Longest substring without repeating characters

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

var lengthOfLongestSubstring = function(s) {
    let maxLength = 0;
    let left = 0;
    const charIndexMap = {};

    for (let char = 0; char < s.length; right++) {
        const currentChar = s[char];

        if (charIndexMap[currentChar] !== undefined && charIndexMap[currentChar] >= left) {
            left = charIndexMap[currentChar] + 1;
        }

        charIndexMap[currentChar] = char;
        const currentLength = char - left + 1;
        maxLength = Math.max(maxLength, currentLength);
    }
    return maxLength;
};

Dry run example

• Initialize maxLength to 0, left to 0, and characterIndexMap to an empty object.

• Loop through each character in the input string.

• char = 0: currentChar = 'a'

  Update characterIndexMap: {'a': 0} Update currentLength to 1 maxLength = 1

• char = 1: currentChar = 'b'

  Update characterIndexMap: {'a': 0, 'b': 1} Update currentLength to 2 maxLength = 2

• char = 2: currentChar = 'c'

  Update characterIndexMap: {'a': 0, 'b': 1, 'c': 2} Update currentLength to 3 maxLength = 3

• char = 3: currentChar = 'a' 'a' is already in characterIndexMap, and its index is 0 (less than left) Update left to 1 (the next index after the previous 'a') Update characterIndexMap: {'a': 3, 'b': 1, 'c': 2} Update currentLength to 3 maxLength remains 3

• char = 4: currentChar = 'b'

  'b' is already in characterIndexMap, and its index is 1 (equal to left) Update left to 2 (the next index after the previous 'b') Update characterIndexMap: {'a': 3, 'b': 4, 'c': 2} Update currentLength to 3 maxLength remains 3

• char = 5: currentChar = 'c'

  'c' is already in characterIndexMap, and its index is 2 (equal to left) Update left to 3 (the next index after the previous 'c') Update characterIndexMap: {'a': 3, 'b': 4, 'c': 5} Update currentLength to 3 maxLength remains 3

• char = 6: currentChar = 'b'

  'b' is already in characterIndexMap, and its index is 4 (greater than left) Update left to 5 (the next index after the previous 'b') Update characterIndexMap: {'a': 3, 'b': 6, 'c': 5} Update currentLength to 2 maxLength remains 3

The loop finishes, and the function returns the final maxLength, which is 3.