leetcode-3-longest-substring-without-repeating-characters.swift

/**

Copyright (c) 2020 David Seek, LLC

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OR OTHER DEALINGS IN THE SOFTWARE.





https://leetcode.com/problems/longest-substring-without-repeating-characters/


Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3. 
Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:

Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3. 
             Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

*/

/**
Big O Annotation
Time complexity O(n) where n is the amount of elements in nums.
Space complexity O(n) where n is the amount of elements in nums.
*/
func lengthOfLongestSubstring(_ s: String) -> Int {
	
    /**
    First we get an array of the string
    to more conveniently access the indicies
    */
    var array = Array(s)

    /**
    We create a set to check for 
    processed character Set.contains 
    in contant time
    */
    var hash = Set<Character>()
    
    // We create a left and a right pointer
    var leftPointer: Int = 0
    var rightPointer: Int = 0
    
    // Our current maximum
    var maximum: Int = 0

    // While we still have characters to process...
    while rightPointer < array.count {
        
        // ... get the left and right character
        let left = array[leftPointer]
        let right = array[rightPointer]
        
        /**
        In the example of "abcabcbb",
        we first get left and right pointers
        on index 0 aka "a".

        Next we check if "a" is already 
        in the Set, no, so insert it.

        Then we move the right pointer forwards.
        "b", not in the list, add it.

        Next. "c" not in the list. Add it.
        Next. "a" already in the list,
        so remove "a" and continue to the next character.

        Once removed, our pointer is still on the 
        2nd "a", so we move it back to the array.
        And move along to the next characters.

        Always updating our maximum to the new total maximum.
        */

        if hash.contains(right) {
            
            hash.remove(left)
            leftPointer += 1
            
        } else {
            
            hash.insert(right)
            rightPointer += 1
            maximum = max(hash.count, maximum)
        }
    }

    return maximum
}