Longest Non-repeating Substring

mediumcodingAlgorithmsSliding WindowStringsHash Map~25 min

Asked at Google, Meta, Stripe

Overview

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

Constraints

0 <= s.length <= 50,000
s consists of English letters, digits, symbols, and spaces.
Must run in O(n) time.

Examples

lengthOfLongestSubstring("abcabcbb");
// => 3  (substring: "abc")

lengthOfLongestSubstring("bbbbb");
// => 1  (substring: "b")

lengthOfLongestSubstring("pwwkew");
// => 3  (substring: "wke")

lengthOfLongestSubstring("");
// => 0

Solution

Reveal solution

function lengthOfLongestSubstring(s) {
  const map = new Map();
  let max = 0, left = 0;

  for (let right = 0; right < s.length; right++) {
    if (map.has(s[right]) && map.get(s[right]) >= left) {
      left = map.get(s[right]) + 1;
    }
    map.set(s[right], right);
    max = Math.max(max, right - left + 1);
  }

  return max;
}

Sliding window with a map storing the last index of each character. When a duplicate is found, jump left past it.

Resources

longest-non-repeating-substring.js

Longest Non-repeating Substring

mediumcodingAlgorithmsSliding Window

Overview

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

Constraints

0 <= s.length <= 50,000
s consists of English letters, digits, symbols, and spaces.
Must run in O(n) time.

Examples

lengthOfLongestSubstring("abcabcbb");
// => 3  (substring: "abc")

lengthOfLongestSubstring("bbbbb");
// => 1  (substring: "b")

lengthOfLongestSubstring("pwwkew");
// => 3  (substring: "wke")

lengthOfLongestSubstring("");
// => 0

Solution

Reveal solution

function lengthOfLongestSubstring(s) {
  const map = new Map();
  let max = 0, left = 0;

  for (let right = 0; right < s.length; right++) {
    if (map.has(s[right]) && map.get(s[right]) >= left) {
      left = map.get(s[right]) + 1;
    }
    map.set(s[right], right);
    max = Math.max(max, right - left + 1);
  }

  return max;
}

Sliding window with a map storing the last index of each character. When a duplicate is found, jump left past it.

Resources

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