Longest Substring Without Repeating Characters

Problem

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

Example 1:

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

Example 2:

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

Example 3:

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

Constraints:

0 <= s.length <= 5 * 10^4
s consists of English letters, digits, symbols and spaces.

Solution

Approach: Sliding Window with Hash Set

The key insight is to use a sliding window technique with a hash set to track characters in the current window. We expand the window by moving the right pointer and shrink it when we encounter a duplicate character.

We maintain:

left: left boundary of the window
char_set: set of characters in the current window
max_length: maximum length found so far

For each character:

If it’s not in the set, add it and update max length
If it’s a duplicate, remove characters from the left until the duplicate is removed

Implementation

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        char_set = set()
        left = 0
        max_length = 0

        for right in range(len(s)):
            # Shrink window until no duplicates
            while s[right] in char_set:
                char_set.remove(s[left])
                left += 1

            # Add current character
            char_set.add(s[right])
            # Update max length
            max_length = max(max_length, right - left + 1)

        return max_length

Alternative (Using Hash Map for Index Tracking):

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        char_index = {}
        left = 0
        max_length = 0

        for right in range(len(s)):
            # If character seen and within window, move left
            if s[right] in char_index and char_index[s[right]] >= left:
                left = char_index[s[right]] + 1

            # Update character's latest index
            char_index[s[right]] = right
            # Update max length
            max_length = max(max_length, right - left + 1)

        return max_length

Complexity Analysis

Time Complexity: O(n), where n is the length of the string. Each character is visited at most twice (once by right pointer, once by left pointer).
Space Complexity: O(min(m, n)), where m is the size of the character set. In the worst case, we store all unique characters in the set.

Key Insights

Sliding Window Pattern: Expand the window with the right pointer and shrink it with the left pointer when we find a duplicate. This maintains the invariant that our window always contains unique characters.
Set vs Map: Using a set is simpler and works well. Using a map to store indices allows us to jump the left pointer directly to the position after the duplicate, potentially skipping unnecessary iterations.
Window Size Calculation: The length of the current window is right - left + 1. We track the maximum of all window sizes.
Edge Cases: Empty string returns 0, single character returns 1, all duplicates returns 1.