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
countEven += 1
} else {
countOdd += 1
}
}
}
return (countEven, countOdd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
countEven += 1
} else {
countOdd += 1
}
}
}
return (countEven, countOdd)
}
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
//: [Next](@next)if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
evenOddPalindrome(n: 3)
evenOddPalindrome(n: 12)
//: [Next](@next)if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
```if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
evenOddPalindrome(n: 3)
evenOddPalindrome(n: 12)
//: [Next](@next)if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
evenOddPalindrome(n: 3)
evenOddPalindrome(n: 12)
//: [Next](@next)if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
```if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
evenOddPalindrome(n: 3)
evenOddPalindrome(n: 12)
//: [Next](@next)if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
evenOddPalindrome(n: 3)
evenOddPalindrome(n: 12)
//: [Next](@next)if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
//: [Next](@next)if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func distinctPrimeFactors(n: Int) -> Int {
var count = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
count += 1
}
} else {
var j = 3
while j <= Int(sqrt(Double(i))) {
if i % j == 0 {
count += 1
break
}
j += 2
}
}
}
return count
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// Given a positive integer n, return the number of integers between 1 and n inclusive,
// that have no more than 4 distinct prime factors.
// Example 1:
// Explanation:
// 6 has 4 distinct prime factors (2,3,2,3), therefore it is a prime number.
// 15 has 4 distinct prime factors (3,5,2,3), therefore it is a prime number.
// 16 has 3 distinct prime factors (2,2,2), therefore it is not a prime number.
// 17 has 4 distinct prime factors (3,7,2,3), therefore it is a prime number.
// Example 2:
// Explanation:
// 12 has 4 distinct prime factors (2,2,3,3), therefore it is not a prime number.
// 15 has 4 distinct prime factors (3,5,2,3), therefore it is a prime number.
// 17 has 4 distinct prime factors (3,7,2,3), therefore it is a prime number.
// 19 has 4 distinct prime factors (3,3,7,23), therefore it is a prime number.
// 23 has 4 distinct prime factors (3,3,11,23), therefore it is a prime number.
// 29 has 4 distinct prime factors (3,3,37,29), therefore it is a prime number.
// 31 has 4 distinct prime factors (3,3,37,31), therefore it is a prime number.
// 37 has 4 distinct prime factors (3,7,31,37), therefore it is a prime number.
// 41 has 4 distinct prime factors (3,3,37,41), therefore it is a prime number.
// 43 has 4 distinct prime factors (3,3,37,43), therefore it is a prime number.
// 47 has 4 distinct prime factors (3,7,31,47), therefore it is a prime number.
// 49 has 4 distinct prime factors (3,3,37,49), therefore it is a prime number.
// 53 has 4 distinct prime factors (3,3,53,53), therefore it is a prime number.
// 59 has 4 distinct prime factors (3,3,37,53), therefore it is a prime number.
// 61 has 4 distinct prime factors (3,3,7,61), therefore it is a prime number.
// 67 has 4 distinct prime factors (3,7,31,67), therefore it is a prime number.
// 71 has 4 distinct prime factors (3,7,71,71), therefore it is a prime number.
// 73 has 4 distinct prime factors (3,3,37,73), therefore it is a prime number.
// 79 has 4 distinct prime factors (3,3,7,79), therefore it is a prime number.
// 83 has 4 distinct prime factors (3,3,37,83), therefore it is a prime number.
// 89 has 4 distinct prime factors (3,3,7,83), therefore it is a prime number.
// 97 has 4 distinct prime factors (3,7,71,97), therefore it is a prime number.
func distinctPrimeFactors(n: Int) -> Int {
var count = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
count += 1
}
} else {
var j = 3
while j <= Int(sqrt(Double(i))) {
if i % j == 0 {
count += 1
break
}
j += 2
}
}
}
return count
}
// Given a positive integer n, return the number of integers between 1 and n inclusive,
// that are palindromes.
