32. Longest Substring Without Repeating Characters

mediumAsked at Plaid

Find the length of the longest substring with no repeated characters. Plaid asks this as a sliding-window baseline before harder webhook-deduplication-window problems.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-20

Source citations

Public interview reports confirming this problem appears in Plaid loops.

Glassdoor (2025)— Plaid SWE II OA — sliding window classic.
Blind (2026)— Plaid backend onsite.

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

Explanation: The answer is "abc", with the length of 3.

Example 2

Input

s = "bbbbb"

Output

Example 3

Input

s = "pwwkew"

Output

Approaches

1. Brute force check every substring

Try every (i, j) pair; check if s[i..j] has duplicates.

Time: O(n^3)
Space: O(n)

function lengthOfLongestSubstring(s) {
  let best = 0;
  for (let i = 0; i < s.length; i++) {
    for (let j = i; j < s.length; j++) {
      if (new Set(s.slice(i, j + 1)).size === j - i + 1) best = Math.max(best, j - i + 1);
    }
  }
  return best;
}

Tradeoff: Cubic. Don't ship.

2. Sliding window with last-seen map

Expand right pointer. When you hit a repeat, jump left to one past the previous occurrence. Track best length.

Time: O(n)
Space: O(k) where k is alphabet

function lengthOfLongestSubstring(s) {
  const last = new Map();
  let left = 0, best = 0;
  for (let right = 0; right < s.length; right++) {
    const c = s[right];
    if (last.has(c) && last.get(c) >= left) {
      left = last.get(c) + 1;
    }
    last.set(c, right);
    best = Math.max(best, right - left + 1);
  }
  return best;
}

Tradeoff: Linear time. The 'last.get(c) >= left' check is critical — without it, a stale entry from before the window pulls left backward.

Plaid-specific tips

Plaid grades this on whether the window stays monotonic — left only moves forward. Bonus signal: mention 'this is the same shape as a webhook-dedup window where we keep only events from the last 60s; when a duplicate arrives, the window jumps past the previous occurrence.'

Common mistakes

Setting left = last.get(c) + 1 unconditionally — pulls left backward if c was last seen before the current window.
Using a Set + slow rebuilds — O(n^2).
Forgetting that right - left + 1 is the current window length.

Follow-up questions

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

Longest substring with at most k distinct characters.
Longest substring with exactly k distinct characters.
Minimum window substring (LC 76) — harder, two-character constraint.

Solve it now

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Output

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FAQ

Why does left only move forward?

Because once a character is removed from the window, the window cannot recover it without violating the no-duplicate constraint. Monotonic left is the optimization.

What's the connection to webhook dedup?

Both maintain a window of recently seen items. When a duplicate arrives, you reset the window boundary past the duplicate. Identical algorithm.

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