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.
class Solution:
    def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
        start = 0
        end = 0
        cost = 0
        best = 0
        while end < len(s):
            while end < len(s) and cost + abs(ord(s[end]) - ord(t[end])) <= maxCost:
                cost += abs(ord(s[end]) - ord(t[end]))
                end += 1
            if start == end:
                start += 1
                end += 1
                continue
            best = max(best, end - start)
            cost -= abs(ord(s[start]) - ord(t[start]))
            start += 1
        return best
class Solution:
    def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
        start = 0
        end = 0
        cost = 0
        best = 0
        while end < len(s):
            while end < len(s) and cost + abs(ord(s[end]) - ord(t[end])) <= maxCost:
                cost += abs(ord(s[end]) - ord(t[end]))
                end += 1
            if start == end:
                start += 1
                end += 1
                continue
            best = max(best, end - start)
            cost -= abs(ord(s[start]) - ord(t[start]))
            start += 1
        return best