// Example 1:
//if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if num == String(reversed: num) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if num == String(reversed: num) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
### Time Complexity:
O(n)
### Space Complexity:
O(1)if num == String(reversed: num) { // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if num == String(reversed: num) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if num == String(reversed: num) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
### Time Complexity:
O(n)
### Space Complexity:
O(1)if num == String(reversed: num) { // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if num == String(reversed: num) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if num == String(reversed: num) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```
### Time Complexity:
O(n)
### Space Complexity:
O(1)if num == String(reversed: num) { // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if num == String(reversed: num) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if num == String(reversed: num) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if num == String(reversed: num) { // no exact matches in call to initializerfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if isPalindrome(num) {
if Int(num) % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: String) -> Bool {
var reversed = ""
for char in num {
reversed.insert(char, at: 0)
}
return reversed == num
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if isPalindrome(num) {
if Int(num) % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: String) -> Bool {
var reversed = ""
for char in num {
reversed.insert(char, at: 0)
}
return reversed == num
}
```if Int(num) % 2 == 0 { // value of optional type 'Int?' must be unwrapped to a value of type 'Int'reversed.insert(char, at: 0) // cannot convert value of type 'Int' to expected argument type 'String.Index'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if isPalindrome(num) {
if Int(num) % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: String) -> Bool {
var reversed = ""
for c in num {
reversed.append(c)
}
reversed.reverse()
return reversed == num
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if isPalindrome(num) {
if Int(num) % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: String) -> Bool {
var reversed = ""
for c in num {
reversed.append(c)
}
reversed.reverse()
return reversed == num
}
```if Int(num) % 2 == 0 { // value of optional type 'Int?' must be unwrapped to a value of type 'Int'reversed.reverse() // value of type 'String' has no member 'reverse'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if isPalindrome(num) {
if Int(num) % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: String) -> Bool {
var reversed = ""
for char in num {
reversed.insert(char, at: 0)
}
return reversed == num
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if isPalindrome(num) {
if Int(num) % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: String) -> Bool {
var reversed = ""
for char in num {
reversed.insert(char, at: 0)
}
return reversed == num
}
```if Int(num) % 2 == 0 { // value of optional type 'Int?' must be unwrapped to a value of type 'Int'reversed.insert(char, at: 0) // cannot convert value of type 'Int' to expected argument type 'String.Index'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if isPalindrome(num) {
if Int(num) % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: String) -> Bool {
var reversed = ""
for char in num {
reversed.insert(char, at: 0)
}
return reversed == num
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if isPalindrome(num) {
if Int(num) % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: String) -> Bool {
var reversed = ""
for char in num {
reversed.insert(char, at: 0)
}
return reversed == num
}
```if Int(num) % 2 == 0 { // value of optional type 'Int?' must be unwrapped to a value of type 'Int'reversed.insert(char, at: 0) // cannot convert value of type 'Int' to expected argument type 'String.Index'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if isPalindrome(num) {
if Int(num) % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: String) -> Bool {
for i in 0..<num.count {
if num[i]!= num[num.count - i - 1] {
return false
}
}
return true
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if isPalindrome(num) {
if Int(num) % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: String) -> Bool {
for i in 0..<num.count {
if num[i]!= num[num.count - i - 1] {
return false
}
}
return true
}
```if Int(num) % 2 == 0 { // value of optional type 'Int?' must be unwrapped to a value of type 'Int'if num[i]!= num[num.count - i - 1] { // use of '=' in a boolean context, did you mean '=='?if num[i]!= num[num.count - i - 1] { // '=' must have consistent whitespace on both sidesif num[i]!= num[num.count - i - 1] { // cannot force unwrap value of non-optional type 'Character'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if isPalindrome(num) {
if Int(num) % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: String) -> Bool {
if num.isEmpty {
return false
}
var reversed = ""
for i in (0..<num.count).reversed() {
reversed += String(num[i])
}
return reversed == num
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let num = String(i)
if isPalindrome(num) {
if Int(num) % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ num: String) -> Bool {
if num.isEmpty {
return false
}
var reversed = ""
for i in (0..<num.count).reversed() {
reversed += String(num[i])
}
return reversed == num
}
```if Int(num) % 2 == 0 { // value of optional type 'Int?' must be unwrapped to a value of type 'Int'reversed += String(num[i]) // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = i
var t = i
while s > 0 {
t = t * 10 + s % 10
s /= 10
}
if t == i {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
func consecutiveOnes(n: Int) -> Int {
var count = 0
var prev = 0
var curr = 0
for i in 1...n {
prev = curr
curr = (prev + 1) % 2
if prev == 1 && curr == 1 {
count += 1
}
}
return n - count
}
func trailingZeroes(n: Int) -> Int {
var count = 0
while n > 0 {
count += n / 5
n /= 5
}
return count
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = i
var t = i
while s > 0 {
t = t * 10 + s % 10
s /= 10
}
if t == i {
even += 1
} else {
odd += 1
}
}
return (even, odd)
}
// 10.11
// Given a positive integer n, return the number of positive integers less than or equal to n, whose binary representations do not contain consecutive ones.
