【leetcode】1208. Get Equal Substrings Within Budget

Topics are as follows:

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.

Outline of Solution: This problem package shell layer, after removing the outer topic is given a positive array of integers, the length of the longest period determined subarray, the subarray requirement is not greater than cost. Solving method is not difficult, referred per_cost [i] is a (s [i] - t [ i]) abs value, cost [i] is (per_cost [0: i]) sum values. For any index i, it is easy to find by a binary search method additionally a cost index j, such that the cost [i: j] <= cost.

code show as below:

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