human
I found an interesting problem on https://leetcode.com/problems/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 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.

I tried it in Python, but could not do it. Can you solve it?

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Can you provide a the Python code you requested?
assistant
#Here is the Python code you requested.

class Solution:
    def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
        currcost=0;i=0;j=0
        if(len(s)==0):
            return 0
        if(len(s)==1):
            if(abs(ord[s[0]])-ord(t[0])>=maxcost):
                return 1
            else:
                return 0
        currcost=0
        ans=0
        state=False
        while(i<len(s) and j<len(s)):
            if(currcost+abs(ord(s[j])-ord(t[j]))<=maxCost):
                currcost+=abs(ord(s[j])-ord(t[j]))
                j+=1
                state=True
            else:
                ans=max(ans,j-i)
                currcost-=abs(ord(s[i])-ord(t[i]))
                i+=1
                state=False
                if(i>=j):
                    j=i
                    currcost=0
        if state:
            ans=max(ans,j-i)
        return ans