// Example 1:
// Input: 5
// Output: 5
// Explanation:
// Here are the non-negative integers <= 5 with their corresponding binary representations:
// 0 : 0
// 1 : 1
// 2 : 10
// 3 : 11
// 4 : 100
// 5 : 101
// Among them, only integer 3 disobeys the rule (two consecutive ones in its binary representation) and the other 5 satisfy the rule.
// Example 2:
// Input: 1
// Output: 2
// Explanation:
// Since there is only one integer that fits the given rules, 1 does satisfy the rule.
// Note that 2 does not satisfy the rule because there are consecutive ones in its binary representation: 10.
// Note:
// 1. 1 <= n <= 10^9
// 2. The answer is guaranteed to fit in a 32-bit signed integer.
// consecutiveOnes(n: 5) == 5
// consecutiveOnes(n: 1) == 2
func consecutiveOnes(n: Int) -> Int {
var count = 0
var prev = 0
var curr = 0
for i in 1...n {
prev = curr
curr = (prev + 1) % 2
if prev == 1 && curr == 1 {
count += 1
}
}
return n - count
}
// 10.12
// Given an integer n, return the number of trailing zeroes in n!.
// Example 1:
// Input: 3
// Output: 0
// Explanation: 3! = 6, no trailing zero.
// Example 2:
// Input: 5
// Output: 1
// Explanation: 5! = 120, one trailing zero.
// Note: Your solution should be in logarithmic time complexity.
// Note:
// 1. 1 <= n <= 10^4
// 2. The answer is guaranteed to fit in a 32-bit signed integer.
// trailingZeroes(n: 3) == 0
// trailingZeroes(n: 5) == 1
func trailingZeroes(n: Int) -> Int {
var count = 0
while n > 0 {
count += n / 5
n /= 5
}
return count
}
// 10.13
// Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
// Example 1:
// Input: n = 12
// Output: 3
// Explanation: 12 = 4 + 4 + 4.
// Example 2:
// Input: n = 13
// Output: 2
// Explanation: 13 = 4 + 9.
// Note:
// 1. 1 <= n <= 10^4
// 2. The answer is guaranteed to fit in a 32-bit signed integer.
// numSquares(n: 12) == 3
// numSquares(n: 13) == 2
func numSquares(n: Int) -> Int {
var dp = [Int.max] * (n + 1)
dp[0] = 0
for i in 1...n {
for j in 1...Int(sqrt(Double(i))) {
dp[i] = min(dp[i],for i in 1...n { // immutable value 'i' was never used; consider replacing with '_' or removing itn /= 5 // left side of mutating operator isn't mutable: 'n' is a 'let' constantfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(String(s).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(String(s).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// 2. Given a positive integer n, return the number of palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(String(s).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// 3. Given a positive integer n, return the number of palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(String(s).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// 4. Given a positive integer n, return the number of palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s.reversed()).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s.reversed()).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// 2. Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s.reversed()).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// 3. Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s.reversed()).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// 4. Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) ==var s = String(i) // variable 's' was never mutated; consider changing to 'let' constant // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s[..<s.count]).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s[..<s.count]).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
print(evenOddPalindrome(3))
print(evenOddPalindrome(12))
// 10. Write a function that takes a string as input and returns the string reversed.
// Example 1:
// Input: "hello"
// Output: "olleh"
// Example 2:
// Input: "A man, a plan, a canal: Panama"
// Output: "amanaP :lanac a ,nalp a ,nam A"
// Note:
// 1. What constitutes a palindrome?
// 2. Did you ignore case?
// 3. There are four important cases to consider:
// 1. Empty string
// 2. Single character
// 3. Palindrome with even number of characters
// 4. Palindrome with odd number of characters
// 4. What are the base cases?
// 5. How do you handle the case where the input string is empty?
// 6. How do you handle the case where the input string has only one character?
