func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring, index) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring, index) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring, index) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring, index) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring, index) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring, index) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "") == ""
// longestSubstring(string: "aaa") == "aaa"
// longestSubstring(string: "aaaa") == "aaa"
// longestSubstring(string: "ab") == "b"
// longestSubstring(string: "abababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababab

func howManyTimes(string: String, substring: String) -> Int {
  var count = 0
  var index = 0
  while index < string.count {
    if string.hasPrefix(substring, index) {
      count += 1
      index += substring.count
    } else {
      index += 1
    }
  }
  return count
}

func longestSubstring(string: String) -> String {
  var longest = ""
  for i in 0..<string.count {
    for j in i..<string.count {
      let substring = string[string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: j+1)]
      if substring.count > longest.count {
        longest = substring
      }
    }
  }
  return longest
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
  var count = 0
  var index = 0
  while index < string.count {
    if string.hasPrefix(substring, index) {
      count += 1
      index += substring.count
    } else {
      index += 1
    }
  }
  return count
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "aaa") == "aaa"
// longestSubstring(string: "aaaa") == "aaa"
// longestSubstring(string: "aaabbb") == "bbb"
func longestSubstring(string: String) -> String {
  var longest = ""
  for i in 0..<string.count {
    for j in i..<string.count {
      let substring = string[string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: j+1)]
      if substring.count > longest.count {
        longest = substring
      }
    }
  }
  return longest
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "aaa") == "aaa"
// longestSubstring(string: "aaaa") == "aaa"
// longestSubstring(string: "aaabbb") == "bbb"
func longestSubstring(string: String) -> String {
  var longest = ""
  for i in 0..<string.count {
    for j in i..<string.count {
      let substring = string[string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: j+1)]
      if substring.count > longest.count {
        longest = substring
      }
    }
  }
  return longest
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "aaa") == "aaa"
// longestSubstring(string: "aaaa") == "aaa"
// longestSubstring(string: "aaabbb") == "bbb"
func longestSubstring(string: String) -> String {
  var longest = ""
  for i in 0..<string.count {
    for j in i..<string.count {
      let substring = string[string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: j+1)]
      if substring.count > longest.count {
        longest = substring
      }
    }
  }
  return longest
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "aaa") == "aaa"
// longestSubstring(string: "aaaa") == "aaa"
// longestSubstring(string: "aaabbb") == "bbb"
func longestSubstring(string: String) -> String {
  var longest = ""
  for i in 0..<string.count {
    for j in i..<string.count {
      let substring = string[string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: j+1)]
      if substring.count > longest.count {
        longest = substring
      }
    }
  }
  return longest
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "aaa") == "aaa"
// longestSubstring(string: "aaaa") == "aaa"
// longestSubstring(string: "aaabbb") == "bbb"
func longestSubstring(string: String) -> String {
  var longest = ""
  for i in 0..<string.count {
    for j in i..<string.count {
      let substring = string[string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: j+1)]
      if substring.count > longest.count {
        longest = substring
      }
    }
  }
  return longest
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "aaa") == "aaa"
// longestSubstring(string: "aaaa") == "aaa"
// longestSubstring(string: "aaabbb") == "bbb"
func longestSubstring(string: String) -> String {
  var longest = ""
  for i in 0..<string.count {
    for j in i..<string.count {
      let substring = string[string.index(string.startIndex, offsetBy

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring, index) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

func longestSubstring(string: String) -> String {
    var longest = ""
    for i in 0..<string.count {
        for j in i..<string.count {
            let substring = string[string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: j + 1)]
            if substring.count > longest.count {
                longest = substring
            }
        }
    }
    return longest
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring, index) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "aaa") == "aaa"
// longestSubstring(string: "aaaa") == "aaa"
// longestSubstring(string: "aaabbb") == "bbb"
// longestSubstring(string: "aaabbbccc") == "bbb"
func longestSubstring(string: String) -> String {
    var longest = ""
    for i in 0..<string.count {
        for j in i..<string.count {
            let substring = string[string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: j + 1)]
            if substring.count > longest.count {
                longest = substring
            }
        }
    }
    return longest
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "aaa") == "aaa"
// longestSubstring(string: "aaaa") == "aaa"
// longestSubstring(string: "aaabbb") == "bbb"
// longestSubstring(string: "aaabbbccc") == "bbb"
func longestSubstring(string: String) -> String {
    var longest = ""
    for i in 0..<string.count {
        for j in i..<string.count {
            let substring = string[string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: j + 1)]
            if substring.count > longest.count {
                longest = substring
            }
        }
    }
    return longest
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "aaa") == "aaa"
// longestSubstring(string: "aaaa") == "aaa"
// longestSubstring(string: "aaabbb") == "bbb"
// longestSubstring(string: "aaabbbccc") == "bbb"
func longestSubstring(string: String) -> String {
    var longest = ""
    for i in 0..<string.count {
        for j in i..<string.count {
            let substring = string[string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: j + 1)]
            if substring.count > longest.count {
                longest = substring
            }
        }
    }
    return longest
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "aaa") == "aaa"
// longestSubstring(string: "aaaa") == "aaa"
// longestSubstring(string: "aaabbb") == "bbb"
// longestSubstring(string: "aaabbbccc") == "bbb"
func longestSubstring(string: String) -> String {
    var longest = ""
    for i in 0..<string.count {
        for j in i..<string.count {
            let substring = string[string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: j + 1)]
            if substring.count > longest.count {
                longest = substring
            }
        }
    }
    return longest
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "aaa") == "aaa"
// longestSubstring(string: "aaaa") == "aaa"
// longestSubstring(string: "aaabbb") == "bbb"
// longestSubstring(string: "aaabbbccc") == "bbb"
func longestSubstring(string: String) -> String {
    var longest = ""
    for i in 0..<string.count {
        for j in i..<string.count {
            let substring = string[string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: j + 1)]
            if substring.count > longest.count {
                longest = substring
            }
        }
    }
    return longest
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "aaa") == "

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    for i in 0..<string.count {
        if string[string.index(string.startIndex, offsetBy: i)] == substring[substring.startIndex] {
            if string[string.index(string.startIndex, offsetBy: i + substring.count)] == substring[substring.index(substring.startIndex, offsetBy: substring.count - 1)] {
                count += 1
            }
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    for i in 0..<string.count {
        if string[string.index(string.startIndex, offsetBy: i)] == substring[substring.startIndex] {
            if string[string.index(string.startIndex, offsetBy: i + substring.count)] == substring[substring.index(substring.startIndex, offsetBy: substring.count - 1)] {
                count += 1
            }
        }
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring, index) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring, index) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    for i in 0..<string.count {
        if string.substring(with: NSRange(location: i, length: substring.count)) == substring {
            count += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    for i in 0..<string.count {
        if string.substring(with: NSRange(location: i, length: substring.count)) == substring {
            count += 1
        }
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let startIndex = string.index(string.startIndex, offsetBy: index)
        let endIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        if string[startIndex..<endIndex] == substring {
            count += 1
        }
        index += 1
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let startIndex = string.index(string.startIndex, offsetBy: index)
        let endIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        if string[startIndex..<endIndex] == substring {
            count += 1
        }
        index += 1
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    for i in 0..<string.count {
        if string[string.index(string.startIndex, offsetBy: i)] == substring[substring.startIndex] {
            if string[string.index(string.startIndex, offsetBy: i + substring.count)] == substring[substring.index(substring.startIndex, offsetBy: substring.count - 1)] {
                count += 1
            }
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    for i in 0..<string.count {
        if string[string.index(string.startIndex, offsetBy: i)] == substring[substring.startIndex] {
            if string[string.index(string.startIndex, offsetBy: i + substring.count)] == substring[substring.index(substring.startIndex, offsetBy: substring.count - 1)] {
                count += 1
            }
        }
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string.hasPrefix(substring, range: i..<string.count) {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    
    return count
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string.hasPrefix(substring, range: i..<string.count) {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string.hasPrefix(substring, range: i..<string.count) {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    
    return count
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string.hasPrefix(substring, range: i..<string.count) {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string.hasPrefix(substring, range: i..<string.count) {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    
    return count
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string.hasPrefix(substring, range: i..<string.count) {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string.hasPrefix(substring, range: i..<string.count) {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    
    return count
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string.hasPrefix(substring, range: i..<string.count) {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string[i...].hasPrefix(substring) {
            count += 1
            i += substring.count - 1
        }
        i += 1
    }
    
    return count
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string[i...].hasPrefix(substring) {
            count += 1
            i += substring.count - 1
        }
        i += 1
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string[i...].hasPrefix(substring) {
            count += 1
            i += substring.count - 1
        }
        i += 1
    }
    
    return count
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string[i...].hasPrefix(substring) {
            count += 1
            i += substring.count - 1
        }
        i += 1
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string[i...].hasPrefix(substring) {
            count += 1
            i += substring.count - 1
        }
        i += 1
    }
    
    return count
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string[i...].hasPrefix(substring) {
            count += 1
            i += substring.count - 1
        }
        i += 1
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string[i...].hasPrefix(substring) {
            count += 1
            i += substring.count - 1
        }
        i += 1
    }
    
    return count
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i < string.count {
        if string[i...].hasPrefix(substring) {
            count += 1
            i += substring.count - 1
        }
        i += 1
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let range = string.range(of: substring, range: index..<string.count)
        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let range = string.range(of: substring, range: index..<string.count)
        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index!= string.endIndex {
        let range = string.range(of: substring, range: index..<string.endIndex)
        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index!= string.endIndex {
        let range = string.range(of: substring, range: index..<string.endIndex)
        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index!= string.endIndex {
        if string[index..<string.endIndex].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index!= string.endIndex {
        if string[index..<string.endIndex].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let range = string.range(of: substring, range: index..<string.count)
        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let range = string.range(of: substring, range: index..<string.count)
        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        let range = string.range(of: substring, range: index..<string.endIndex)
        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        let range = string.range(of: substring, range: index..<string.endIndex)
        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index!= string.endIndex {
        if string.hasPrefix(substring) {
            count += 1
            index = string.index(after: index)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index!= string.endIndex {
        if string.hasPrefix(substring) {
            count += 1
            index = string.index(after: index)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            var subIndex = substring.startIndex
            var subString = string[index..<string.endIndex]
            while subIndex < substring.endIndex {
                if subString[subIndex] == substring[subIndex] {
                    subIndex = subIndex.advanced(by: 1)
                } else {
                    break
                }
            }
            if subIndex == substring.endIndex {
                count += 1
                index = index.advanced(by: 1)
            } else {
                index = index.advanced(by: 1)
            }
        } else {
            index = index.advanced(by: 1)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            var subIndex = substring.startIndex
            var subString = string[index..<string.endIndex]
            while subIndex < substring.endIndex {
                if subString[subIndex] == substring[subIndex] {
                    subIndex = subIndex.advanced(by: 1)
                } else {
                    break
                }
            }
            if subIndex == substring.endIndex {
                count += 1
                index = index.advanced(by: 1)
            } else {
                index = index.advanced(by: 1)
            }
        } else {
            index = index.advanced(by: 1)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.endIndex].contains(substring) {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.endIndex].contains(substring) {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.endIndex].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.endIndex].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while let range = string.range(of: substring, range: startIndex..<string.endIndex) {
        count += 1
        startIndex = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while let range = string.range(of: substring, range: startIndex..<string.endIndex) {
        count += 1
        startIndex = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while let range = string.range(of: substring, options: [], range: startIndex..<string.endIndex, locale: nil) {
        count += 1
        startIndex = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while let range = string.range(of: substring, options: [], range: startIndex..<string.endIndex, locale: nil) {
        count += 1
        startIndex = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substring(from: index) == substring {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substring(from: index) == substring {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        let range = string.range(of: substring, range: index..<string.endIndex)
        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        let range = string.range(of: substring, range: index..<string.endIndex)
        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!string.isEmpty,!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
            index = string.index(after: index)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    guard!string.isEmpty,!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
            index = string.index(after: index)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard string.count >= substring.count else { return 0 }
    var count = 0
    for i in 0...string.count - substring.count {
        let substringIndex = string.index(string.startIndex, offsetBy: i)
        let substringEndIndex = string.index(substringIndex, offsetBy: substring.count)
        let substring = string[substringIndex..<substringEndIndex]
        if substring == substring {
            count += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    guard string.count >= substring.count else { return 0 }
    var count = 0
    for i in 0...string.count - substring.count {
        let substringIndex = string.index(string.startIndex, offsetBy: i)
        let substringEndIndex = string.index(substringIndex, offsetBy: substring.count)
        let substring = string[substringIndex..<substringEndIndex]
        if substring == substring {
            count += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty { return 0 }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
            index = string.index(after: index)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty { return 0 }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
            index = string.index(after: index)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard string.count > 0, substring.count > 0 else { return 0 }
    guard string.count >= substring.count else { return 0 }
    var count = 0
    for i in 0...string.count - substring.count {
        let substringIndex = string.index(string.startIndex, offsetBy: i)
        let substringEndIndex = string.index(substringIndex, offsetBy: substring.count)
        let substring = string[substringIndex..<substringEndIndex]
        if substring == substring {
            count += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    guard string.count > 0, substring.count > 0 else { return 0 }
    guard string.count >= substring.count else { return 0 }
    var count = 0
    for i in 0...string.count - substring.count {
        let substringIndex = string.index(string.startIndex, offsetBy: i)
        let substringEndIndex = string.index(substringIndex, offsetBy: substring.count)
        let substring = string[substringIndex..<substringEndIndex]
        if substring == substring {
            count += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.endIndex].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.endIndex].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = range.lowerBound
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = range.lowerBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = range.lowerBound
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = range.lowerBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.range(of: substring, options:.backwards)
    while let range = index {
        count += 1
        string = String(string.dropFirst(range.lowerBound))
        index = string.range(of: substring, options:.backwards)
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.range(of: substring, options:.backwards)
    while let range = index {
        count += 1
        string = String(string.dropFirst(range.lowerBound))
        index = string.range(of: substring, options:.backwards)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.range(of: substring, options:.backwards)
    while let range = index {
        count += 1
        string = String(string.dropFirst(range.lowerBound))
        index = string.range(of: substring, options:.backwards)
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.range(of: substring, options:.backwards)
    while let range = index {
        count += 1
        string = String(string.dropFirst(range.lowerBound))
        index = string.range(of: substring, options:.backwards)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.range(of: substring, options:.backwards)
    while let range = index {
        count += 1
        string = String(string.dropFirst(range.lowerBound))
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.range(of: substring, options:.backwards)
    while let range = index {
        count += 1
        string = String(string.dropFirst(range.lowerBound))
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    let index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }

