3. Longest Substring Without Repeating Characters

Summary

Solution Approach	Explanation (1-minute)	Time Complexity	Space Complexity
Brute Force	Check all possible substrings and verify each one has unique characters using nested loops.	\(O(n^2)\)	\(O(min(m,n))\)
Sliding Window (Two Pointers)	Use left and right pointers with a set to track characters. When duplicate found, shrink window from left until duplicate removed.	\(O(n)\)	\(O(min(m,n))\)
Sliding Window (Length Tracking)	Similar approach but tracks substring length directly instead of using two pointers. Removes characters from the beginning when duplicates found.	\(O(n)\)	\(O(min(m,n))\)

Implementation

PythonJavaScriptGo

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        max_length = 0
        chars = set()
        left = 0
        for right in range(len(s)):
            while s[right] in chars:
                chars.remove(s[left])
                left += 1
            chars.add(s[right])
            max_length = max(
                max_length,
                len(s[left:right+1])
            )
        return max_length

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        max_length = 0
        char_set = set()
        len_substr = 0
        for i in range(len(s)):
            while s[i] in char_set:
                char_set.remove(s[i-len_substr])
                len_substr -= 1
            char_set.add(s[i])
            len_substr += 1
            max_length = max(
                max_length,
                len_substr
            )
        return max_length