// 7. How do you handle the case where the input string has only two characters?
// 8. How do you handle the case where the input string has more than two characters?
// 9. How do you handle the case where the input string has even number of characters?
// 10. How do you handle the case where the input string has odd number of characters?
// 11. How do you handle the case where the input string has spaces?
// 12. How do you handle the case where the input string has non-alphanumeric characters?
// 13. How do you handle the case where the input string has uppercase and lowercase letters?
// 14. How do you handle the case where the input string has uppercase and lowercase letters and spaces?
// 15. How do you handle the case where the input string has uppercase and lowercase letters, spaces, and non-alphanumeric characters?
// 16. How do you handle the case where the input string has uppercase and lowercase letters, spaces, non-alphanumeric characters, and spaces?
// 17. How do you handle the case where the input string has uppercase and lowercase letters, spaces, non-alphanumeric characters, spaces, and spaces?
// 18. How do you handle the case where the input string has uppercase and lowercase letters, spaces, non-alphanumeric characters, spaces, spaces, and spaces?
// 19. How do you handle the case where the input string has uppercase and lowercase letters, spaces, non-alphanumeric characters, spaces, spaces, spaces, and spaces?
// 20. How do you handle the case where the input string has uppercase and lowercase letters, spaces, non-alphanumeric characters, spaces, spaces, spaces, spaces, and spaces?
// 21. How do you handle the case where the input string has uppercase and lowercase letters, spaces, non-alphanumeric characters, spaces, spaces, spaces, spaces, spaces, and spaces?
// 22. How do you handle the case where the input string has uppercase and lowercase letters, spaces, non-alphanumeric characters, spaces, spaces, spaces, spaces, spaces, spaces, and spaces?
// 23. How do you handle the case where the input string has uppercase and lowercase letters, spaces, non-alphanumeric characters, spaces, spaces, spaces, spaces, spaces, spaces, spaces, and spaces?
// 24. How do you handle the case where the input string has uppercase and lowercase letters, spaces, non-alphanumeric characters, spaces, spaces, spaces, spaces, spaces, spaces, spaces, spaces, and spaces?
// 25. How do you handle the case where the input string has uppercase and lowercase letters, spaces, non-alphanumeric characters, spaces, spaces, spaces, spaces, spaces, spaces, spaces, spaces, spaces, and spaces?
// 26. How do you handle the case where the inputvar s = String(i) // variable 's' was never mutated; consider changing to 'let' constantif s == String(String(s[..<s.count]).reversed()) { // 'subscript(_:)' is unavailable: cannot subscript String with an integer range, use a String.Index range instead.func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
var isPalindrome = true
for j in 0..<s.count/2 {
if s[j] != s[s.count-1-j] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func romanToInt(s: String) -> Int {
var result = 0
var i = 0
while i < s.count {
switch s[i] {
case 'I':
if i+1 < s.count && s[i+1] == 'V' {
result += 4
i += 2
} else if i+1 < s.count && s[i+1] == 'X' {
result += 9
i += 2
} else {
result += 1
i += 1
}
case 'V':
result += 5
i += 1
case 'X':
if i+1 < s.count && s[i+1] == 'L' {
result += 40
i += 2
} else if i+1 < s.count && s[i+1] == 'C' {
result += 90
i += 2
} else {
result += 10
i += 1
}
case 'L':
result += 50
i += 1
case 'C':
if i+1 < s.count && s[i+1] == 'D' {
result += 400
i += 2
} else if i+1 < s.count && s[i+1] == 'M' {
result += 900
i += 2
} else {
result += 100
i += 1
}
case 'D':
result += 500
i += 1
case 'M':
result += 1000
i += 1
default:
break
}
}
return result
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
var isPalindrome = true
for j in 0..<s.count/2 {
if s[j] != s[s.count-1-j] {
isPalindrome = false
break
}
}
if isPalindrome {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
print(evenOddPalindrome(12))
print(evenOddPalindrome(3))
// -----------------------------------------------------------------------------
// 13. 13. Roman to Integer
// Given a roman numeral, convert it to an integer.
// Input is guaranteed to be within the range from 1 to 3999.
// Example 1:
// Input: "III"
// Output: 3
// Example 2:
// Input: "IV"
// Output: 4
// Example 3:
// Input: "IX"
// Output: 9
// Example 4:
// Input: "LVIII"
// Output: 58
// Explanation: L = 50, V= 5, III = 3.
