Longest Repeating Substring After Replacements

mediumcodingAlgorithmsSliding WindowStrings~30 min

Asked at Google, Meta

Overview

Given a string s consisting of uppercase English letters and an integer k, you can choose any character and change it to any other uppercase letter at most k times. Return the length of the longest substring containing the same letter after performing at most k replacements.

Constraints

1 <= s.length <= 100,000
s consists of only uppercase English letters.
0 <= k <= s.length
Must run in O(n) time.

Examples

characterReplacement("ABAB", 2);
// => 4  (replace both A's or both B's → "AAAA" or "BBBB")

characterReplacement("AABABBA", 1);
// => 4  (replace one B at index 3 → "AAAAABA", substring "AAAA")

Solution

Reveal solution

function characterReplacement(s, k) {
  const count = {};
  let left = 0, maxFreq = 0, result = 0;

  for (let right = 0; right < s.length; right++) {
    count[s[right]] = (count[s[right]] || 0) + 1;
    maxFreq = Math.max(maxFreq, count[s[right]]);

    // window size - most frequent char count = replacements needed
    while ((right - left + 1) - maxFreq > k) {
      count[s[left]]--;
      left++;
    }

    result = Math.max(result, right - left + 1);
  }

  return result;
}

Sliding window: track the frequency of each character in the window. The number of replacements needed equals window size minus the most frequent character's count. Shrink from the left when replacements exceed k.

Resources

longest-repeating-substring-after-replacements.js

Longest Repeating Substring After Replacements

mediumcodingAlgorithmsSliding Window

Overview

Given a string s consisting of uppercase English letters and an integer k, you can choose any character and change it to any other uppercase letter at most k times. Return the length of the longest substring containing the same letter after performing at most k replacements.

Constraints

1 <= s.length <= 100,000
s consists of only uppercase English letters.
0 <= k <= s.length
Must run in O(n) time.

Examples

characterReplacement("ABAB", 2);
// => 4  (replace both A's or both B's → "AAAA" or "BBBB")

characterReplacement("AABABBA", 1);
// => 4  (replace one B at index 3 → "AAAAABA", substring "AAAA")

Solution

Reveal solution

function characterReplacement(s, k) {
  const count = {};
  let left = 0, maxFreq = 0, result = 0;

  for (let right = 0; right < s.length; right++) {
    count[s[right]] = (count[s[right]] || 0) + 1;
    maxFreq = Math.max(maxFreq, count[s[right]]);

    // window size - most frequent char count = replacements needed
    while ((right - left + 1) - maxFreq > k) {
      count[s[left]]--;
      left++;
    }

    result = Math.max(result, right - left + 1);
  }

  return result;
}

Sliding window: track the frequency of each character in the window. The number of replacements needed equals window size minus the most frequent character's count. Shrink from the left when replacements exceed k.

Resources

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