1208. Get Equal Substrings Within Budget

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 ith character of s to ith 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:

• $1<=s.length<=10^5$
• t.length == s.length
• $0<=maxCost<=10^6$
• s and t consist of only lowercase English letters.

From: LeetCode
Link: 1208. Get Equal Substrings Within Budget

Solution:

Ideas:

Use a sliding window: expand the right pointer while adding character costs, and shrink the left pointer whenever the total cost exceeds maxCost, tracking the maximum valid window length.

Code:

#include <stdlib.h>

int equalSubstring(char* s, char* t, int maxCost) {
    int n = 0;
    while (s[n] != '\0') n++;

    int l = 0;
    int cost = 0;
    int ans = 0;

    for (int r = 0; r < n; r++) {
        cost += abs(s[r] - t[r]);

        while (cost > maxCost) {
            cost -= abs(s[l] - t[l]);
            l++;
        }

        int len = r - l + 1;
        if (len > ans) ans = len;
    }

    return ans;
}