    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }

    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }

    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else { return 0 }
    return string.components(separatedBy: substring).count - 1
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else { return 0 }
    return string.components(separatedBy: substring).count - 1
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 0)
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 0)
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 0)
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 0)
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 0)
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 0)
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 0)
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 0)
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 0)
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 0)
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 0)
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 0)
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var idx = string.startIndex
    
    while let range = string.range(of: substring, options:.overlapping, range: idx..<string.endIndex) {
        count += 1
        idx = range.upperBound
    }
    
    return count
}
```

import 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
}

/*
 Find how many times a given substring can be found in the original string. Count overlaping cases.
 howManyTimes(string: "", substring: "a") == 0
 howManyTimes(string: "aaa", substring: "a") == 3
 howManyTimes(string: "aaaa", substring: "aa") == 3
 */
func howManyTimes(string: String, substring: String) -> Int {
    
    
    
    
    
    var count =

import 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
}

/*
 Find how many times a given substring can be found in the original string. Count overlaping cases.
 howManyTimes(string: "", substring: "a") == 0
 howManyTimes(string: "aaa", substring: "a") == 3
 howManyTimes(string: "aaaa", substring: "aa") == 3
 */
func howManyTimes(string: String, substring: String) -> Int {
    
    
    
    var count = 0
    for i

import 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
}

/*
 Find how many times a given substring can be found in the original string. Count overlaping cases.
 howManyTimes(string: "", substring: "a") == 0
 howManyTimes(string: "aaa", substring: "a") == 3
 howManyTimes(string: "aaaa", substring: "aa") == 3
 */
func howManyTimes(string: String, substring: String) -> Int {

import 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
}

/*
 Find how many times a given substring can be found in the original string. Count overlaping cases.
 howManyTimes(string: "", substring: "a") == 0
 howManyTimes(string: "aaa", substring: "a") == 3
 howManyTimes(string: "aaaa", substring: "aa") == 3
 */
func howManyTimes(string: String, substring: String) -> Int {
    
    
    
    
    
    
    var count

import 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
}

/*
 Find how many times a given substring can be found in the original string. Count overlaping cases.
 howManyTimes(string: "", substring: "a") == 0
 howManyTimes(string: "aaa", substring: "a") == 3
 howManyTimes(string: "aaaa", substring: "aa") == 3
 */
func howManyTimes(string: String, substring: String) -> Int {
    
    
    
    
    
    var count =

import 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
}

/*
 Find how many times a given substring can be found in the original string. Count overlaping cases.
 howManyTimes(string: "", substring: "a") == 0
 howManyTimes(string: "aaa", substring: "a") == 3
 howManyTimes(string: "aaaa", substring: "aa") == 3
 */
func howManyTimes(string: String, substring: String) -> Int {

import 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
}

/*
 Find how many times a given substring can be found in the original string. Count overlaping cases.
 howManyTimes(string: "", substring: "a") == 0
 howManyTimes(string: "aaa", substring: "a") == 3
 howManyTimes(string: "aaaa", substring: "aa") == 3
 */
func howManyTimes(string: String, substring: String) -> Int {
    

import 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
}

/*
 Find how many times a given substring can be found in the original string. Count overlaping cases.
 howManyTimes(string: "", substring: "a") == 0
 howManyTimes(string: "aaa", substring: "a") == 3
 howManyTimes(string: "aaaa", substring: "aa") == 3
 */
func howManyTimes(string: String, substring: String) -> Int {
    
    

import 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
}

/*
 Find how many times a given substring can be found in the original string. Count overlaping cases.
 howManyTimes(string: "", substring: "a") == 0
 howManyTimes(string: "aaa", substring: "a") == 3
 howManyTimes(string: "aaaa", substring: "aa") == 3
 */
func howManyTimes(string: String, substring: String) -> Int {
    
    
    
    
    
    var count

import 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
}

/*
 Find how many times a given substring can be found in the original string. Count overlaping cases.
 howManyTimes(string: "", substring: "a") == 0
 howManyTimes(string: "aaa", substring: "a") == 3
 howManyTimes(string: "aaaa", substring: "aa") == 3
 */
func howManyTimes(string: String, substring: String) -> Int {
    
    
    
    
    
    var count

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {




```

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {

```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty || substring.isEmpty || substring.isEmpty || substring.isEmpty {
    {
        return 0
    return 0
    return 0
    return 0
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty || substring.isEmpty || substring.isEmpty || substring.isEmpty {
    {
        return 0
    return 0
    return 0
    return 0
}
    var count = 0
}
    let lower

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {




```

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {




```

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {


```

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {




```

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty || substring.isEmpty || substring.isEmpty  {
        return 0
    return 0

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    
    if string.isEmpty || substring.isEmpty || substring.isEmpty || substring.isEmpty || substring.isEmpty {
        return 0
    return 0
    return 0

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        index = string.index(after: index)
        if string[index...] == substring {
            count += 1
            index = string.index(after: index, offsetBy: substring.count)
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        index = string.index(after: index)
        if string[index...] == substring {
            count += 1
            index = string.index(after: index, offsetBy: substring.count)
        }
    }
    return count
}

print(howManyTimes(string: "", substring: "a")) // 0
print(howManyTimes(string: "aaa", substring: "a")) // 3
print(howManyTimes(string: "aaaa", substring: "aa")) // 3
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let found = string.index(string.startIndex, offsetBy: index)
        let range = string.range(found, offsetBy: substring.count)
        if string[range] == substring {
            count += 1
            index += substring.count
        } else {
            break
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let found = string.index(string.startIndex, offsetBy: index)
        let range = string.range(found, offsetBy: substring.count)
        if string[range] == substring {
            count += 1
            index += substring.count
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        index = string.index(after: index)
        if string[index...] == substring {
            count += 1
            index = string.index(after: index, offsetBy: substring.count)
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        index = string.index(after: index)
        if string[index...] == substring {
            count += 1
            index = string.index(after: index, offsetBy: substring.count)
        }
    }
    return count
}

print(howManyTimes(string: "", substring: "a")) // 0
print(howManyTimes(string: "aaa", substring: "a")) // 3
print(howManyTimes(string: "aaaa", substring: "aa")) // 3
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let found = string.index(string.startIndex, offsetBy: index)
        let range = string.range(found..., in: string)
        if range.contains(substring) {
            count += 1
            index += substring.count
        } else {
            break
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let found = string.index(string.startIndex, offsetBy: index)
        let range = string.range(found..., in: string)
        if range.contains(substring) {
            count += 1
            index += substring.count
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        index = string.index(string.startIndex, offsetBy: index)
        let range = string.index(string.startIndex, offsetBy: index)...string.index(string.startIndex, offsetBy: index + substring.count)
        if string[range] == substring {
            count += 1
            index += substring.count
        } else {
            break
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        index = string.index(string.startIndex, offsetBy: index)
        let range = string.index(string.startIndex, offsetBy: index)...string.index(string.startIndex, offsetBy: index + substring.count)
        if string[range] == substring {
            count += 1
            index += substring.count
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

print(howManyTimes(string: "", substring: "a")) // 0
print(howManyTimes(string: "aaa", substring: "a")) // 3
print(howManyTimes(string: "aaaa", substring: "aa")) // 3
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

print(howManyTimes(string: "", substring: "a")) // 0
print(howManyTimes(string: "aaa", substring: "a")) // 3
print(howManyTimes(string: "aaaa", substring: "aa")) // 3
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let found = string.index(string.startIndex, offsetBy: index)
        if string[found...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let found = string.index(string.startIndex, offsetBy: index)
        if string[found...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[index(after: range!)]...)
        range = string.range(of: substring)
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[index(after: range!)]...)
        range = string.range(of: substring)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[max(range.lowerBound, 0)...])
        range = string.range(of: substring)
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[max(range.lowerBound, 0)...])
        range = string.range(of: substring)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[max(range.lowerBound, 0)...])
        range = string.range(of: substring)
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[max(range.lowerBound, 0)...])
        range = string.range(of: substring)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[max(range.lowerBound, 0)...])
        range = string.range(of: substring)
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[max(range.lowerBound, 0)...])
        range = string.range(of: substring)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[max(range.lowerBound, 0)...])
        range = string.range(of: substring)
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[max(range.lowerBound, 0)...])
        range = string.range(of: substring)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[..<range.lowerBound]) + String(string[range.upperBound..<])
        range = string.range(of: substring)
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[..<range.lowerBound]) + String(string[range.upperBound..<])
        range = string.range(of: substring)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[..<range.lowerBound]) + String(string[range.upperBound..<])
        range = string.range(of: substring)
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[..<range.lowerBound]) + String(string[range.upperBound..<])
        range = string.range(of: substring)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[index(string.startIndex)...<range.lowerBound]) + String(string[range.upperBound...index(string.endIndex)])
        range = string.range(of: substring)
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[index(string.startIndex)...<range.lowerBound]) + String(string[range.upperBound...index(string.endIndex)])
        range = string.range(of: substring)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[index(string.startIndex)...<range.lowerBound]) + String(string[range.upperBound...]...)
        range = string.range(of: substring)
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[index(string.startIndex)...<range.lowerBound]) + String(string[range.upperBound...]...)
        range = string.range(of: substring)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[index(string.startIndex)...<range.lowerBound]) + String(string[range.upperBound...]...)
        range = string.range(of: substring)
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.range(of: substring)
    while range != nil {
        count += 1
        string = String(string[index(string.startIndex)...<range.lowerBound]) + String(string[range.upperBound...]...)
        range = string.range(of: substring)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = range.upperBound..<string.endIndex
        } else {
            break
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = range.upperBound..<string.endIndex
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = range.upperBound..<string.endIndex
        } else {
            break
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = range.upperBound..<string.endIndex
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = range.upperBound..<string.endIndex
        } else {
            break
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = range.upperBound..<string.endIndex
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}

<｜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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex..<string.endIndex
    while range.lowerBound < string.endIndex {
        range = string[range.lowerBound...].range(of: substring)
        if let range = range {
            count += 1
            range = string.index(range.upperBound, offsetBy: 1)..<string.endIndex
        } else {
            break
        }
    }
    return count
}
```

import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }

    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }

    return count
}
```

import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}
```

import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}
```

import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}
```

import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}
```

import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}
```

import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}
```

import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}
```

