[LeetCode] 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 <= 105
t.length == s.length
0 <= maxCost <= 106
s and t consist of only lowercase English letters.

尽可能使字符串相等。

给你两个长度相同的字符串，s 和 t。

将 s 中的第 i 个字符变到 t 中的第 i 个字符需要 |s[i] - t[i]| 的开销（开销可能为 0），也就是两个字符的 ASCII 码值的差的绝对值。

用于变更字符串的最大预算是 maxCost。在转化字符串时，总开销应当小于等于该预算，这也意味着字符串的转化可能是不完全的。

如果你可以将 s 的子字符串转化为它在 t 中对应的子字符串，则返回可以转化的最大长度。

如果 s 中没有子字符串可以转化成 t 中对应的子字符串，则返回 0。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/get-equal-substrings-within-budget
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思路

思路是滑动窗口，而且可以用模板。模板参见 76 题。可以消耗的开销一共是 maxCost，对于每个相同位置上的字符的 ASCII 码的差值，可以通过先移动 end 指针不断消耗 maxCost，直到 maxCost == 0。当 maxCost == 0 的时候，可以试图移动 start 指针，过程中记录 res 的最大值即可。

复杂度

时间O(n)
空间O(1)

代码

Java实现

class Solution {
    public int equalSubstring(String s, String t, int maxCost) {
        int len = s.length();
        int res = 0;
        int start = 0;
        int end = 0;
        while (end < len) {
            maxCost -= Math.abs(s.charAt(end) - t.charAt(end));
            end++;
            while (maxCost < 0) {
                maxCost += Math.abs(s.charAt(start) - t.charAt(start));
                start++;
            }
            res = Math.max(res, end - start);
        }
        return res;
    }
}