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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1201_1300.s1208_get_equal_substrings_within_budget;

// #Medium #String #Binary_Search #Prefix_Sum #Sliding_Window
// #2022_03_09_Time_7_ms_(73.41%)_Space_43.3_MB_(45.62%)

/**
 * 1208 - Get Equal Substrings Within Budget\.
 *
 * Medium
 *
 * You are given two strings `s` and `t` of the same length and an integer `maxCost`.
 *
 * You want to change `s` to `t`. Changing the i^th character of `s` to i^th character of `t` costs `|s[i] - t[i]|` (i.e., the absolute difference between the ASCII values of the characters).
 *
 * Return _the maximum length of a substring of_ `s` _that can be changed to be the same as the corresponding substring of_ `t` _with a cost less than or equal to_ `maxCost`. If there is no substring from `s` that can be changed to its corresponding substring from `t`, return `0`.
 *
 * **Example 1:**
 *
 * **Input:** s = "abcd", t = "bcdf", maxCost = 3
 *
 * **Output:** 3
 *
 * **Explanation:** "abc" of s can change to "bcd". That costs 3, so the maximum length is 3.
 *
 * **Example 2:**
 *
 * **Input:** s = "abcd", t = "cdef", maxCost = 3
 *
 * **Output:** 1
 *
 * **Explanation:** Each character in s costs 2 to change to character in t, so the maximum length is 1.
 *
 * **Example 3:**
 *
 * **Input:** s = "abcd", t = "acde", maxCost = 0
 *
 * **Output:** 1
 *
 * **Explanation:** You cannot make any change, so the maximum length is 1.
 *
 * **Constraints:**
 *
 * *   1 <= s.length <= 10⁵
 * *   `t.length == s.length`
 * *   0 <= maxCost <= 10⁶
 * *   `s` and `t` consist of only lowercase English letters.
**/
public class Solution {
    public int equalSubstring(String s, String t, int maxCost) {
        int start = 0;
        int end = 0;
        int currCost = 0;
        int maxLength = Integer.MIN_VALUE;
        while (end < s.length()) {
            currCost += Math.abs(s.charAt(end) - t.charAt(end));
            while (currCost > maxCost) {
                currCost -= Math.abs(s.charAt(start) - t.charAt(start));
                start++;
            }
            if (end - start + 1 > maxLength) {
                maxLength = Math.max(end - start + 1, maxLength);
            }
            end++;
        }
        return maxLength;
    }
}