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32. Longest Substring Without Repeating Characters

mediumAsked at Reddit

Find the longest substring without repeating characters. Reddit asks this to test sliding-window technique — the same pattern used to find the longest run of non-duplicate IPs hitting an endpoint (a fraud-detection primitive).

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Reddit loops.

  • Glassdoor (2026-Q1)— Reddit phone-screen sliding-window canonical.
  • LeetCode Discuss (2025-11)— Reported as Reddit trust-and-safety warm-up.

Problem

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

Constraints

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

Examples

Example 1

Input
s = "abcabcbb"
Output
3

Explanation: "abc"

Example 2

Input
s = "bbbbb"
Output
1

Example 3

Input
s = "pwwkew"
Output
3

Explanation: "wke"

Approaches

1. Brute force all substrings

Check every substring for uniqueness.

Time
O(n^3)
Space
O(min(n, alphabet))
function lengthOfLongestSubstring(s) {
  let best = 0;
  for (let i = 0; i < s.length; i++) {
    for (let j = i; j < s.length; j++) {
      const set = new Set(s.slice(i, j + 1));
      if (set.size === j - i + 1) best = Math.max(best, j - i + 1);
    }
  }
  return best;
}

Tradeoff: Cubic. Mention only as the anti-pattern.

2. Sliding window with hash map (optimal)

Maintain a window [l, r] of unique chars. When you see a duplicate at index i, jump l to max(l, lastSeen[c] + 1).

Time
O(n)
Space
O(min(n, alphabet))
function lengthOfLongestSubstring(s) {
  const lastSeen = new Map();
  let l = 0, best = 0;
  for (let r = 0; r < s.length; r++) {
    const c = s[r];
    if (lastSeen.has(c) && lastSeen.get(c) >= l) {
      l = lastSeen.get(c) + 1;
    }
    lastSeen.set(c, r);
    best = Math.max(best, r - l + 1);
  }
  return best;
}

Tradeoff: Linear single pass. The 'jump l forward' trick avoids the O(n) shift-by-one of naive window.

Reddit-specific tips

Reddit interviewers grade on the jump-l-forward trick (vs. shrinking one step at a time). Bonus signal: mention that the same pattern handles 'longest run of non-duplicate IPs' (a vote-fraud primitive) when keyed by IP instead of char.

Common mistakes

  • Decrementing one step at a time — degrades to O(n^2).
  • Forgetting the 'lastSeen.get(c) >= l' check (matters when char appeared before but outside the window).
  • Computing window size as r - l (off by one).

Follow-up questions

An interviewer at Reddit may pivot to one of these next:

  • Longest substring with at most k distinct chars (LC 340).
  • Longest substring with at most 2 distinct chars (LC 159).
  • Minimum window substring (LC 76).

Solve it now

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Output

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FAQ

Why use lastSeen index instead of a set?

Set lets you detect duplicates but not jump the left pointer to the right place — you'd have to shrink one at a time.

Could we use an int array of size 128?

Yes — bounded ASCII makes this even faster. Initialize to -1.

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