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32. Longest Substring Without Repeating Characters

mediumAsked at Salesforce

Find the length of the longest substring without repeating characters. Salesforce uses this as the canonical sliding-window introduction — they grade on left-pointer correctness.

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

Source citations

Public interview reports confirming this problem appears in Salesforce loops.

  • Glassdoor (2026-Q1)— Salesforce uses this as a baseline check for sliding-window comfort.
  • Blind (2025-11)— Recurring on backend onsite loops.

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; track the longest without duplicates.

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

Tradeoff: TLE on 5*10^4. Salesforce wants O(n).

2. Sliding window with last-seen map

Map last-seen index of each char. Maintain window [l, r]. When s[r] was seen at idx >= l, jump l to idx + 1.

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

Tradeoff: Single pass. The `>= l` check is critical — earlier duplicates outside the window must be ignored.

Salesforce-specific tips

Salesforce explicitly tests the `lastSeen.get(c) >= l` clause — a common bug is using `if (lastSeen.has(c))` alone, which causes l to JUMP BACKWARD when the same char appears far outside the current window. Bonus signal: mention this 'don't backtrack' principle generalizes to all sliding-window problems.

Common mistakes

  • Jumping l backward when an old duplicate appears — corrupts the window.
  • Computing length as `r - l` instead of `r - l + 1` — off by one.
  • Resetting the map on each iteration — wipes useful history.

Follow-up questions

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

  • Longest Substring with At Most K Distinct (LC 340).
  • Longest Repeating Character Replacement (LC 424).
  • Minimum Window Substring (LC 76).

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Output

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FAQ

Why the `>= l` clause?

Because a duplicate from before the current window start shouldn't move l. Without the check, l could move backward, breaking the monotone-window invariant.

What if the alphabet is unicode-wide?

Map scales to O(unique chars). For dense ASCII you could use a 128-element array; for full Unicode, Map is the right choice.

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