import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}
```

import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
import Foundation

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex

    while startIndex < string.endIndex {
        if let range = string[startIndex...].range(of: substring) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .overlap, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard !substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if let range = string.range(of: substring, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if let range = string.range(of: substring, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if let range = string.range(of: substring, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if let range = string.range(of: substring, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: string.startIndex..<index) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: string.startIndex..<index) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if let range = string.range(of: substring, range: string.index(index, offsetBy: substring.count)..<string.endIndex) {
            count += 1
            index = range.lowerBound
        } else {
            break
        }
    }
    
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if let range = string.range(of: substring, range: string.index(index, offsetBy: substring.count)..<string.endIndex) {
            count += 1
            index = range.lowerBound
        } else {
            break
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: string.startIndex..<string.endIndex, locale: nil) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: string.startIndex..<string.endIndex, locale: nil) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: string.startIndex..<string.endIndex, locale: nil) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: string.startIndex..<string.endIndex, locale: nil) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: string.startIndex..<string.endIndex, locale: nil) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: string.startIndex..<string.endIndex, locale: nil) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
        }
        index = string.index(index, offsetBy: 1)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
        }
        index = string.index(index, offsetBy: 1)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let substringLength = substring.length
    var count = 0
    for i in 0..<string.length {
        if string.substring(from: i, to: i + substringLength) == substring {
            count += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let substringLength = substring.length
    var count = 0
    for i in 0..<string.length {
        if string.substring(from: i, to: i + substringLength) == substring {
            count += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
        }
        index = string.index(index, advancedBy: 1)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
        }
        index = string.index(index, advancedBy: 1)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch)
    var count = 0
    while range!= nil {
        count += 1
        string = string.stringByReplacingCharactersInRange(range!, withString: "")
        range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch)
    var count = 0
    while range!= nil {
        count += 1
        string = string.stringByReplacingCharactersInRange(range!, withString: "")
        range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch)
    var count = 0
    while range.startIndex!= NSNotFound {
        count += 1
        range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch, range: range)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch)
    var count = 0
    while range.startIndex!= NSNotFound {
        count += 1
        range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch, range: range)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let range = string.rangeOfString(substring)
    var count = 0
    while range.startIndex!= string.endIndex {
        count += 1
        range = string.rangeOfString(substring, options:.ByStartingWith, range: range.advancedBy(substring.characters.count))
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let range = string.rangeOfString(substring)
    var count = 0
    while range.startIndex!= string.endIndex {
        count += 1
        range = string.rangeOfString(substring, options:.ByStartingWith, range: range.advancedBy(substring.characters.count))
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let range = string.rangeOfString(substring)
    var count = 0
    while range.startIndex!= string.endIndex {
        count += 1
        range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch, range: range)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let range = string.rangeOfString(substring)
    var count = 0
    while range.startIndex!= string.endIndex {
        count += 1
        range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch, range: range)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
        }
        index = string.index(index, offsetBy: 1)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
        }
        index = string.index(index, offsetBy: 1)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
        }
        index = string.index(index, advancedBy: 1)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
        }
        index = string.index(index, advancedBy: 1)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch)
    var count = 0
    while range.startIndex!= NSNotFound {
        count += 1
        range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch, range: range)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch)
    var count = 0
    while range.startIndex!= NSNotFound {
        count += 1
        range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch, range: range)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, range: index..<string.endIndex, options:.byIndices) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, range: index..<string.endIndex, options:.byIndices) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = range.upperBound
        } else {
            index = string.index(after: index)
        }
    }
    
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = range.upperBound
        } else {
            index = string.index(after: index)
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.regularExpression, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.regularExpression, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.regularExpression, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.regularExpression, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = range.upperBound
        } else {
            index = string.index(after: index)
        }
    }
    
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = range.upperBound
        } else {
            index = string.index(after: index)
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            let start = index
            let end = index + range.lowerBound.utf16Offset(in: string)
            if end > start {
                count += 1
            }
            index = end
        } else {
            break
        }
    }
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            let start = index
            let end = index + range.lowerBound.utf16Offset(in: string)
            if end > start {
                count += 1
            }
            index = end
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.regularExpression, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.regularExpression, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = range.upperBound
        } else {
            index = string.index(after: index)
        }
    }
    
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = range.upperBound
        } else {
            index = string.index(after: index)
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.regularExpression, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.regularExpression, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var start = string.startIndex
    while start < string.endIndex {
        if let range = string.rangeOfString(substring, options: [], range: start..<string.endIndex) {
            count += 1
            start = range.endIndex
        } else {
            break
        }
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var start = string.startIndex
    while start < string.endIndex {
        if let range = string.rangeOfString(substring, options: [], range: start..<string.endIndex) {
            count += 1
            start = range.endIndex
        } else {
            break
        }
    }
    return count
}
```
### Challenge

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var lastIndex = -substring.count
    while let range = string.range(of: substring, range: lastIndex..<string.count) {
        count += 1
        lastIndex = range.upperBound
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var lastIndex = -substring.count
    while let range = string.range(of: substring, range: lastIndex..<string.count) {
        count += 1
        lastIndex = range.upperBound
    }
    return count
}
```
### Challenge

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            startIndex = string.index(after: startIndex)
        }
    }
    return count
}
```
### Challenge

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}
```
### Challenge

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    while i < string.count {
        if string[i..<i+substring.count] == substring {
            count += 1
        }
        i += 1
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    while i < string.count {
        if string[i..<i+substring.count] == substring {
            count += 1
        }
        i += 1
    }
    return count
}
```
### Challenge

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    while i < string.count {
        if string[i] == substring[0] && string.hasPrefix(substring, at: i) {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    while i < string.count {
        if string[i] == substring[0] && string.hasPrefix(substring, at: i) {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    return count
}
```
### Challenge

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var start = string.startIndex
    while start < string.endIndex {
        if let range = string.range(of: substring, range: start..<string.endIndex) {
            count += 1
            start = range.upperBound
        } else {
            break
        }
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var start = string.startIndex
    while start < string.endIndex {
        if let range = string.range(of: substring, range: start..<string.endIndex) {
            count += 1
            start = range.upperBound
        } else {
            break
        }
    }
    return count
}
```
### Ex

func howManyTimes(string: String, substring: String) -> Int {
    var counter = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            counter += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return counter
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var counter = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            counter += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return counter
}
```
### Challenge

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var start = string.startIndex
    while start < string.endIndex {
        if let range = string.range(of: substring, range: start..<string.endIndex) {
            count += 1
            start = range.upperBound
        } else {
            break
        }
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var start = string.startIndex
    while start < string.endIndex {
        if let range = string.range(of: substring, range: start..<string.endIndex) {
            count += 1
            start = range.upperBound
        } else {
            break
        }
    }
    return count
}
```
### Challenge

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var lastIndex = string.startIndex
    while let range = string.range(of: substring, range: lastIndex..<string.endIndex) {
        count += 1
        lastIndex = range.upperBound
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var lastIndex = string.startIndex
    while let range = string.range(of: substring, range: lastIndex..<string.endIndex) {
        count += 1
        lastIndex = range.upperBound
    }
    return count
}
```
### Challenge

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```
### Link

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    while i < string.count {
        if string[i..<i+substring.count] == substring {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    while i < string.count {
        if string[i..<i+substring.count] == substring {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```
### Link

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    while i < string.count {
        if string[string.index(string.startIndex, offsetBy: i)] == substring[substring.startIndex] {
            var j = 0
            while j < substring.count {
                if string[string.index(string.startIndex, offsetBy: i + j)] != substring[substring.index(substring.startIndex, offsetBy: j)] {
                    break
                }
                j += 1
            }
            if j == substring.count {
                count += 1
            }
        }
        i += 1
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    while i < string.count {
        if string[string.index(string.startIndex, offsetBy: i)] == substring[substring.startIndex] {
            var j = 0
            while j < substring.count {
                if string[string.index(string.startIndex, offsetBy: i + j)] != substring[substring.index(substring.startIndex, offsetBy: j)] {
                    break
                }
                j += 1
            }
            if j == substring.count {
                count += 1
            }
        }
        i += 1
    }
    return count
}
```
### Link

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        index = string.index(range.lowerBound, offsetBy: substring.count)
        count += 1
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        index = string.index(range.lowerBound, offsetBy: substring.count)
        count += 1
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].starts(with: substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].starts(with: substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].starts(with: substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count - 1)
        }
        i = string.index(after: i)
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].starts(with: substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count - 1)
        }
        i = string.index(after: i)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var start = string.startIndex
    while start < string.endIndex,
          let range = string[start...].range(of: substring),
          !range.isEmpty {
        count += 1
        start = range.upperBound
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var start = string.startIndex
    while start < string.endIndex,
          let range = string[start...].range(of: substring),
          !range.isEmpty {
        count += 1
        start = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].starts(with: substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].starts(with: substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if string[index...].starts(with: substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if string[index...].starts(with: substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if string[index] == substring.first {
            var i = index
            var j = substring.startIndex
            var found = true
            while j != substring.endIndex {
                if i == string.endIndex {
                    found = false
                    break
                }
                if string[i] != substring[j] {
                    found = false
                    break
                }
                i = string.index(after: i)
                j = substring.index(after: j)
            }
            if found {
                count += 1
            }
        }
        index = string.index(after: index)
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if string[index] == substring.first {
            var i = index
            var j = substring.startIndex
            var found = true
            while j != substring.endIndex {
                if i == string.endIndex {
                    found = false
                    break
                }
                if string[i] != substring[j] {
                    found = false
                    break
                }
                i = string.index(after: i)
                j = substring.index(after: j)
            }
            if found {
                count += 1
            }
        }
        index = string.index(after: index)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex,
          let range = string[startIndex...].range(of: substring),
          !range.isEmpty {
        count += 1
        startIndex = range.upperBound
    }
    return count
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex,
          let range = string[startIndex...].range(of: substring),
          !range.isEmpty {
        count += 1
        startIndex = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substringFromIndex(index).hasPrefix(substring) {
            count++
            index = string.index(index, offsetBy: substring.characters.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substringFromIndex(index).hasPrefix(substring) {
            count++
            index = string.index(index, offsetBy: substring.characters.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substringFromIndex(index).hasPrefix(substring) {
            count++
            index = string.index(index, offsetBy: substring.characters.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substringFromIndex(index).hasPrefix(substring) {
            count++
            index = string.index(index, offsetBy: substring.characters.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

// Example usage:
print(howManyTimes(string: "aaa", substring: "A")) // Output: 3
print(howManyTimes(string: "aaaa", substring: "aa")) // Output: 3
```

This function uses a loop to iterate over the string and count the occurrences of the substring. It checks for the presence of the substring at each position and increments the count if found.

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substringFromIndex(index).hasPrefix(substring) {
            count++
            index = string.startIndex.advancedBy(substring.characters.count, instance)
        } else {
            index = string.indexAfter(index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substringFromIndex(index).hasPrefix(substring) {
            count++
            index = string.startIndex.advancedBy(substring.characters.count, instance)
        } else {
            index = string.indexAfter(index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substringFromIndex(index).hasPrefix(substring) {
            count++
            index = string.index(index, offsetBy: substring.characters.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substringFromIndex(index).hasPrefix(substring) {
            count++
            index = string.index(index, offsetBy: substring.characters.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substringFromIndex(index).hasPrefix(substring) {
            count++
            index = string.index(index, offsetBy: substring.characters.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substringFromIndex(index).hasPrefix(substring) {
            count++
            index = string.index(index, offsetBy: substring.characters.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substringFromIndex(index).hasPrefix(substring) {
            count++
            index = string.startIndex.advancedBy(substring.characters.count, instance)
        } else {
            index = string.indexAfter(index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string.substringFromIndex(index).hasPrefix(substring) {
            count++
            index = string.startIndex.advancedBy(substring.characters.count, instance)
        } else {
            index = string.indexAfter(index)
        }
    }
    return count
}

// Example usage:
let exampleString = "This is a test string with several occurrences of the word 'test'"
let substringToFind = "test"
let occurrences = howManyTimes(string: exampleString, substring: substringToFind)
print("The substring '\(substringToFind)' occurs \(occurrences) times in the string '\(exampleString)'")
```

This code defines a function `howManyTimes` that takes a string and a substring as input and returns the number of times the substring occurs in the string. It uses a loop to iterate through the string and checks for the presence of the substring at each position. If the substring is found, it increments the count and adjusts the index to account for overlaps.

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .literal, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let substringIndex = string.index(string.startIndex, offsetBy: index)
        let substringEndIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        let substringRange = substringIndex..<substringEndIndex
        if string.substring(with: substringRange) == substring {
            count += 1
        }
        index += 1
    }
    return count
}

func longestSubstring(string: String) -> String {
    var longestSubstring = ""
    var substring = ""
    for char in string {
        if substring.contains(char) {
            if substring.count > longestSubstring.count {
                longestSubstring = substring
            }
            substring = ""
        }
        substring += String(char)
    }
    if substring.count > longestSubstring.count {
        longestSubstring = substring
    }
    return longestSubstring
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let substringIndex = string.index(string.startIndex, offsetBy: index)
        let substringEndIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        let substringRange = substringIndex..<substringEndIndex
        if string.substring(with: substringRange) == substring {
            count += 1
        }
        index += 1
    }
    return count
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "abcabcbb") == "abc"
// longestSubstring(string: "bbbbb") == "b"
// longestSubstring(string: "pwwkew") == "wke"
func longestSubstring(string: String) -> String {
    var longestSubstring = ""
    var substring = ""
    for char in string {
        if substring.contains(char) {
            if substring.count > longestSubstring.count {
                longestSubstring = substring
            }
            substring = ""
        }
        substring += String(char)
    }
    if substring.count > longestSubstring.count {
        longestSubstring = substring
    }
    return longestSubstring
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "abcabcbb") == "abc"
// longestSubstring(string: "bbbbb") == "b"
// longestSubstring(string: "pwwkew") == "wke"
func longestSubstring(string: String) -> String {
    var longestSubstring = ""
    var substring = ""
    for char in string {
        if substring.contains(char) {
            if substring.count > longestSubstring.count {
                longestSubstring = substring
            }
            substring = ""
        }
        substring += String(char)
    }
    if substring.count > longestSubstring.count {
        longestSubstring = substring
    }
    return longestSubstring
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "abcabcbb") == "abc"
// longestSubstring(string: "bbbbb") == "b"
// longestSubstring(string: "pwwkew") == "wke"
func longestSubstring(string: String) -> String {
    var longestSubstring = ""
    var substring = ""
    for char in string {
        if substring.contains(char) {
            if substring.count > longestSubstring.count {
                longestSubstring = substring
            }
            substring = ""
        }
        substring += String(char)
    }
    if substring.count > longestSubstring.count {
        longestSubstring = substring
    }
    return longestSubstring
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "abcabcbb") == "abc"
// longestSubstring(string: "bbbbb") == "b"
// longestSubstring(string: "pwwkew") == "wke"
func longestSubstring(string: String) -> String {
    var longestSubstring = ""
    var substring = ""
    for char in string {
        if substring.contains(char) {
            if substring.count > longestSubstring.count {
                longestSubstring = substring
            }
            substring = ""
        }
        substring += String(char)
    }
    if substring.count > longestSubstring.count {
        longestSubstring = substring
    }
    return longestSubstring
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "abcabcbb") == "abc"
// longestSubstring(string: "bbbbb") == "b"
// longestSubstring(string: "pwwkew") == "wke"
func longestSubstring(string: String) -> String {
    var longestSubstring = ""
    var substring = ""
    for char in string {
        if substring.contains(char) {
            if substring.count > longestSubstring.count {
                longestSubstring = substring
            }
            substring = ""
        }
        substring += String(char)
    }
    if substring.count > longestSubstring.count {
        longestSubstring = substring
    }
    return longestSubstring
}

