g0401_0500.s0424_longest_repeating_character_replacement.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0401_0500.s0424_longest_repeating_character_replacement;

// #Medium #String #Hash_Table #Sliding_Window #Level_1_Day_12_Sliding_Window/Two_Pointer
// #2022_07_16_Time_5_ms_(95.15%)_Space_43.4_MB_(48.28%)

/**
 * 424 - Longest Repeating Character Replacement\.
 *
 * Medium
 *
 * 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. 
 *
 * **Constraints:**
 *
 * *   1 <= s.length <= 10⁵
 * *   `s` consists of only uppercase English letters.
 * *   `0 <= k <= s.length
**/
public class Solution {
    public int characterReplacement(String s, int k) {
        int left = 0;
        int right = 0;
        int len = s.length();
        int[] count = new int[256];
        char[] sArr = s.toCharArray();
        int currMax = 0;
        int maxLen = 0;
        char curr;
        while (right < len) {
            curr = sArr[right];
            count[curr]++;
            currMax = Math.max(currMax, count[curr]);
            if (right - left + 1 <= currMax + k) {
                maxLen = Math.max(maxLen, right - left + 1);
            }
            while (right - left + 1 > currMax + k) {
                curr = sArr[left];
                count[curr]--;
                left++;
            }
            right++;
        }
        return maxLen;
    }
}