Get equal substrings within budget [Sliding Window]

Time: O(N); Space: O(1); medium

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.

Notes:

  • 1 <= len(s), len(t) <= 10^5

  • 0 <= maxCost <= 10^6

  • s and t only contain lower case English letters.

Hint:

Calculate the differences between a[i] and b[i].

[3]:

class Solution1(object):
    def equalSubstring(self, s, t, maxCost):
        """
        :type s: str
        :type t: str
        :type maxCost: int
        :rtype: int
        """
        left = 0
        for right in range(len(s)):
            maxCost -= abs(ord(s[right]) - ord(t[right]))
            if maxCost < 0:
                maxCost += abs(ord(s[left]) - ord(t[left]))
                left += 1
        return (right + 1) - left

[2]:

sol = Solution1()
s = "abcd"
t = "bcdf"
maxCost = 3
assert sol.equalSubstring(s, t, maxCost) == 3

s = "abcd"
t = "cdef"
maxCost = 3
assert sol.equalSubstring(s, t, maxCost) == 1

s = "abcd"
t = "acde"
maxCost = 0
assert sol.equalSubstring(s, t, maxCost) == 1