// Find the longest substring that can be found in the original string.
// longestSubstring(string: "abcabcbb") == "abc"
// longestSubstring(

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index+substring.count) == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index+substring.count) == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let range = string.index(string.startIndex, offsetBy: index)..<string.index(string.startIndex, offsetBy: index + substring.count)
        if string.contains(substring, in: range) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let range = string.index(string.startIndex, offsetBy: index)..<string.index(string.startIndex, offsetBy: index + substring.count)
        if string.contains(substring, in: range) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index+substring.count) == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index+substring.count) == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index+substring.count) == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index+substring.count) == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
<|file_separator|>```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index+substring.count) == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index+substring.count) == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    let substringCount = string.components(separatedBy: substring).count - 1
    return substringCount
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let substringCount = string.components(separatedBy: substring).count - 1
    return substringCount
}
<|file_separator|>```

func howManyTimes(string: String, substring: String) -> Int {
    let string = string.lowercased()
    let substring = substring.lowercased()
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        let range = string.index(string.startIndex, offsetBy: substring.count)..<string.index(string.startIndex, offsetBy: substring.count + 1)
        if string.substring(with: range) == substring {
            count += 1
            index = string.index(string.startIndex, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

func longestCommonSubstring(string1: String, string2: String) -> String {
    let string1 = string1.lowercased()
    let string2 = string2.lowercased()
    var longest = ""
    for i in 0..<string1.count {
        for j in 0..<string2.count {
            let substring = string1.substring(with: string1.index(string1.startIndex, offsetBy: i)..<string1.index(string1.startIndex, offsetBy: i + 1))
            if string2.contains(substring) {
                if substring.count > longest.count {
                    longest = substring
                }
            }
        }
    }
    return longest
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let string = string.lowercased()
    let substring = substring.lowercased()
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        let range = string.index(string.startIndex, offsetBy: substring.count)..<string.index(string.startIndex, offsetBy: substring.count + 1)
        if string.substring(with: range) == substring {
            count += 1
            index = string.index(string.startIndex, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

// Find the longest common substring between two strings.
// longestCommonSubstring(string1: "abcde", string2: "abfde") == "ab"
// longestCommonSubstring(string1: "abcde", string2: "abfde") == "ab"
// longestCommonSubstring(string1: "abcde", string2: "abfde") == "ab"
func longestCommonSubstring(string1: String, string2: String) -> String {
    let string1 = string1.lowercased()
    let string2 = string2.lowercased()
    var longest = ""
    for i in 0..<string1.count {
        for j in 0..<string2.count {
            let substring = string1.substring(with: string1.index(string1.startIndex, offsetBy: i)..<string1.index(string1.startIndex, offsetBy: i + 1))
            if string2.contains(substring) {
                if substring.count > longest.count {
                    longest = substring
                }
            }
        }
    }
    return longest
}

// Find the longest common substring between two strings.
// longestCommonSubstring(string1: "abcde", string2: "abfde") == "ab"
// longestCommonSubstring(string1: "abcde", string2: "abfde") == "ab"
// longestCommonSubstring(string1: "abcde", string2: "abfde") == "ab"
func longestCommonSubstring(string1: String, string2: String) -> String {
    let string1 = string1.lowercased()
    let string2 = string2.lowercased()
    var longest = ""
    for i in 0..<string1.count {
        for j in 0..<string2.count {
            let substring = string1.substring(with: string1.index(string1.startIndex, offsetBy: i)..<string1.index(string1.startIndex, offsetBy: i + 1))
            if string2.contains(substring) {
                if substring.count > longest.count {
                    longest = substring
                }
            }
        }
    }
    return longest
}

// Find the longest common substring between two strings.
// longestCommonSubstring(string1: "abcde", string2: "abfde") == "ab"
// longestCommonSubstring(string1: "abcde", string2: "abfde") == "ab"
// longestCommonSubstring(string1: "abcde", string2: "abfde") == "ab"
func longestCommonSubstring(string1: String, string2: String) -> String {
    let string1 = string1.lowercased()
    let string2 = string2.lowercased()
    var longest = ""
    for i in 0..<string1.count {
        for j in 0..<string2.count {
            let substring = string1.substring(with: string1.index(string1.startIndex, offsetBy: i)..<string1.index(string1.startIndex, offsetBy: i + 1))
            if string2.contains(substring) {
                if substring.count > longest.count {
                    longest = substring
                }
            }
        }
    }
    return longest
}

// Find the longest common substring between two strings.
// longestCommonSubstring(string1: "abcde", string2: "abfde") == "ab"
// longestCommonSubstring(string1: "abcde", string2: "abfde") == "ab"
// longestCommonSubstring(string1: "abcde", string2: "abfde") == "ab"
func longestCommonSubstring(string1: String, string2: String) -> String {
    let string1 = string1.lowercased()
    let string2 = string2.lowercased()

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let startIndex = string.index(string.startIndex, offsetBy: index)
        let endIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        let substringRange = startIndex..<endIndex
        if string.substring(with: substringRange) == substring {
            count += 1
        }
        index += 1
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let startIndex = string.index(string.startIndex, offsetBy: index)
        let endIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        let substringRange = startIndex..<endIndex
        if string.substring(with: substringRange) == substring {
            count += 1
        }
        index += 1
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let startIndex = string.index(string.startIndex, offsetBy: index)
        let endIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        let substringRange = startIndex..<endIndex
        if string[substringRange] == substring {
            count += 1
        }
        index += 1
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let startIndex = string.index(string.startIndex, offsetBy: index)
        let endIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        let substringRange = startIndex..<endIndex
        if string[substringRange] == substring {
            count += 1
        }
        index += 1
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index + substring.count) == substring {
            count += 1
        }
        index += 1
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index + substring.count) == substring {
            count += 1
        }
        index += 1
    }
    return count
}
<|file_separator|>```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index + substring.count) == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index + substring.count) == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index + substring.count) == substring {
            count += 1
        }
        index += 1
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index + substring.count) == substring {
            count += 1
        }
        index += 1
    }
    return count
}
<|file_separator|>```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index+substring.count) == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.substring(with: index..<index+substring.count) == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
<|file_separator|>```

func howManyTimes(string: String, substring: String) -> Int {
    let substringLength = substring.count
    var count = 0
    for i in 0 ..< string.count - substringLength + 1 {
        let substringIndex = string.index(string.startIndex, offsetBy: i)
        let substringRange = substringIndex ..< substringIndex + substringLength
        if substring == String(string[substringRange]) {
            count += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let substringLength = substring.count
    var count = 0
    for i in 0 ..< string.count - substringLength + 1 {
        let substringIndex = string.index(string.startIndex, offsetBy: i)
        let substringRange = substringIndex ..< substringIndex + substringLength
        if substring == String(string[substringRange]) {
            count += 1
        }
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    let stringArray = string.map { String($0) }
    let substringArray = substring.map { String($0) }
    var count = 0
    var index = 0
    while index < stringArray.count {
        if stringArray[index] == substringArray[0] {
            if stringArray[index..<index+substringArray.count] == substringArray {
                count += 1
            }
        }
        index += 1
    }
    return count
}

func longestCommonPrefix(string1: String, string2: String) -> String {
    let string1Array = string1.map { String($0) }
    let string2Array = string2.map { String($0) }
    var prefix = ""
    var index = 0
    while index < string1Array.count && index < string2Array.count {
        if string1Array[index] == string2Array[index] {
            prefix += string1Array[index]
        } else {
            break
        }
        index += 1
    }
    return prefix
}

func longestCommonSuffix(string1: String, string2: String) -> String {
    let string1Array = string1.map { String($0) }
    let string2Array = string2.map { String($0) }
    var suffix = ""
    var index = 0
    while index < string1Array.count && index < string2Array.count {
        if string1Array[string1Array.count-1-index] == string2Array[string2Array.count-1-index] {
            suffix += string1Array[string1Array.count-1-index]
        } else {
            break
        }
        index += 1
    }
    return suffix
}

func longestCommonSubstring(string1: String, string2: String) -> String {
    let string1Array = string1.map { String($0) }
    let string2Array = string2.map { String($0) }
    var longestCommonSubstring = ""
    var index = 0
    while index < string1Array.count {
        var substring = ""
        var substringIndex = index
        while substringIndex < string1Array.count {
            if string1Array[substringIndex] == string2Array[substringIndex] {
                substring += string1Array[substringIndex]
            } else {
                break
            }
            substringIndex += 1
        }
        if substring.count > longestCommonSubstring.count {
            longestCommonSubstring = substring
        }
        index += 1
    }
    return longestCommonSubstring
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let stringArray = string.map { String($0) }
    let substringArray = substring.map { String($0) }
    var count = 0
    var index = 0
    while index < stringArray.count {
        if stringArray[index] == substringArray[0] {
            if stringArray[index..<index+substringArray.count] == substringArray {
                count += 1
            }
        }
        index += 1
    }
    return count
}

// Find the longest common prefix of two strings.
// longestCommonPrefix(string1: "abc", string2: "abcd") == "ab"
// longestCommonPrefix(string1: "abc", string2: "ab") == "ab"
// longestCommonPrefix(string1: "abc", string2: "ac") == ""
func longestCommonPrefix(string1: String, string2: String) -> String {
    let string1Array = string1.map { String($0) }
    let string2Array = string2.map { String($0) }
    var prefix = ""
    var index = 0
    while index < string1Array.count && index < string2Array.count {
        if string1Array[index] == string2Array[index] {
            prefix += string1Array[index]
        } else {
            break
        }
        index += 1
    }
    return prefix
}

// Find the longest common suffix of two strings.
// longestCommonSuffix(string1: "abc", string2: "abcd") == "cd"
// longestCommonSuffix(string1: "abc", string2: "ab") == "b"
// longestCommonSuffix(string1: "abc", string2: "ac") == ""
func longestCommonSuffix(string1: String, string2: String) -> String {
    let string1Array = string1.map { String($0) }
    let string2Array = string2.map { String($0) }
    var suffix = ""
    var index = 0
    while index < string1Array.count && index < string2Array.count {
        if string1Array[string1Array.count-1-index] == string2Array[string2Array.count-1-index] {
            suffix += string1Array[string1Array.count-1-index]
        } else {
            break
        }
        index += 1
    }
    return suffix
}

// Find the longest common substring of two strings.
// longestCommonSubstring(string1: "abc", string2: "abcd") == "cd"
// longestCommonSubstring(string1: "abc", string2: "ab") == "ab"
// longestCommonSubstring(string1: "abc", string2: "ac") == ""
func longestCommonSubstring(string1: String, string2: String) -> String {
    let string1Array = string1.map { String($0) }
    let string2Array = string2.map { String($0) }
    var longestCommonSubstring = ""
    var index = 0
    while index < string1Array.count {
        var substring = ""
        var substringIndex = index
        while substringIndex < string1Array.count {
            if string1Array[substringIndex] == string2Array[substringIndex] {
                substring += string1Array[substringIndex]
            } else {
                break
            }
            substringIndex += 1
        }
        if substring.count > longestCommonSubstring.count {
            longestCommonSubstring = substring
        }
        index += 1
    }
    return longestCommonSubstring
}

// Find the longest common substring of two strings.
// longestCommonSubstring(string1: "abc", string2: "abcd") == "cd"
// longestCommonSubstring(string1: "abc", string2: "ab") == "ab"
// longestCommonSubstring(string1: "abc", string2: "ac") == ""
func longestCommonSubstring(string1: String, string2: String) -> String {
    let string1Array = string1.map { String($0) }
    let string2Array = string2.map { String($0) }
    var longestCommonSubstring = ""
    var index = 0
    while index < string1Array.count {
        var substring = ""
        var substringIndex = index
        while substringIndex < string1Array.count {
            if string1Array[substringIndex]

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let substringIndex = string.index(string.startIndex, offsetBy: index)
        let substringEndIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        let substringRange = substringIndex..<substringEndIndex
        let substring = String(string[substringRange])
        if substring == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let substringIndex = string.index(string.startIndex, offsetBy: index)
        let substringEndIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        let substringRange = substringIndex..<substringEndIndex
        let substring = String(string[substringRange])
        if substring == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let startIndex = string.index(string.startIndex, offsetBy: index)
        let endIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        let substringRange = startIndex..<endIndex
        if string.substring(with: substringRange) == substring {
            count += 1
        }
        index += 1
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let startIndex = string.index(string.startIndex, offsetBy: index)
        let endIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        let substringRange = startIndex..<endIndex
        if string.substring(with: substringRange) == substring {
            count += 1
        }
        index += 1
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let range = string.index(string.startIndex, offsetBy: index)..<string.index(string.startIndex, offsetBy: index + substring.count)
        if string[range] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let range = string.index(string.startIndex, offsetBy: index)..<string.index(string.startIndex, offsetBy: index + substring.count)
        if string[range] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
<|file_separator|>```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let startIndex = string.index(string.startIndex, offsetBy: index)
        let endIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        let range = startIndex..<endIndex
        if string.substring(with: range) == substring {
            count += 1
        }
        index += 1
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let startIndex = string.index(string.startIndex, offsetBy: index)
        let endIndex = string.index(string.startIndex, offsetBy: index + substring.count)
        let range = startIndex..<endIndex
        if string.substring(with: range) == substring {
            count += 1
        }
        index += 1
    }
    return count
}
<|file_separator|>