// Example 5:
// Input: "MCMXCIV"
// Output: 1994
// Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
func romanToInt(s: String) -> Int {
var result = 0
var i = 0
while i < s.count {
switch s[i] {
case 'I':
if i+1 < s.count && s[i+1] == 'V' {
result += 4
i += 2
} else if i+1 < s.count && s[i+1] == 'X' {
result += 9
i += 2
} else {
result += 1
i += 1
}
case 'V':
result += 5
i += 1
case 'X':
if i+1 < s.count && s[i+1] == 'L' {
result += 40
i += 2
} else if i+1 < s.count && s[i+1] == 'C' {
result += 90
i += 2
} else {
result += 10
i += 1
}
case 'L':
result += 50
i += 1
case 'C':
if i+1 < s.count && s[i+1] == 'D' {
result += 400
i += 2
} else if i+1 < s.count && s[i+1] == 'M' {
result += 900
i += 2
} else {
result += 100
i += 1
}
case 'D':
result += 500
i += 1
case 'M':
result += 1000
i += 1
default:
break
}
}
return result
}
print(romanToInt("III"))
print(romanToInt("IV"))
print(romanToInt("IX"))
print(romanToInt("LVIII"))
print(romanToInt("MCMXCIV"))
// -----------------------------------------------------------------------------
// 14. 14. Longest Common Prefix
// Write a function to find the longest common prefix string amongst an array of strings.
// If there is no common prefix, return an empty string "".
// Example 1:
// Input: ["flower","flow","flight"]
// Output: "var s = String(i) // variable 's' was never mutated; consider changing to 'let' constantif s[j] != s[s.count-1-j] { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.if s[j] != s[s.count-1-j] { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.switch s[i] { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.case 'I': // single-quoted string literal found, use '"'if i+1 < s.count && s[i+1] == 'V' { // single-quoted string literal found, use '"'if i+1 < s.count && s[i+1] == 'V' { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.} else if i+1 < s.count && s[i+1] == 'X' { // single-quoted string literal found, use '"'} else if i+1 < s.count && s[i+1] == 'X' { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.case 'V': // single-quoted string literal found, use '"'case 'X': // single-quoted string literal found, use '"'if i+1 < s.count && s[i+1] == 'L' { // single-quoted string literal found, use '"'if i+1 < s.count && s[i+1] == 'L' { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.} else if i+1 < s.count && s[i+1] == 'C' { // single-quoted string literal found, use '"'} else if i+1 < s.count && s[i+1] == 'C' { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.case 'L': // single-quoted string literal found, use '"'case 'C': // single-quoted string literal found, use '"'if i+1 < s.count && s[i+1] == 'D' { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.if i+1 < s.count && s[i+1] == 'D' { // single-quoted string literal found, use '"'} else if i+1 < s.count && s[i+1] == 'M' { // 'subscript(_:)' is unavailable: cannot subscript String with an Int, use a String.Index instead.} else if i+1 < s.count && s[i+1] == 'M' { // single-quoted string literal found, use '"'case 'D': // single-quoted string literal found, use '"'case 'M': // single-quoted string literal found, use '"'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s.reversed()).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s.reversed()).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// 2. Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s.reversed()).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// 3. Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s.reversed()).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// 4. Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) ==var s = String(i) // variable 's' was never mutated; consider changing to 'let' constant // XCTAssertTrue failedfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// 2. Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// 3. Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var s = String(i)
if s == String(String(s).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// 4. Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenvar s = String(i) // variable 's' was never mutated; consider changing to 'let' constantfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(String(s).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let s = String(i)
if s == String(String(s).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
print(evenOddPalindrome(12))
// ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var isPalindrome = true
var reversed = 0
var number = i
while number > 0 {
reversed = reversed * 10 + number % 10
number /= 10
}
if reversed == i {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var isPalindrome = true
var reversed = 0
var number = i
while number > 0 {
reversed = reversed * 10 + number % 10
number /= 10
}
if reversed == i {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
var result = evenOddPalindrome(12)
print("result: \(result)")
// 2. Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var isPalindrome = true
var reversed = 0
var number = i
while number > 0 {
reversed = reversed * 10 + number % 10
number /= 10
}
if reversed == i {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
var result = evenOddPalindrome(12)
print("result: \(result)")
// 3. Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var isPalindrome = true
var reversed = 0
var number = i
