leetcode1208. Get Equal Substrings Within Budget

Posted 2020-11-15 seyjs

题目如下：

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.

解题思路：本题包了一层壳，去掉外表后题目是给定一个正整数组成的数组，求出最长的一段子数组的长度，要求子数组的和不大于cost。解题方法也不难，记per_cost[i]为abs(s[i] - t[i])的值，cost[i]为sum(per_cost[0:i])的值。对于任意一个下标i，很容易通过二分查找的方法找出cost中另外一个下标j，使得cost[i:j] <= cost。

代码如下：

class Solution(object):
    def equalSubstring(self, s, t, maxCost):
        """
        :type s: str
        :type t: str
        :type maxCost: int
        :rtype: int
        """
        cost = []
        amount = 0
        per_cost = []
        for cs,ct in zip(s,t):
            amount += abs(ord(cs) - ord(ct))
            cost.append(amount)
            per_cost.append(abs(ord(cs) - ord(ct)))
        #cost.sort()
        #print cost
        #print per_cost
        import bisect
        res = -float(‘inf‘)
        for i in range(len(cost)):
            inx = bisect.bisect_right(cost,cost[i] + maxCost - per_cost[i])
            res = max(res,inx - i)
        return res