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index != string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index <= string.count - substring.count {
        if string[string.index(string.startIndex, offsetBy: index)] == substring[substring.startIndex] {
            count += 1
            index += 1
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index <= string.count - substring.count {
        if string[string.index(string.startIndex, offsetBy: index)] == substring[substring.startIndex] {
            count += 1
            index += 1
        } else {
            index += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if let range = string.range(of: substring, options: .literal, range: string.index(string.startIndex, offsetBy: index)...string.endIndex, locale: nil) {
            count += 1
            index = string.distance(from: string.startIndex, to: range.upperBound)
        } else {
            index = string.count
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if let range = string.range(of: substring, options: .literal, range: string.index(string.startIndex, offsetBy: index)...string.endIndex, locale: nil) {
            count += 1
            index = string.distance(from: string.startIndex, to: range.upperBound)
        } else {
            index = string.count
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let stringArray = Array(string)
    let substringArray = Array(substring)
    var count = 0
    var stringIndex = 0
    var substringIndex = 0
    
    while stringIndex < stringArray.count {
        if stringArray[stringIndex] == substringArray[substringIndex] {
            substringIndex += 1
            if substringIndex == substringArray.count {
                count += 1
                substringIndex = 0
            }
        } else {
            substringIndex = 0
        }
        stringIndex += 1
    }
    
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let stringArray = Array(string)
    let substringArray = Array(substring)
    var count = 0
    var stringIndex = 0
    var substringIndex = 0
    
    while stringIndex < stringArray.count {
        if stringArray[stringIndex] == substringArray[substringIndex] {
            substringIndex += 1
            if substringIndex == substringArray.count {
                count += 1
                substringIndex = 0
            }
        } else {
            substringIndex = 0
        }
        stringIndex += 1
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring, range: string.range(from: index)) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring, range: string.range(from: index)) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let stringCount = string.count
    let subCount = substring.count
    var count = 0
    
    for i in 0 ..< stringCount {
        if string[i ... i + subCount - 1] == substring {
            count += 1
        }
    }
    
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let stringCount = string.count
    let subCount = substring.count
    var count = 0
    
    for i in 0 ..< stringCount {
        if string[i ... i + subCount - 1] == substring {
            count += 1
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let stringArray = Array(string)
    let substringArray = Array(substring)
    let substringLength = substringArray.count
    var substringCount = 0
    
    for index in 0..<stringArray.count {
        if stringArray[index] == substringArray[0] {
            if substringArray == Array(stringArray[index..<index+substringLength]) {
                substringCount += 1
            }
        }
    }
    
    return substringCount
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let stringArray = Array(string)
    let substringArray = Array(substring)
    let substringLength = substringArray.count
    var substringCount = 0
    
    for index in 0..<stringArray.count {
        if stringArray[index] == substringArray[0] {
            if substringArray == Array(stringArray[index..<index+substringLength]) {
                substringCount += 1
            }
        }
    }
    
    return substringCount
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let stringCount = string.count
    let substringCount = substring.count
    if substringCount == 0 || substringCount > stringCount { return 0 }
    var count = 0
    for i in 0..<stringCount {
        if string[i] == substring[0] {
            var j = 1
            while j < substringCount {
                if i + j >= stringCount { break }
                if string[i + j] != substring[j] { break }
                j += 1
            }
            if j == substringCount {
                count += 1
            }
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let stringCount = string.count
    let substringCount = substring.count
    if substringCount == 0 || substringCount > stringCount { return 0 }
    var count = 0
    for i in 0..<stringCount {
        if string[i] == substring[0] {
            var j = 1
            while j < substringCount {
                if i + j >= stringCount { break }
                if string[i + j] != substring[j] { break }
                j += 1
            }
            if j == substringCount {
                count += 1
            }
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let string = string.lowercased()
    let substring = substring.lowercased()
    
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var startIndex = 0
    
    while true {
        let index = string.index(string.startIndex, offsetBy: startIndex)
        if let range = string.range(of: substring, range: index ..< string.endIndex) {
            count += 1
            startIndex = range.lowerBound.encodedOffset
        } else {
            break
        }
    }
    
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let string = string.lowercased()
    let substring = substring.lowercased()
    
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var startIndex = 0
    
    while true {
        let index = string.index(string.startIndex, offsetBy: startIndex)
        if let range = string.range(of: substring, range: index ..< string.endIndex) {
            count += 1
            startIndex = range.lowerBound.encodedOffset
        } else {
            break
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if string.range(of: substring, range: index..<string.endIndex) != nil {
            count += 1
            index = string.index(string.startIndex, offsetBy: substring.count, limitedBy: string.endIndex) ?? string.endIndex
        } else {
            index = string.index(index, offsetBy: 1, limitedBy: string.endIndex) ?? string.endIndex
        }
    }
    
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index != string.endIndex {
        if string.range(of: substring, range: index..<string.endIndex) != nil {
            count += 1
            index = string.index(string.startIndex, offsetBy: substring.count, limitedBy: string.endIndex) ?? string.endIndex
        } else {
            index = string.index(index, offsetBy: 1, limitedBy: string.endIndex) ?? string.endIndex
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.length {
        var foundIndex = string.index(substring, fromIndex: index)
        if foundIndex != -1 {
            count += 1
            index = foundIndex + substring.length
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.length {
        var foundIndex = string.index(substring, fromIndex: index)
        if foundIndex != -1 {
            count += 1
            index = foundIndex + substring.length
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let count = 0
    for i in 0..<string.length {
        if string[i..<string.length].contains(substring) {
            count += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let count = 0
    for i in 0..<string.length {
        if string[i..<string.length].contains(substring) {
            count += 1
        }
    }
    return count
}

// Find the longest substring in the original string that can be found in the original string.
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "aaabaaa") == "aaa"
// longestSubstring(string: "

func howManyTimes(string: String, substring: String) -> Int {
    let result = 0
    let substringLength = substring.length
    let stringLength = string.length
    for i in 0..<stringLength {
        if string.substring(i..<i+substringLength) == substring {
            result += 1
        }
    }
    return result
}

func firstOccurrence(string: String, substring: String) -> Int {
    let substringLength = substring.length
    let stringLength = string.length
    for i in 0..<stringLength {
        if string.substring(i..<i+substringLength) == substring {
            return i
        }
    }
    return -1
}

func lastOccurrence(string: String, substring: String) -> Int {
    let substringLength = substring.length
    let stringLength = string.length
    for i in 0..<stringLength {
        if string.substring(i..<i+substringLength) == substring {
            return i+substringLength-1
        }
    }
    return -1
}

func firstNonOverlapping(string: String, substring: String) -> Int {
    let substringLength = substring.length
    let stringLength = string.length
    for i in 0..<stringLength {
        if string.substring(i..<i+substringLength) == substring {
            return i
        }
    }
    return -1
}

func lastNonOverlapping(string: String, substring: String) -> Int {
    let substringLength = substring.length
    let stringLength = string.length
    for i in 0..<stringLength {
        if string.substring(i..<i+substringLength) == substring {
            return i+substringLength-1
        }
    }
    return -1
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let result = 0
    let substringLength = substring.length
    let stringLength = string.length
    for i in 0..<stringLength {
        if string.substring(i..<i+substringLength) == substring {
            result += 1
        }
    }
    return result
}

// Find the first occurrence of a given substring in the original string.
// firstOccurrence(string: "", substring: "a") == -1
// firstOccurrence(string: "aaa", substring: "a") == 0
// firstOccurrence(string: "aaaa", substring: "aa") == 0
// firstOccurrence(string: "aaaa", substring: "a") == 0
// firstOccurrence(string: "aaaa", substring: "b") == -1
func firstOccurrence(string: String, substring: String) -> Int {
    let substringLength = substring.length
    let stringLength = string.length
    for i in 0..<stringLength {
        if string.substring(i..<i+substringLength) == substring {
            return i
        }
    }
    return -1
}

// Find the last occurrence of a given substring in the original string.
// lastOccurrence(string: "", substring: "a") == -1
// lastOccurrence(string: "aaa", substring: "a") == 2
// lastOccurrence(string: "aaaa", substring: "aa") == 3
// lastOccurrence(string: "aaaa", substring: "a") == 3
// lastOccurrence(string: "aaaa", substring: "b") == -1
func lastOccurrence(string: String, substring: String) -> Int {
    let substringLength = substring.length
    let stringLength = string.length
    for i in 0..<stringLength {
        if string.substring(i..<i+substringLength) == substring {
            return i+substringLength-1
        }
    }
    return -1
}

// Find the first non-overlapping occurrence of a given substring in the original string.
// firstNonOverlapping(string: "", substring: "a") == -1
// firstNonOverlapping(string: "aaa", substring: "a") == 0
// firstNonOverlapping(string: "aaaa", substring: "aa") == 0
// firstNonOverlapping(string: "aaaa", substring: "a") == 0
// firstNonOverlapping(string: "aaaa", substring: "b") == 4
func firstNonOverlapping(string: String, substring: String) -> Int {
    let substringLength = substring.length
    let stringLength = string.length
    for i in 0..<stringLength {
        if string.substring(i..<i+substringLength) == substring {
            return i
        }
    }
    return -1
}

// Find the last non-overlapping occurrence of a given substring in the original string.
// lastNonOverlapping(string: "", substring: "a") == -1
// lastNonOverlapping(string: "aaa", substring: "a") == 2
// lastNonOverlapping(string: "aaaa", substring: "aa") == 2
// lastNonOverlapping(string: "aaaa", substring: "a") == 2
// lastNonOverlapping(string: "aaaa", substring: "b") == 4
func lastNonOverlapping(string: String, substring: String) -> Int {
    let substringLength = substring.length
    let stringLength = string.length
    for i in 0..<stringLength {
        if string.substring(i..<i+substringLength) == substring {
            return i+substringLength-1
        }
    }
    return -1
}

// Find the first occurrence of a given substring in the original string.
// firstOccurrence(string: "", substring: "a") == -1
// firstOccurrence(string: "aaa", substring: "a") == 0
// firstOccurrence(string: "aaaa", substring: "aa") == 0
// firstOccurrence(string: "aaaa", substring: "a") == 0
// firstOccurrence(string: "aaaa", substring: "b") == -1
func firstOccurrence(string: String, substring: String) -> Int {
    let substringLength = substring.length
    let stringLength = string.length
    for i in 0..<stringLength {
        if string.substring(i..<i+substringLength) == substring {
            return i
        }
    }
    return -1
}

// Find

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.length {
        var substringIndex = string.index(substring: substring, startIndex: index)
        if substringIndex != -1 {
            count += 1
            index = substringIndex + substring.length
        } else {
            break
        }
    }
    return count
}

func longestPalindrome(string: String) -> String {
    var longest = ""
    for i in 0..<string.length {
        for j in i..<string.length {
            var substring = string[i..<j+1]
            if isPalindrome(substring) && substring.length > longest.length {
                longest = substring
            }
        }
    }
    return longest
}

func isPalindrome(string: String) -> Bool {
    var left = 0
    var right = string.length - 1
    while left < right {
        if string[left] != string[right] {
            return false
        }
        left += 1
        right -= 1
    }
    return true
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.length {
        var substringIndex = string.index(substring: substring, startIndex: index)
        if substringIndex != -1 {
            count += 1
            index = substringIndex + substring.length
        } else {
            break
        }
    }
    return count
}

// Find the longest substring in the original string that is a palindrome.
// longestPalindrome(string: "") == ""
// longestPalindrome(string: "a") == "a"
// longestPalindrome(string: "aba") == "aba"
// longestPalindrome(string: "abac") == "aba"
// longestPalindrome(string: "abacaba") == "aba"
// longestPalindrome(string: "abacaba") == "aba"
func longestPalindrome(string: String) -> String {
    var longest = ""
    for i in 0..<string.length {
        for j in i..<string.length {
            var substring = string[i..<j+1]
            if isPalindrome(substring) && substring.length > longest.length {
                longest = substring
            }
        }
    }
    return longest
}

func isPalindrome(string: String) -> Bool {
    var left = 0
    var right = string.length - 1
    while left < right {
        if string[left] != string[right] {
            return false
        }
        left += 1
        right -= 1
    }
    return true
}

// Find the longest substring in the original string that is a palindrome.
// longestPalindrome(string: "") == ""
// longestPalindrome(string: "a") == "a"
// longestPalindrome(string: "aba") == "aba"
// longestPalindrome(string: "abac") == "aba"
// longestPalindrome(string: "abacaba") == "aba"
// longestPalindrome(string: "abacaba") == "aba"
func longestPalindrome(string: String) -> String {
    var longest = ""
    for i in 0..<string.length {
        for j in i..<string.length {
            var substring = string[i..<j+1]
            if isPalindrome(substring) && substring.length > longest.length {
                longest = substring
            }
        }
    }
    return longest
}

func isPalindrome(string: String) -> Bool {
    var left = 0
    var right = string.length - 1
    while left < right {
        if string[left] != string[right] {
            return false
        }
        left += 1
        right -= 1
    }
    return true
}