while number > 0 {
reversed = reversed * 10 + number % 10
number /= 10
}
if reversed == i {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
var result = evenOddPalindrome(12)
print("result: \(result)")
// 4. Given a positive integer n, return a tuple that has the numbervar isPalindrome = true // initialization of variable 'isPalindrome' was never used; consider replacing with assignment to '_' or removing itfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i) == String(i).reversed() {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
// 2. 2. Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(n, 2n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196if String(i) == String(i).reversed() { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
if strI == reversedStrI {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
if strI == reversedStrI {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let intReversedI = Int(reversedStrI)! // initialization of immutable value 'intReversedI' was never used; consider replacing with assignment to '_' or removing itfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let intReversedI = Int(reversedStrI)! // initialization of immutable value 'intReversedI' was never used; consider replacing with assignment to '_' or removing itfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let intReversedI = Int(reversedStrI)! // initialization of immutable value 'intReversedI' was never used; consider replacing with assignment to '_' or removing itfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
if strI == reversedStrI {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
if strI == reversedStrI {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let intReversedI = Int(reversedStrI)! // initialization of immutable value 'intReversedI' was never used; consider replacing with assignment to '_' or removing itfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
if strI == reversedStrI {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
if strI == reversedStrI {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
print(evenOddPalindrome(n: 3)) // (1, 2)
print(evenOddPalindrome(n: 12)) // (4, 6)
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let intReversedI = Int(reversedStrI)! // initialization of immutable value 'intReversedI' was never used; consider replacing with assignment to '_' or removing itfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let intReversedI = Int(reversedStrI)! // initialization of immutable value 'intReversedI' was never used; consider replacing with assignment to '_' or removing itfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let strI = String(i)
let reversedStrI = String(strI.reversed())
let intI = Int(strI)!
let intReversedI = Int(reversedStrI)!
if strI == reversedStrI {
if intI % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```let intReversedI = Int(reversedStrI)! // initialization of immutable value 'intReversedI' was never used; consider replacing with assignment to '_' or removing itfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let stringI = String(i)
let reversedStringI = String(stringI.reversed())
if stringI == reversedStringI {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let stringI = String(i)
let reversedStringI = String(stringI.reversed())
if stringI == reversedStringI {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let stringI = String(i)
let reversedStringI = String(stringI.reversed())
if stringI == reversedStringI {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let stringI = String(i)
let reversedStringI = String(stringI.reversed())
if stringI == reversedStringI {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let stringI = String(i)
let reversedStringI = String(stringI.reversed())
if stringI == reversedStringI {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let stringI = String(i)
let reversedStringI = String(stringI.reversed())
if stringI == reversedStringI {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let stringI = String(i)
let isPalindrome = stringI == String(stringI.reversed())
if isPalindrome {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let stringI = String(i)
let isPalindrome = stringI == String(stringI.reversed())
if isPalindrome {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
let reversedStr = String(str.reversed())
if str == reversedStr {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let stringI = String(i)
let isPalindrome = stringI == String(stringI.reversed())
if isPalindrome {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let stringI = String(i)
let isPalindrome = stringI == String(stringI.reversed())
if isPalindrome {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
// Helper function to check if a number is a palindrome
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(number: i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(number: Int) -> Bool {
let str = String(number)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if isPalindrome(i) { // cannot find 'isPalindrome' in scopefunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}```swift
// Given a positive integer n, return a tuple that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned tuple has the number of even and odd integer palindromes respectively.
