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Longest Substring Without Repeating Characters LC #3 Medium Sliding Window · HashSet
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 length 3.
Example 2
Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with length 3. Note that "pwke" is a subsequence, not a substring.
Constraints: 0 ≤ s.length ≤ 5×10⁴ | s consists of English letters, digits, symbols, and spaces
Try Examples
Custom:
Approach
Variable sliding window [l..r] + HashSet — expand r, shrink l on duplicate
Move r right. If s[r] is already in the window, remove s[l] and advance l until unique again. Each character enters and leaves the set at most once → O(n) time.
String Window
l (left)
r (right)
in window
duplicate
best window
HashSet (window chars)
{ empty }
maxLen
0
Ready
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Variables
l
0
r
—
winLen
—
maxLen
0
set
∅
Step Logic
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Ready
0 / 0
Select an example above and press Play.
Algorithm
1
Init empty Set, l = 0, maxLen = 0
2
For each r from 0 to n−1: expand window right
3
While set.contains(s[r]): remove s[l], l++ (shrink left)
4
Add s[r] to set; maxLen = max(maxLen, r − l + 1)
5
Return maxLen after scanning all positions
Time
O(n)
Space
O(min(n, α))
Why This Works

A variable sliding window always holds a valid substring (no repeats). When s[r] duplicates a char already in the window, we shrink from the left until the duplicate is gone — the smallest valid window ending at r.

The HashSet gives O(1) membership checks. Each index enters and leaves the window at most once, so total work is O(n) despite the inner while loop.

maxLen tracks the best window seen so far. After each valid expansion at r, window length r − l + 1 is a candidate answer.