LeetCode 1156: Swap For Longest Repeated Character Substring Solution

Master LeetCode problem 1156 (Swap For Longest Repeated Character Substring), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1156. Swap For Longest Repeated Character Substring

Medium

Hash Table String Sliding Window

Problem Explanation

Explanation

To solve this problem, we can iterate through the input string and keep track of the longest substring with repeated characters that can be obtained by swapping two characters. We can use a sliding window approach to find the longest substring with repeated characters.

Initialize variables:
- maxLen to store the maximum length of the repeated character substring.
- maxFreq to store the maximum frequency of any character in the current window.
- count to store the total count of characters in the current window.
- window to store the frequency of characters in the current window.
- start to mark the start of the current window.
Iterate through the string:
- For each character in the string:
  - Increment the frequency of the character in the window.
  - Update maxFreq with the maximum frequency encountered so far.
  - Update count with the total count of characters in the current window.
  - While the condition count - maxFreq > k is met:
    - Decrement the frequency of the character at index start.
    - Increment start to move the window.
  - Update maxLen with the maximum of maxLen and count.
Return the maxLen.

Solution Code

class Solution {
    public int characterReplacement(String s, int k) {
        int[] window = new int[26];
        int maxLen = 0, maxFreq = 0, start = 0, count = 0;

        for (int end = 0; end < s.length(); end++) {
            maxFreq = Math.max(maxFreq, ++window[s.charAt(end) - 'A']);
            count++;

            while (count - maxFreq > k) {
                window[s.charAt(start) - 'A']--;
                start++;
            }

            maxLen = Math.max(maxLen, count);
        }

        return maxLen;
    }
}

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