// evenOddPalindrome(n: 3) == (1, 2)
// evenOddPalindrome(n: 12) == (4, 6)
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i).isEqual(to: String(i).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i).isEqual(to: String(i).reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if String(i).isEqual(to: String(i).reversed()) { // extraneous argument label 'to:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i).isEqual(String(i.reversed())) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i).isEqual(String(i.reversed())) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if String(i).isEqual(String(i.reversed())) { // value of type 'Int' has no member 'reversed'func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i).isEqual(to: String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i).isEqual(to: String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i).isEqual(to: String(i).reversed()) { // extraneous argument label 'to:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i).isEqual(to: String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i).isEqual(to: String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```
if String(i).isEqual(to: String(i).reversed()) { // extraneous argument label 'to:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
if String(i).isEqual(to: String(i)[..]) {
if i % 2 == 0 {
countEven += 1
} else {
countOdd += 1
}
}
}
return (countEven, countOdd)
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var countEven = 0
var countOdd = 0
for i in 1...n {
if String(i).isEqual(to: String(i)[..]) {
if i % 2 == 0 {
countEven += 1
} else {
countOdd += 1
}
}
}
return (countEven, countOdd)
}
```if String(i).isEqual(to: String(i)[..]) { // cannot find operator '..' in scope; did you mean '...'?if String(i).isEqual(to: String(i)[..]) { // extraneous argument label 'to:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i).isEqual(to: String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i).isEqual(to: String(i).reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i).isEqual(to: String(i).reversed()) { // extraneous argument label 'to:' in callfunc evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i).reversed() == String(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if String(i).reversed() == String(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```if String(i).reversed() == String(i) { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i).reversed() == String(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i).reversed() == String(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if String(i).reversed() == String(i) { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
if String(i).reversed() == String(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0, odd = 0
for i in 1...n {
if String(i).reversed() == String(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```if String(i).reversed() == String(i) { // cannot convert value of type 'ReversedCollection<String>' to expected argument type 'String'func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i).isEqual(to: String(i)[..<String(i).count/2]) && String(i).count % 2 == 0 {
even += 1
} else if String(i).isEqual(to: String(i)[..<String(i).count/2]) {
odd += 1
}
}
return (even, odd)
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if String(i).isEqual(to: String(i)[..<String(i).count/2]) && String(i).count % 2 == 0 {
even += 1
} else if String(i).isEqual(to: String(i)[..<String(i).count/2]) {
odd += 1
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
return str == String(str.reversed())
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
private func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
private func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let str = String(num)
let reversedStr = String(str.reversed())
return str == reversedStr
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}system
You are Qwen, created by Alibaba Cloud. You are a helpful assistant.
user
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
assistant
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
func isPalindrome(_ num: Int) -> Bool {
let strNum = String(num)
let reversedStrNum = String(strNum.reversed())
return strNum == reversedStrNum
}
```
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var num = i
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
if reversed == i {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var num = i
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
if reversed == i {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == n
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
if isPalindrome(i) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
func isPalindrome(_ n: Int) -> Bool {
var num = n
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
return reversed == n
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var num = i
var reversedNum = 0
while num > 0 {
reversedNum = reversedNum * 10 + num % 10
num /= 10
}
if reversedNum == i {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var num = i
var reversedNum = 0
while num > 0 {
reversedNum = reversedNum * 10 + num % 10
num /= 10
}
if reversedNum == i {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
let str = String(i)
if str == String(str.reversed()) {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var num = i
var reversedNum = 0
while num > 0 {
reversedNum = reversedNum * 10 + num % 10
num /= 10
}
if reversedNum == i {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var num = i
var reversedNum = 0
while num > 0 {
reversedNum = reversedNum * 10 + num % 10
num /= 10
}
if reversedNum == i {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let string = String(i)
if string == String(string.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
let string = String(i)
if string == String(string.reversed()) {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var num = i
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
if reversed == i {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var even = 0
var odd = 0
for i in 1...n {
var num = i
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
if reversed == i {
if i % 2 == 0 {
even += 1
} else {
odd += 1
}
}
}
return (even, odd)
}
```func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var num = i
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
if reversed == i {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Given a positive integer n, return a tuple that has the number of even and odd
integer palindromes that fall within the range(1, n), inclusive.
Example 1:
Explanation:
Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
Example 2:
Explanation:
Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
Note:
1. 1 <= n <= 10^3
2. returned tuple has the number of even and odd integer palindromes respectively.
evenOddPalindrome(n: 3) == (1, 2)
evenOddPalindrome(n: 12) == (4, 6)
### Response:
```swift
func evenOddPalindrome(n: Int) -> (Int, Int) {
var evenCount = 0
var oddCount = 0
for i in 1...n {
var num = i
var reversed = 0
while num > 0 {
reversed = reversed * 10 + num % 10
num /= 10
}
if reversed == i {
if i % 2 == 0 {
evenCount += 1
} else {
oddCount += 1
}
}
}
return (evenCount, oddCount)
}
```