1208

1208. Get Equal Substrings Within Budget (Medium)

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:

• 1 <= s.length, t.length <= 10^5
• 0 <= maxCost <= 10^6
• s and t only contain lower case English letters.

Companies:
Traveloka

Related Topics:
String, Binary Search, Sliding Window, Prefix Sum

Solution 1. Sliding Window

Check out "C++ Maximum Sliding Window Cheatsheet Template!".

Shrinkable Sliding Window:

// OJ: https://leetcode.com/problems/get-equal-substrings-within-budget/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int equalSubstring(string s, string t, int maxCost) {
        int i = 0, j = 0, N = s.size(), cost = 0, ans = 0;
        for (; j < N; ++j) {
            cost += abs(s[j] - t[j]);
            for (; cost > maxCost; ++i) cost -= abs(s[i] - t[i]);
            ans = max(ans, j - i + 1);
        }
        return ans;
    }
};

Non-shrinkable Sliding Window:

// OJ: https://leetcode.com/problems/get-equal-substrings-within-budget/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int equalSubstring(string s, string t, int maxCost) {
        int i = 0, j = 0, N = s.size(), cost = 0;
        for (; j < N; ++j) {
            cost += abs(s[j] - t[j]);
            if (cost > maxCost) {
                cost -= abs(s[i] - t[i]);
                ++i;
            }
        }
        return j - i;
    }
};

Discuss

https://leetcode.com/problems/get-equal-substrings-within-budget/discuss/1529234/C%2B%2B-Sliding-Window-(%2B-Cheat-Sheet)