3. Longest Substring Without Repeating Characters

mediumAsked at Apple

Longest Substring Without Repeating Characters is the canonical sliding-window question and Apple's most-repeated string medium. Expand the right pointer; when a duplicate appears, jump the left pointer to last_index_of_dup + 1. The hash-map-of-char-to-last-index variant is what Apple grades on.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-19

Source citations

Public interview reports confirming this problem appears in Apple loops.

Glassdoor (2026-Q1)— Apple SWE phone-screen reports list this as the most-repeated sliding-window medium.
Blind (2025-12)— Apple ICT3/ICT4 reports cite Longest Substring without Repeating Characters as the canonical string medium.

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

Explanation: Note that the answer must be a substring, 'pwke' is a subsequence and not a substring.

Approaches

1. Sliding window with set, advance left one step at a time

Expand right, adding chars to a set. When a duplicate appears, remove chars from left until the duplicate is gone.

Time: O(n)
Space: O(min(n, alphabet))

function lengthOfLongestSubstring(s) {
  const seen = new Set();
  let left = 0;
  let best = 0;
  for (let right = 0; right < s.length; right++) {
    while (seen.has(s[right])) {
      seen.delete(s[left]);
      left++;
    }
    seen.add(s[right]);
    best = Math.max(best, right - left + 1);
  }
  return best;
}

Tradeoff: Linear amortized — each char enters and exits the window at most once. The inner while loop can advance left many times in a single right step, but the total left advances across the whole run is bounded by n.

2. Sliding window with last-seen map, jump left (optimal)

Track last seen index per char. When the current char was seen at index >= left, jump left to last_seen + 1.

Time: O(n)
Space: O(min(n, alphabet))

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

Tradeoff: Each right step is O(1) work — no inner loop. The 'jump' replaces the iterative shrink. Apple prefers this version because it's strictly fewer comparisons and the lastSeen >= left check is what makes it correct without resetting the map.

Apple-specific tips

Apple interviewers grade this on the lastSeen >= left invariant. Say it out loud: 'I use a map of char to its last-seen index. A duplicate only matters if it was seen INSIDE the current window — that is, at an index >= left.' That guard is what lets you skip resetting the map between windows and keeps the algorithm truly linear.

Common mistakes

Forgetting the lastSeen.get(ch) >= left check — jumps left backward when an old duplicate is outside the window.
Using left = lastSeen.get(ch) (off by one) instead of left = lastSeen.get(ch) + 1.
Returning the window contents instead of its length.

Follow-up questions

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

Longest Substring with At Most K Distinct Characters (LC 340) — same window, frequency map.
Longest Substring with At Most Two Distinct Characters (LC 159).
Minimum Window Substring (LC 76) — variable window with target counts.

Solve it now

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Output

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FAQ

Set + shrink or map + jump?

Same O(n) but the map+jump version does strictly fewer operations and is what Apple expects to see on a senior interview.

Does the alphabet size matter?

For the bound — the space is O(min(n, alphabet)) — yes. For correctness, no; the map handles arbitrary characters.

Free learning resources

Curated free links for this problem.

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