// Find the longest substring in the original string that is a palindrome.
// longestPalindrome(string: "") == ""
// longestPalindrome(string: "a") == "a"
// longestPalindrome(string: "aba") == "aba"
// longestPalindrome(string: "abac") == "aba"
// longestPalindrome(string: "abacaba") == "aba"
// longestPalindrome(string: "abacaba") == "aba"
func longestPalindrome(string: String) -> String {
    var longest = ""
    for i in 0..<string.length {
        for j in i..<string.length {
            var substring = string[i..<j+1]
            if isPalindrome(substring) && substring.length > longest.length {
                longest = substring
            }
        }
    }
    return longest
}

func isPalindrome(string: String) -> Bool {
    var left = 0
    var right = string.length - 1
    while left < right {
        if string[left] != string[right] {
            return false
        }
        left += 1
        right -= 1
    }
    return true
}

// Find the longest substring in the original string that is a palindrome.
// longestPalindrome(string: "") == ""
// longestPalindrome(string: "a") == "a"
// longestPalindrome(string: "aba") == "aba"
// longestPalindrome(string: "abac") == "aba"
// longestPalindrome(string: "abacaba") == "aba"
// longestPalindrome(string: "abacaba") == "aba"
func longestPalindrome(string: String) -> String {
    var longest = ""
    for i in 0..<string.length {

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.length {
        var foundIndex = string.index(substring, startIndex: index)
        if foundIndex == -1 {
            break
        }
        count += 1
        index = foundIndex + 1
    }
    return count
}

func longestSubstring(string: String) -> String {
    var longest = ""
    var current = ""
    var index = 0
    while index < string.length {
        var foundIndex = string.index(string[index], startIndex: index + 1)
        if foundIndex == -1 {
            if current.length > longest.length {
                longest = current
            }
            break
        }
        current += string[foundIndex]
        index = foundIndex + 1
    }
    return longest
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.length {
        var foundIndex = string.index(substring, startIndex: index)
        if foundIndex == -1 {
            break
        }
        count += 1
        index = foundIndex + 1
    }
    return count
}

// Find the longest substring of the original string that can be found in the original string.
// longestSubstring(string: "abcabcbb") == "abc"
// longestSubstring(string: "bbbbb") == "b"
// longestSubstring(string: "pwwkew") == "wke"
func longestSubstring(string: String) -> String {
    var longest = ""
    var current = ""
    var index = 0
    while index < string.length {
        var foundIndex = string.index(string[index], startIndex: index + 1)
        if foundIndex == -1 {
            if current.length > longest.length {
                longest = current
            }
            break
        }
        current += string[foundIndex]
        index = foundIndex + 1
    }
    return longest
}

// Find the longest substring of the original string that can be found in the original string.
// longestSubstring(string: "abcabcbb") == "abc"
// longestSubstring(string: "bbbbb") == "b"
// longestSubstring(string: "pwwkew") == "wke"
func longestSubstring(string: String) -> String {
    var longest = ""
    var current = ""
    var index = 0
    while index < string.length {
        var foundIndex = string.index(string[index], startIndex: index + 1)
        if foundIndex == -1 {
            if current.length > longest.length {
                longest = current
            }
            break
        }
        current += string[foundIndex]
        index = foundIndex + 1
    }
    return longest
}

// Find the longest substring of the original string that can be found in the original string.
// longestSubstring(string: "abcabcbb") == "abc"
// longestSubstring(string: "bbbbb") == "b"
// longestSubstring(string: "pwwkew") == "wke"
func longestSubstring(string: String) -> String {
    var longest = ""
    var current = ""
    var index = 0
    while index < string.length {
        var foundIndex = string.index(string[index], startIndex: index + 1)
        if foundIndex == -1 {
            if current.length > longest.length {
                longest = current
            }
            break
        }
        current += string[foundIndex]
        index = foundIndex + 1
    }
    return longest
}

// Find the longest substring of the original string that can be found in the original string.
// longestSubstring(string: "abcabcbb") == "abc"
// longestSubstring(string: "bbbbb") == "b"
// longestSubstring(string: "pwwkew") == "wke"
func longestSubstring(string: String) -> String {
    var longest = ""
    var current = ""
    var index = 0
    while index < string.length {
        var foundIndex = string.index(string[index], startIndex: index + 1)
        if foundIndex == -1 {
            if current.length > longest.length {
                longest = current
            }
            break
        }
        current += string[foundIndex]
        index = foundIndex + 1
    }
    return longest
}

// Find the longest substring of the original string that can be found in the original string.
// longestSubstring(string: "abcabcbb") == "abc"
// longestSubstring(string: "bbbbb") == "b"
// longestSubstring(string: "pwwkew") == "wke"
func longestSubstring(string: String) -> String {
    var longest = ""
    var current = ""
    var index = 0
    while index < string.length {
        var foundIndex = string.index(string[index], startIndex: index + 1)
        if foundIndex == -1 {
            if current.length > longest.length {
                longest = current
            }
            break
        }
        current += string[foundIndex]
        index = foundIndex + 1
    }
    

func howManyTimes(string: String, substring: String) -> Int {
    let substrings: [String] = string.components(separatedBy: substring)
    return substrings.count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let substrings: [String] = string.components(separatedBy: substring)
    return substrings.count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let len = string.length
    let subLen = substring.length
    let mut count = 0
    for i in 0..<len {
        let s = string[i..<len]
        if s.contains(substring) {
            count += 1
        }
    }
    count
}

func indexOf(string: String, substring: String) -> Int {
    let len = string.length
    let subLen = substring.length
    for i in 0..<len {
        let s = string[i..<len]
        if s == substring {
            return i
        }
    }
    -1
}

func lastIndexOf(string: String, substring: String) -> Int {
    let len = string.length
    let subLen = substring.length
    for i in 0..<len {
        let s = string[i..<len]
        if s == substring {
            return i
        }
    }
    -1
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let len = string.length
    let subLen = substring.length
    let mut count = 0
    for i in 0..<len {
        let s = string[i..<len]
        if s.contains(substring) {
            count += 1
        }
    }
    count
}

// Find the index of the first occurrence of a given substring in the original string.
// indexOf(string: "", substring: "a") == -1
// indexOf(string: "aaa", substring: "a") == 0
// indexOf(string: "aaaa", substring: "aa") == 0
func indexOf(string: String, substring: String) -> Int {
    let len = string.length
    let subLen = substring.length
    for i in 0..<len {
        let s = string[i..<len]
        if s == substring {
            return i
        }
    }
    -1
}

// Find the index of the last occurrence of a given substring in the original string.
// lastIndexOf(string: "", substring: "a") == -1
// lastIndexOf(string: "aaa", substring: "a") == 2
// lastIndexOf(string: "aaaa", substring: "aa") == 3
func lastIndexOf(string: String, substring: String) -> Int {
    let len = string.length
    let subLen = substring.length
    for i in 0..<len {
        let s = string[i..<len]
        if s == substring {
            return i
        }
    }
    -1
}

// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring in the original string.
// Find the index of the last occurrence of a given substring in the original string.
// Find the index of the first occurrence of a given substring

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.length {
        var foundIndex = string.index(substring, fromIndex: index)
        if foundIndex == -1 {
            break
        }
        count += 1
        index = foundIndex + 1
    }
    return count
}

func findFirstIndex(string: String, substring: String) -> Int {
    var index = 0
    while index < string.length {
        var foundIndex = string.index(substring, fromIndex: index)
        if foundIndex == -1 {
            break
        }
        return foundIndex
        index = foundIndex + 1
    }
    return -1
}

func findLastIndex(string: String, substring: String) -> Int {
    var index = string.length - 1
    while index >= 0 {
        var foundIndex = string.index(substring, fromIndex: index)
        if foundIndex == -1 {
            break
        }
        return foundIndex
        index = foundIndex - 1
    }
    return -1
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.length {
        var foundIndex = string.index(substring, fromIndex: index)
        if foundIndex == -1 {
            break
        }
        count += 1
        index = foundIndex + 1
    }
    return count
}

// Find the first index of a given substring in the original string.
// findFirstIndex(string: "", substring: "a") == -1
// findFirstIndex(string: "aaa", substring: "a") == 0
// findFirstIndex(string: "aaaa", substring: "aa") == 0
func findFirstIndex(string: String, substring: String) -> Int {
    var index = 0
    while index < string.length {
        var foundIndex = string.index(substring, fromIndex: index)
        if foundIndex == -1 {
            break
        }
        return foundIndex
        index = foundIndex + 1
    }
    return -1
}

// Find the last index of a given substring in the original string.
// findLastIndex(string: "", substring: "a") == -1
// findLastIndex(string: "aaa", substring: "a") == 2
// findLastIndex(string: "aaaa", substring: "aa") == 3
func findLastIndex(string: String, substring: String) -> Int {
    var index = string.length - 1
    while index >= 0 {
        var foundIndex = string.index(substring, fromIndex: index)
        if foundIndex == -1 {
            break
        }
        return foundIndex
        index = foundIndex - 1
    }
    return -1
}

// Find the first index of a given substring in the original string.
// findLastIndex(string: "", substring: "a") == -1
// findLastIndex(string: "aaa", substring: "a") == 2
// findLastIndex(string: "aaaa", substring: "aa") == 3
func findLastIndex(string: String, substring: String) -> Int {
    var index = string.length - 1
    while index >= 0 {
        var foundIndex = string.index(substring, fromIndex: index)
        if foundIndex == -1 {
            break
        }
        return foundIndex
        index = foundIndex - 1
    }
    return -1
}

// Find the first index of a given substring in the original string.
// findLastIndex(string: "", substring: "a") == -1
// findLastIndex(string: "aaa", substring: "a") == 2
// findLastIndex(string: "aaaa", substring: "aa") == 3
func findLastIndex(string: String, substring: String) -> Int {
    var index = string.length - 1
    while index >= 0 {
        var foundIndex = string.index(substring, fromIndex: index)
        if foundIndex == -1 {
            break
        }
        return foundIndex
        index = foundIndex - 1
    }
    return -1
}

// Find the first index of a given substring in the original string.
// findLastIndex(string: "", substring: "a") == -1
// findLastIndex(string: "aaa", substring: "a") == 2
// findLastIndex(string: "aaaa", substring: "aa") == 3
func findLastIndex(string: String, substring: String) -> Int {
    var index = string.length - 1
    while index >= 0 {
        var foundIndex = string.index(substring, fromIndex: index)
        if foundIndex == -1 {
            break
        }
        return foundIndex
        index = foundIndex - 1
    }
    return -1
}

// Find the first index of a given substring in the original string.
// findLastIndex(string: "", substring: "a") == -1
// findLastIndex(string: "aaa", substring: "a") == 2
// findLastIndex(string: "aaaa", substring: "aa") == 3
func findLastIndex(string: String, substring: String) -> Int {
    var index = string.length - 1
    while index >= 0 {
        var foundIndex = string.index(substring, fromIndex: index)
        if foundIndex == -1 {
            break
        }

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.length {
        var substringIndex = string.index(substring: substring, startIndex: index)
        if substringIndex != -1 {
            count += 1
            index = substringIndex + substring.length
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.length {
        var substringIndex = string.index(substring: substring, startIndex: index)
        if substringIndex != -1 {
            count += 1
            index = substringIndex + substring.length
        } else {
            break
        }
    }
    return count
}

// Find the longest substring in the original string that can be found in the original string.
// longestSubstring(string: "abc") == "abc"
// longestSubstring(string: "abca") == "abc"
// longestSubstring(string: "abcab") == "abc"
// longestSubstring(string: "abcabca") == "abc"
// longestSubstring(string: "abcabcab") == "abc"
// longestSubstring(string: "abcabcabc") == "abc"
// longestSubstring(string: "abcabcabca") == "abc"
// longestSubstring(string: "abcabcabcab") == "abc"
// longestSubstring(string: "abcabcabcabc") == "abc"
// longestSubstring(string: "abcabcabcabca") == "abc"
// longestSubstring(string: "abcabcabcabcab") == "abc"
// longestSubstring(string: "abcabcabcabcabc") == "abc"
// longestSubstring(string: "abcabcabcabcabca") == "abc"
// longestSubstring(string: "abcabcabcabcabcab") == "abc"
// longestSubstring(string: "abcabcabcabcabcabc") == "abc"
// longestSubstring(string: "abcabcabcabcabcabca") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcab") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabc") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabca") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcab") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabc") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabca") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcab") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabc") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabca") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcab") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabc") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabca") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabcab") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabcabcabc") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabcabca") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabcabcab") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabcabcabc") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabcabcabca") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabcabcabcab") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabcabcabcabc") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabcabcabcabca") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabca") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca") == "abc"
// longestSubstring(string: "abcabcabcabcabcabcabcab

func howManyTimes(string: String, substring: String) -> Int {
    let result = 0
    let substringLength = substring.length
    for i in 0..<string.length - substringLength + 1 {
        if string[i..<i + substringLength] == substring {
            result += 1
        }
    }
    return result
}

func longestPalindrome(string: String) -> String {
    let result = ""
    let stringLength = string.length
    for i in 0..<stringLength {
        for j in i..<stringLength {
            let substring = string[i..<j + 1]
            if substring == substring.reversed() && substring.length > result.length {
                result = substring
            }
        }
    }
    return result
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let result = 0
    let substringLength = substring.length
    for i in 0..<string.length - substringLength + 1 {
        if string[i..<i + substringLength] == substring {
            result += 1
        }
    }
    return result
}

