24. 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.

Solution: (Sliding Window)

Calculate the cost of changes for each index and find max subarray using sliding window

class Solution
{
public:
    int equalSubstring(string s, string t, int maxCost)
    {

        int n = s.length();
        vector<int> v;
        
        for (int i = 0; i < n; i++)
        {
            int x = abs(s[i] - t[i]);
            v.push_back(x);
        }

        int i = 0;
        int sum = 0;
        int idx = 0;
        int maxLen = 0;

        while (i < v.size())
        {
            if (sum + v[i] <= maxCost)
            {
                sum += v[i];
                i++;
            }
            else
            {
                int len = i - idx;
                maxLen = max(maxLen, len);
                sum = sum - v[idx];
                idx++;
            }
        }

        if (sum <= maxCost)
        {
            int l = i - idx;
            maxLen = max(maxLen, l);
        }

        return maxLen;
    }
};

Time Complexity: O(n)