1208. 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 froms that can be changed to its corresponding substring from t, return 0.

Example 1:

Input:s = "abcd", t = "bcdf", maxCost = 3Output:3Explanation: "abc" of s can change to "bcd". That costs 3, so the maximum length is 3.

Example 2:

Input:s = "abcd", t = "cdef", maxCost = 3Output:1Explanation: Each character in s costs 2 to change to charactor int, so the maximum length is 1.

Example 3:

Input:s = "abcd", t = "acde", maxCost = 0Output:1Explanation: 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 andt only contain lower case English letters.

思路：

用队列记录连续子数组，当队列中的数据的和大于最大值时，反复出队，直到队列中的数值小于等于最大值或者队列为空。

细节1:出队时，判断队列为空。

细节2:若队列为空，push当前值数据时，需要判断位置的差是否小于等于最大值。示例代码如下：

intequalSubstring(strings,stringt,intmaxCost) {

    vector<int> diffval(s.size() +1,0);

    for(inti =1; i <= s.size(); i++) {

        diffval[i] =abs(s[i-1]-t[i-1]);

    }

    queue<int>dp;

    intcur =0;

    intresult =0;

    for(inti =1; i <= s.size(); i++) {

        intdiff = diffval[i];

        if(diff + cur > maxCost) {

            while(diff + cur > maxCost && dp.size()) {

                    intval = dp.front();

                    cur -= diffval[val];

                    dp.pop();

            }

            if(diff + cur <= maxCost) {

                dp.push(i);

                cur += diff;

            }

        }else{

            dp.push(i);

            cur += diff;

        }

        result =max(result, (int)dp.size());

    }

    return result;

}