Skip to content

424. Longest Repeating Character Replacement

You are given a string s and an integer k. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most k times.

Return the length of the longest substring containing the same letter you can get after performing the above operations.

Example 1:

Input: s = "ABAB", k = 2
Output: 4
Explanation: Replace the two 'A's with two 'B's or vice versa.

Example 2:

Input: s = "AABABBA", k = 1
Output: 4
Explanation: Replace the one 'A' in the middle with 'B' and form "AABBBBA".
The substring "BBBB" has the longest repeating letters, which is 4.
There may exists other ways to achieve this answer too.

Constraints:

  • 1 <= s.length <= 105
  • s consists of only uppercase English letters.
  • 0 <= k <= s.length

Solution:

class Solution {
    public int characterReplacement(String s, int k) {
        int[] freq = new int[26];
        int slow = 0;
        int maxFreq = 0;
        int result = 0;

        for (int fast = 0; fast < s.length(); fast++){
            int cur = s.charAt(fast) - 'A';

            freq[cur] = freq[cur] + 1;

            maxFreq = Math.max(maxFreq, freq[cur]);


            // check valid
            Boolean isValid = false;
            if (fast - slow + 1 - maxFreq <=  k){
                // I can replace k 
                isValid = true;
            }

            // remove slow
            if (!isValid){
                int removeSlow = s.charAt(slow) - 'A';

                freq[removeSlow] = freq[removeSlow] - 1;

                slow++;
            }


            // update result;
            result = Math.max(fast - slow + 1, result);

        }
        return result;


    }
}
// TC: O(n)
// SC: O(26)

class Solution {
    public int characterReplacement(String s, int k) {
        int slow = 0;
        int[] freqMap = new int[26];
        int maxFreq = 0;
        int result = 0;

        for (int fast = 0; fast < s.length(); fast++){
            int cur = s.charAt(fast) - 'A';

            freqMap[cur]++;

            maxFreq = Math.max(maxFreq, freqMap[cur]);

            Boolean isValid = false;
            if (fast - slow + 1 - maxFreq <= k){
                isValid = true;
            }

            if (!isValid){
                int outgoingChar = s.charAt(slow) - 'A';

                freqMap[outgoingChar]--;

                slow++;
            }
            result = fast + 1 - slow;
        }

        return result;
    }
}

// TC: O(n)
// SC: O(26)