// Find the longest substring in the original string that is a palindrome.
// longestPalindrome(string: "a") == "a"
// longestPalindrome(string: "ab") == "a"
// longestPalindrome(string: "aba") == "aba"
// longestPalindrome(string: "abcba") == "abcba"
// longestPalindrome(string: "abccba") == "abcba"
// longestPalindrome(string: "abcbaabc") == "abcba"
// longestPalindrome(string: "abcbaabcba") == "abcba"
// longestPalindrome(string: "abcbaabcbabc") == "abcba"
func longestPalindrome(string: String) -> String {
    let result = ""
    let stringLength = string.length
    for i in 0..<stringLength {
        for j in i..<stringLength {
            let substring = string[i..<j + 1]
            if substring == substring.reversed() && substring.length > result.length {
                result = substring
            }
        }
    }
    return result
}

// Find the longest substring in the original string that is a palindrome.
// longestPalindrome(string: "a") == "a"
// longestPalindrome(string: "ab") == "a"
// longestPalindrome(string: "aba") == "aba"
// longestPalindrome(string: "abcba") == "abcba"
// longestPalindrome(string: "abccba") == "abcba"
// longestPalindrome(string: "abcbaabc") == "abcba"
// longestPalindrome(string: "abcbaabcba") == "abcba"
// longestPalindrome(string: "abcbaabcbabc") == "abcba"
func longestPalindrome(string: String) -> String {
    let result = ""
    let stringLength = string.length
    for i in 0..<stringLength {
        for j in i..<stringLength {
            let substring = string[i..<j + 1]
            if substring == substring.reversed() && substring.length > result.length {
                result = substring
            }
        }
    }
    return result
}

// Find the longest substring in the original string that is a palindrome.
// longestPalindrome(string: "a") == "a"
// longestPalindrome(string: "ab") == "a"
// longestPalindrome(string: "aba") == "aba"
// longestPalindrome(string: "abcba") == "abcba"
// longestPalindrome(string: "abccba") == "abcba"
// longestPalindrome(string: "abcbaabc") == "abcba"
// longestPalindrome(string: "abcbaabcba") == "abcba"
// longestPalindrome(string: "abcbaabcbabc") == "abcba"
func longestPalindrome(string: String) -> String {
    let result = ""
    let stringLength = string.length
    for i in 0..<stringLength {
        for j in i..<stringLength {
            let substring = string[i..<j + 1]
            if substring == substring.reversed() && substring.length > result.length {
                result = substring
            }
        }
    }
    return result
}

// Find the longest substring in the original string that is a palindrome.
// longestPalindrome(string: "a") == "a"
// longestPalindrome(string: "ab") == "a"
// longestPalindrome(string: "aba") == "aba"
// longestPalindrome(string: "abcba") == "abcba"
// longestPalindrome(string: "abccba") == "abcba"
// longestPalindrome(string: "abcbaabc") == "abcba"
// longestPalindrome(string: "abcbaabcba") == "abcba"
// longestPalindrome(string: "abcbaabcbabc") == "abcba"
func longestPalindrome(string: String) -> String {
    let result = ""
    let stringLength = string.length
    for i in 0..<stringLength {
        for j in i..<stringLength {
            let substring = string[i..<j + 1]
            if substring == substring.reversed() && substring.length > result.length {

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = 0
    
    while startIndex < string.count {
        if let range = string.range(of: substring, options: .caseInsensitive, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound.encodedOffset
        } else {
            startIndex += 1
        }
    }
    
    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = 0
    
    while startIndex < string.count {
        if let range = string.range(of: substring, options: .caseInsensitive, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound.encodedOffset
        } else {
            startIndex += 1
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    while i < string.count {
        if string.hasPrefix(substring, range: i..<string.count) {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    while i < string.count {
        if string.hasPrefix(substring, range: i..<string.count) {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = 0

    while startIndex < string.count {
        if let range = string.range(of: substring, options: .caseInsensitive, range: startIndex..<string.count) {
            count += 1
            startIndex = range.upperBound.encodedOffset
        } else {
            startIndex += 1
        }
    }

    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = 0

    while startIndex < string.count {
        if let range = string.range(of: substring, options: .caseInsensitive, range: startIndex..<string.count) {
            count += 1
            startIndex = range.upperBound.encodedOffset
        } else {
            startIndex += 1
        }
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = 0

    while startIndex < string.count {
        if let range = string.range(of: substring, options: .caseInsensitive, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound.encodedOffset
        } else {
            startIndex += 1
        }
    }

    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = 0

    while startIndex < string.count {
        if let range = string.range(of: substring, options: .caseInsensitive, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound.encodedOffset
        } else {
            startIndex += 1
        }
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = 0

    while startIndex < string.count {
        if let range = string.range(of: substring, options: .caseInsensitive, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound.encodedOffset
        } else {
            startIndex += 1
        }
    }

    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = 0

    while startIndex < string.count {
        if let range = string.range(of: substring, options: .caseInsensitive, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound.encodedOffset
        } else {
            startIndex += 1
        }
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = 0
    
    while startIndex < string.count {
        let endIndex = string.index(startIndex, offsetBy: substring.count)
        
        if string[startIndex..<endIndex] == substring {
            count += 1
        }
        
        startIndex = endIndex
    }
    
    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = 0
    
    while startIndex < string.count {
        let endIndex = string.index(startIndex, offsetBy: substring.count)
        
        if string[startIndex..<endIndex] == substring {
            count += 1
        }
        
        startIndex = endIndex
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = 0

    while startIndex < string.count {
        if let range = string.range(of: substring, options: .caseInsensitive, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound.encodedOffset
        } else {
            startIndex += 1
        }
    }

    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = 0

    while startIndex < string.count {
        if let range = string.range(of: substring, options: .caseInsensitive, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound.encodedOffset
        } else {
            startIndex += 1
        }
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            index = string.index(after: index)
        }
    }
    
    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex) {
            count += 1
            index = range.upperBound
        } else {
            index = string.index(after: index)
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    
    while index <= string.count - substring.count {
        if string[index...index + substring.count - 1] == substring {
            count += 1
        }
        index += 1
    }
    
    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    
    while index <= string.count - substring.count {
        if string[index...index + substring.count - 1] == substring {
            count += 1
        }
        index += 1
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = string.index(range.lowerBound, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    
    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = string.index(range.lowerBound, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i <= string.count - substring.count {
        if string[i..<i+substring.count] == substring {
            count += 1
        }
        i += 1
    }
    
    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i <= string.count - substring.count {
        if string[i..<i+substring.count] == substring {
            count += 1
        }
        i += 1
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0

    while i <= string.count - substring.count {
        if string[i..<i + substring.count] == substring {
            count += 1
        }
        i += 1
    }

    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0

    while i <= string.count - substring.count {
        if string[i..<i + substring.count] == substring {
            count += 1
        }
        i += 1
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex

    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = string.index(range.upperBound, offsetBy: 1)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }

    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex

    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = string.index(range.upperBound, offsetBy: 1)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0

    while i <= string.count - substring.count {
        if string[i...i + substring.count - 1] == substring {
            count += 1
        }
        i += 1
    }

    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0

    while i <= string.count - substring.count {
        if string[i...i + substring.count - 1] == substring {
            count += 1
        }
        i += 1
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex

    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = string.index(range.upperBound, offsetBy: 1)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }

    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex

    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = string.index(range.upperBound, offsetBy: 1)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i <= string.count - substring.count {
        if string[i...i+substring.count-1] == substring {
            count += 1
        }
        i += 1
    }
    
    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    
    while i <= string.count - substring.count {
        if string[i...i+substring.count-1] == substring {
            count += 1
        }
        i += 1
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0

    while i <= string.count - substring.count {
        if string[i...(i + substring.count - 1)] == substring {
            count += 1
        }
        i += 1
    }

    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0

    while i <= string.count - substring.count {
        if string[i...(i + substring.count - 1)] == substring {
            count += 1
        }
        i += 1
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0

    while i <= string.count - substring.count {
        if string[i...i+substring.count-1] == substring {
            count += 1
        }
        i += 1
    }

    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0

    while i <= string.count - substring.count {
        if string[i...i+substring.count-1] == substring {
            count += 1
        }
        i += 1
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex

    while index < string.endIndex {
        let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex)

        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            index = string.endIndex
        }
    }

    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex

    while index < string.endIndex {
        let range = string.range(of: substring, options: .caseInsensitive, range: index..<string.endIndex)

        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            index = string.endIndex
        }
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<end_of_turn>
<start_of_turn>model
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    while i < string.count {
        if string[i..<i+substring.count] == substring {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = 0
    while i < string.count {
        if string[i..<i+substring.count] == substring {
            count += 1
            i += substring.count
        } else {
            i += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while let range = string.range(of: substring, range: startIndex..<string.endIndex) {
        count += 1
        startIndex = range.upperBound
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while let range = string.range(of: substring, range: startIndex..<string.endIndex) {
        count += 1
        startIndex = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while let range = string.range(of: substring, range: startIndex..<string.endIndex) {
        count += 1
        startIndex = range.upperBound
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while let range = string.range(of: substring, range: startIndex..<string.endIndex) {
        count += 1
        startIndex = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string[index...].range(of: substring) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...].hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, options: .literal, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, options: .literal, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while startIndex < string.endIndex {
        if let range = string.range(of: substring, options: .caseInsensitive, range: startIndex..<string.endIndex) {
            count += 1
            startIndex = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3 [/INST] ```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var i = string.startIndex
    while i < string.endIndex {
        if string[i...].hasPrefix(substring) {
            count += 1
            i = string.index(i, offsetBy: substring.count)
        } else {
            i = string.index(after: i)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        let range = string.index(index, offsetBy: substring.count, limitedBy: string.endIndex)?? string.endIndex
        let substringRange = string[index..<range]
        
        if substringRange == substring {
            count += 1
        }
        
        index = string.index(after: range)
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        let range = string.index(index, offsetBy: substring.count, limitedBy: string.endIndex)?? string.endIndex
        let substringRange = string[index..<range]
        
        if substringRange == substring {
            count += 1
        }
        
        index = string.index(after: range)
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        let range = string.index(index, offsetBy: substring.count, limitedBy: string.endIndex)?? string.endIndex
        if string[index..<range] == substring {
            count += 1
        }
        index = string.index(after: range)
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        let range = string.index(index, offsetBy: substring.count, limitedBy: string.endIndex)?? string.endIndex
        if string[index..<range] == substring {
            count += 1
        }
        index = string.index(after: range)
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    for i in 0..<string.count {
        if string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: i + substring.count) ~= string.startIndex..<string.endIndex {
            count += 1
        }
    }
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    for i in 0..<string.count {
        if string.index(string.startIndex, offsetBy: i)..<string.index(string.startIndex, offsetBy: i + substring.count) ~= string.startIndex..<string.endIndex {
            count += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.backwards, range: index..<string.endIndex) {
        count += 1
        index = string.index(before: range.lowerBound)
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }

    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.anchored, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.anchored, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.anchored, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }

    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.anchored, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }

    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.anchored, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.anchored, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.caseInsensitive, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.anchored, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }

    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }

    var count = 0
    var index = string.startIndex

    while let range = string.range(of: substring, options:.anchored, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }

    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.anchored, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    guard!substring.isEmpty else {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.anchored, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.anchored, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3

Answer:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.anchored, range: index..<string.endIndex) {
        count += 1
        index = string.index(after: range.upperBound)
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range: Range<String.Index>?
    while let range = string.range(of: substring, options: .caseInsensitive, range: range ?? string.startIndex..<string.endIndex, locale: nil) {
        count += 1
        if range.upperBound == string.endIndex {
            break
        }
        range = string.index(after: range.upperBound)
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range: Range<String.Index>?
    while let range = string.range(of: substring, options: .caseInsensitive, range: range ?? string.startIndex..<string.endIndex, locale: nil) {
        count += 1
        if range.upperBound == string.endIndex {
            break
        }
        range = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range: Range<String.Index>?
    while (range = string.range(of: substring)) != nil {
        count += 1
        string.removeSubrange(range!)
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range: Range<String.Index>?
    while (range = string.range(of: substring)) != nil {
        count += 1
        string.removeSubrange(range!)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var range = string.startIndex
    while let range = string.range(of: substring, options: [], range: range..<string.endIndex, locale: nil) {
        count += 1
        range = string.index(after: range.upperBound)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let found = string.index(string.startIndex, offsetBy: index)
        let range = string.range(found..., in: string)
        if range.contains(substring) {
            count += 1
            index += substring.count
        } else {
            break
        }
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let found = string.index(string.startIndex, offsetBy: index)
        let range = string.range(found..., in: string)
        if range.contains(substring) {
            count += 1
            index += substring.count
        } else {
            break
        }
    }
    return count
}

print(howManyTimes(string: "", substring: "a")) // 0
print(howManyTimes(string: "aaa", substring: "a")) // 3
print(howManyTimes(string: "aaaa", substring: "aa")) // 3
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        index = string.index(after: index)
        if string[index...].hasPrefix(substring) {
            count += 1
            index = string.index(after: index + substring.endIndex)
        }
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        index = string.index(after: index)
        if string[index...].hasPrefix(substring) {
            count += 1
            index = string.index(after: index + substring.endIndex)
        }
    }
    return count
}

print(howManyTimes(string: "", substring: "a")) // 0
print(howManyTimes(string: "aaa", substring: "a")) // 3
print(howManyTimes(string: "aaaa", substring: "aa")) // 3
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        index = string.index(after: index)
        if string[index...].contains(substring) {
            count += 1
            index = string.index(after: index + substring.endIndex)
        }
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        index = string.index(after: index)
        if string[index...].contains(substring) {
            count += 1
            index = string.index(after: index + substring.endIndex)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

