Get Equal Substrings Within Budget

You are given two strings s and t of the same length. 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]| that is, the absolute difference between the ASCII values of the characters.

You are also given an integer maxCost.

Return the maximum length of a substring of s that can be changed to be the same as the corresponding substring of twith 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 charactor in t, so the maximum length is 1.

Example 3:

Input: s = "abcd", t = "acde", maxCost = 0
Output: 1
Explanation: You can't make any change, so the maximum length is 1.

 

Constraints:


Solution:

class Solution {
    public int equalSubstring(String s, String t, int maxCost) {
        int n = s.length();
        int max = 0;
        int curLen = 0;
        int left = 0;
        int right = 0;
        int curCost = maxCost;
        while (right < n) {
            char a = s.charAt(right);
            char b = t.charAt(right);
            if (a == b) {
                curLen ++;
            } else {
                while (left < right && curCost - Math.abs(a - b) < 0) {
                    curCost += Math.abs(s.charAt(left) - t.charAt(left));
                    left ++;
                    curLen --;
                }
                if (curCost - Math.abs(a - b) >= 0) {
                    // System.out.println(a + ", " + b + ", " + Math.abs(a - b));
                    curCost -= Math.abs(a - b);
                    curLen ++;
                } else {
                    left ++;
                    curLen = 0;
                    curCost = maxCost;
                }
            }
            max = Math.max(curLen, max);
            right ++;
        }
        return max;
    }
}