424. Longest Repeating Character Replacement

package LeetCode_424

/**
 * 424. Longest Repeating Character Replacement
 * https://leetcode.com/problems/longest-repeating-character-replacement/description/
 *
 * Given a string s that consists of only uppercase English letters, you can perform at most k operations on that string.
In one operation, you can choose any character of the string and change it to any other uppercase English character.
Find the length of the longest sub-string containing all repeating letters you can get after performing the above operations.
Note:
Both the string's length and k will not exceed 10^4.

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.
 * */
class Solution {
    /*
    * solution: Sliding Window,
    * window's data:
    *   maxCount: the number of occurrence of the character that occur most in the window;
    *   window: the number of occurrence of each character in the window;
    * */
    fun characterReplacement(s: String, k: Int): Int {
        var i = 0
        var j = 0
        var maxCount = 0
        var maxLen = 0
        val map = IntArray(26)
        while (j < s.length) {
            val index = s[j] - 'A'//because only uppercase
            map[index]++
            /*
            for example:
            * 0 1 2 3
            * A A B A
            * j-i+1=4
            * so the maxCount is 3
            * */
            maxCount = Math.max(maxCount, map[index])
            /*
            * j-i+1-maxCount: represent the number of character in the window that are not the char that occur the most == the
            * number of replacements we need to do
            * */
            while (j - i + 1 - maxCount > k) {
                map[s[i] - 'A']--
                //move the left pointer to right
                i++
            }
            maxLen = Math.max(maxLen, j - i + 1)
            j++
        }
        //println(maxLen)
        return maxLen
    }
}