print(howManyTimes(string: "", substring: "a")) // 0
print(howManyTimes(string: "aaa", substring: "a")) // 3
print(howManyTimes(string: "aaaa", substring: "aa")) // 3
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        index = string.index(after: index)
        if string[index...].contains(substring) {
            count += 1
            index = string.index(after: index, offsetBy: substring.count)
        }
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        index = string.index(after: index)
        if string[index...].contains(substring) {
            count += 1
            index = string.index(after: index, offsetBy: substring.count)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let count = string.filter { $0.isEqual(substring) }.count
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    let count = string.filter { $0.isEqual(substring) }.count
    return count
}

print(howManyTimes(string: "", substring: "a")) // 0
print(howManyTimes(string: "aaa", substring: "a")) // 3
print(howManyTimes(string: "aaaa", substring: "aa")) // 3
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        index = string.index(after: index)
        if string[index...].starts(with: substring) {
            count += 1
            index = string.index(after: index + substring.endIndex)
        }
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        index = string.index(after: index)
        if string[index...].starts(with: substring) {
            count += 1
            index = string.index(after: index + substring.endIndex)
        }
    }
    return count
}

print(howManyTimes(string: "", substring: "a")) // 0
print(howManyTimes(string: "aaa", substring: "a")) // 3
print(howManyTimes(string: "aaaa", substring: "aa")) // 3
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

print(howManyTimes(string: "", substring: "a")) // 0
print(howManyTimes(string: "aaa", substring: "a")) // 3
print(howManyTimes(string: "aaaa", substring: "aa")) // 3
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        index = string.index(string.startIndex, offsetBy: index)
        if string[index...].starts(with: substring) {
            count += 1
            index += substring.count
        } else {
            break
        }
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        index = string.index(string.startIndex, offsetBy: index)
        if string[index...].starts(with: substring) {
            count += 1
            index += substring.count
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3
### Response:
```swift
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let foundIndex = string.index(string.startIndex, offsetBy: index)
        if string[foundIndex...] == substring {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            var subIndex = substring.startIndex
            var subString = string[index..<string.endIndex]
            while subIndex < substring.endIndex {
                if subString[subIndex] == substring[subIndex] {
                    subIndex = subIndex.advancedBy(1)
                } else {
                    break
                }
            }
            if subIndex == substring.endIndex {
                count++
                index = index.advancedBy(1)
            } else {
                index = index.advancedBy(1)
            }
        } else {
            index = index.advancedBy(1)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            var subIndex = substring.startIndex
            var subString = string[index..<string.endIndex]
            while subIndex < substring.endIndex {
                if subString[subIndex] == substring[subIndex] {
                    subIndex = subIndex.advancedBy(1)
                } else {
                    break
                }
            }
            if subIndex == substring.endIndex {
                count++
                index = index.advancedBy(1)
            } else {
                index = index.advancedBy(1)
            }
        } else {
            index = index.advancedBy(1)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let range = string.range(of: substring, range: index..<string.count)
        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        let range = string.range(of: substring, range: index..<string.count)
        if let range = range {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index!= string.endIndex {
        if string.hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index!= string.endIndex {
        if string.hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            var subIndex = substring.startIndex
            var subString = string[index..<string.endIndex]
            while subIndex < substring.endIndex {
                if subString[subIndex] == substring[subIndex] {
                    subIndex = subIndex.advancedBy(1)
                } else {
                    break
                }
            }
            if subIndex == substring.endIndex {
                count++
                index = index.advancedBy(1)
            } else {
                index = index.advancedBy(1)
            }
        } else {
            index = index.advancedBy(1)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            var subIndex = substring.startIndex
            var subString = string[index..<string.endIndex]
            while subIndex < substring.endIndex {
                if subString[subIndex] == substring[subIndex] {
                    subIndex = subIndex.advancedBy(1)
                } else {
                    break
                }
            }
            if subIndex == substring.endIndex {
                count++
                index = index.advancedBy(1)
            } else {
                index = index.advancedBy(1)
            }
        } else {
            index = index.advancedBy(1)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            var i = index
            var j = substring.startIndex
            while i < string.endIndex && j < substring.endIndex {
                if string[i] == substring[j] {
                    i += 1
                    j += 1
                } else {
                    break
                }
            }
            if j == substring.endIndex {
                count += 1
                index = i
            } else {
                index += 1
            }
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            var i = index
            var j = substring.startIndex
            while i < string.endIndex && j < substring.endIndex {
                if string[i] == substring[j] {
                    i += 1
                    j += 1
                } else {
                    break
                }
            }
            if j == substring.endIndex {
                count += 1
                index = i
            } else {
                index += 1
            }
        } else {
            index += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = 0
    while index < string.count {
        if string.hasPrefix(substring) {
            count += 1
            index += substring.count
        } else {
            index += 1
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index!= string.endIndex {
        if string.hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index!= string.endIndex {
        if string.hasPrefix(substring) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            var index2 = index
            var index3 = substring.startIndex
            while index2 < string.endIndex && index3 < substring.endIndex && string[index2] == substring[index3] {
                index2 = string.index(after: index2)
                index3 = substring.index(after: index3)
            }
            if index3 == substring.endIndex {
                count += 1
            }
        }
        index = string.index(after: index)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            var index2 = index
            var index3 = substring.startIndex
            while index2 < string.endIndex && index3 < substring.endIndex && string[index2] == substring[index3] {
                index2 = string.index(after: index2)
                index3 = substring.index(after: index3)
            }
            if index3 == substring.endIndex {
                count += 1
            }
        }
        index = string.index(after: index)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            var i = index
            var j = substring.startIndex
            while i < string.endIndex && j < substring.endIndex && string[i] == substring[j] {
                i = string.index(after: i)
                j = substring.index(after: j)
            }
            if j == substring.endIndex {
                count += 1
            }
        }
        index = string.index(after: index)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            var i = index
            var j = substring.startIndex
            while i < string.endIndex && j < substring.endIndex && string[i] == substring[j] {
                i = string.index(after: i)
                j = substring.index(after: j)
            }
            if j == substring.endIndex {
                count += 1
            }
        }
        index = string.index(after: index)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            let range = string[index...].range(of: substring)
            if let range = range {
                count += 1
                index = range.lowerBound
            }
        }
        index = string.index(after: index)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring[substring.startIndex] {
            let range = string[index...].range(of: substring)
            if let range = range {
                count += 1
                index = range.lowerBound
            }
        }
        index = string.index(after: index)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(index, offsetBy: substring.count)] == substring {
            count += 1
        }
        index = string.index(index, offsetBy: 1)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(index, offsetBy: substring.count)] == substring {
            count += 1
        }
        index = string.index(index, offsetBy: 1)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while let range = string.range(of: substring, options: [], range: startIndex..<string.endIndex, locale: nil) {
        count += 1
        startIndex = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while let range = string.range(of: substring, options: [], range: startIndex..<string.endIndex, locale: nil) {
        count += 1
        startIndex = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while let range = string.range(of: substring, options: [], range: startIndex..<string.endIndex, locale: nil) {
        count += 1
        startIndex = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var startIndex = string.startIndex
    while let range = string.range(of: substring, options: [], range: startIndex..<string.endIndex, locale: nil) {
        count += 1
        startIndex = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...] == substring {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...] == substring {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(index, offsetBy: substring.count)] == substring {
            count += 1
        }
        index = string.index(index, offsetBy: 1)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(index, offsetBy: substring.count)] == substring {
            count += 1
        }
        index = string.index(index, offsetBy: 1)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while let range = string.range(of: substring, options: [], range: index..<string.endIndex, locale: nil) {
        count += 1
        index = range.upperBound
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!string.isEmpty &&!substring.isEmpty else {
        return 0
    }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    guard!string.isEmpty &&!substring.isEmpty else {
        return 0
    }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if let range = string.range(of: substring, range: index..<string.endIndex, options:.caseInsensitive) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if let range = string.range(of: substring, range: index..<string.endIndex, options:.caseInsensitive) {
            count += 1
            index = range.upperBound
        } else {
            break
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!string.isEmpty &&!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
            index = string.index(after: index)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    guard!string.isEmpty &&!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
            index = string.index(after: index)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!string.isEmpty &&!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
            index = string.index(after: index)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    guard!string.isEmpty &&!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
            index = string.index(after: index)
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    guard!string.isEmpty &&!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring.first! {
            let range = index..<string.index(index, offsetBy: substring.count)
            if string[range] == substring {
                count += 1
                index = range.upperBound
            } else {
                index = string.index(after: index)
            }
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    guard!string.isEmpty &&!substring.isEmpty else { return 0 }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring.first! {
            let range = index..<string.index(index, offsetBy: substring.count)
            if string[range] == substring {
                count += 1
                index = range.upperBound
            } else {
                index = string.index(after: index)
            }
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string.range(of: substring, range: index..<string.endIndex) {
        index = range.upperBound
        count += 1
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string.range(of: substring, range: index..<string.endIndex) {
        index = range.upperBound
        count += 1
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring.first! {
            let range = index..<string.index(index, offsetBy: substring.count)
            if string[range] == substring {
                count += 1
            }
        }
        index = string.index(after: index)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index] == substring.first! {
            let range = index..<string.index(index, offsetBy: substring.count)
            if string[range] == substring {
                count += 1
            }
        }
        index = string.index(after: index)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...].hasPrefix(substring) {
            index = string.index(index, offsetBy: substring.count)
            count += 1
        } else {
            index = string.index(after: index)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index...].hasPrefix(substring) {
            index = string.index(index, offsetBy: substring.count)
            count += 1
        } else {
            index = string.index(after: index)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string.range(of: substring, range: index..<string.endIndex) {
        index = range.upperBound
        count += 1
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.startIndex
    while index < string.endIndex, let range = string.range(of: substring, range: index..<string.endIndex) {
        index = range.upperBound
        count += 1
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    var count = 0
    var index = string.startIndex
    while index < string.endIndex {
        if string[index..<string.index(after: index)] == substring {
            count += 1
        }
        index = string.index(after: index)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: index...) {
            count += 1
        }
        index = string.index(index, offsetBy: 1)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: index...) {
            count += 1
        }
        index = string.index(index, offsetBy: 1)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch)
    var count = 0
    while range!= nil {
        count += 1
        string = string.stringByReplacingCharactersInRange(range!, withString: "")
        range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch)
    var count = 0
    while range!= nil {
        count += 1
        string = string.stringByReplacingCharactersInRange(range!, withString: "")
        range = string.rangeOfString(substring, options: NSStringCompareOptions.CaseInsensitiveSearch)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: substring.characters.count)
    while index!= string.endIndex {
        count += 1
        index = string.index(index, advancedBy: substring.characters.count)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: substring.characters.count)
    while index!= string.endIndex {
        count += 1
        index = string.index(index, advancedBy: substring.characters.count)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
        }
        index = string.index(index, offsetBy: 1)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, offsetBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
        }
        index = string.index(index, offsetBy: 1)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: substring.characters.count)
    while index!= string.endIndex {
        count += 1
        index = string.index(index, advancedBy: substring.characters.count)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: substring.characters.count)
    while index!= string.endIndex {
        count += 1
        index = string.index(index, advancedBy: substring.characters.count)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
        }
        index = string.index(index, advancedBy: 1)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
        }
        index = string.index(index, advancedBy: 1)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: 1)
    while index!= string.endIndex {
        if string.hasPrefix(substring, range: Range<String.Index>(start: index, end: index + substring.count)) {
            count += 1
            index = string.index(index, offsetBy: substring.count)
        } else {
            index = string.index(index, offsetBy: 1)
        }
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let count = string.components(separatedBy: substring).count - 1
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let count = string.components(separatedBy: substring).count - 1
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: substring.characters.count)
    while index!= string.endIndex {
        count += 1
        index = string.index(index, advancedBy: substring.characters.count)
    }
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    var count = 0
    var index = string.index(string.startIndex, advancedBy: substring.characters.count)
    while index!= string.endIndex {
        count += 1
        index = string.index(index, advancedBy: substring.characters.count)
    }
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    let count = string.components(separatedBy: substring).count - 1
    return count
}

```swift
// Find how many times a given substring can be found in the original string. Count overlaping cases.
// howManyTimes(string: "", substring: "a") == 0
// howManyTimes(string: "aaa", substring: "a") == 3
// howManyTimes(string: "aaaa", substring: "aa") == 3
func howManyTimes(string: String, substring: String) -> Int {
    let count = string.components(separatedBy: substring).count - 1
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.regularExpression, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while let range = string.range(of: substring, options:.regularExpression, range: index..<string.endIndex) {
        count += 1
        index = range.upperBound
    }
    
    return count
}
```

func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = range.upperBound
        } else {
            index = string.index(after: index)
        }
    }
    
    return count
}

<|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.
Find how many times a given substring can be found in the original string. Count overlaping cases.
howManyTimes(string: "", substring: "a") == 0
howManyTimes(string: "aaa", substring: "a") == 3
howManyTimes(string: "aaaa", substring: "aa") == 3<|im_end|>
<|im_start|>assistant
```swift
func howManyTimes(string: String, substring: String) -> Int {
    if string.isEmpty || substring.isEmpty {
        return 0
    }
    
    var count = 0
    var index = string.startIndex
    
    while index < string.endIndex {
        if let range = string[index...].range(of: substring) {
            count += 1
            index = range.upperBound
        } else {
            index = string.index(after: index)
        }
    }
    
    return count
}
```