Get Equal Substrings Within Budget
<p>You are given two strings <code>s</code> and <code>t</code> of the same length and an integer <code>maxCost</code>.</p>

<p>You want to change <code>s</code> to <code>t</code>. Changing the <code>i<sup>th</sup></code> character of <code>s</code> to <code>i<sup>th</sup></code> character of <code>t</code> costs <code>|s[i] - t[i]|</code> (i.e., the absolute difference between the ASCII values of the characters).</p>

<p>Return <em>the maximum length of a substring of </em><code>s</code><em> that can be changed to be the same as the corresponding substring of </em><code>t</code><em> with a cost less than or equal to </em><code>maxCost</code>. If there is no substring from <code>s</code> that can be changed to its corresponding substring from <code>t</code>, return <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcd&quot;, t = &quot;bcdf&quot;, maxCost = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> &quot;abc&quot; of s can change to &quot;bcd&quot;.
That costs 3, so the maximum length is 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcd&quot;, t = &quot;cdef&quot;, maxCost = 3
<strong>Output:</strong> 1
<strong>Explanation:</strong> Each character in s costs 2 to change to character in t,  so the maximum length is 1.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcd&quot;, t = &quot;acde&quot;, maxCost = 0
<strong>Output:</strong> 1
<strong>Explanation:</strong> You cannot make any change, so the maximum length is 1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>t.length == s.length</code></li>
	<li><code>0 &lt;= maxCost &lt;= 10<sup>6</sup></code></li>
	<li><code>s</code> and <code>t</code> consist of only lowercase English letters.</li>
</ul>
String; Binary Search; Prefix Sum; Sliding Window
class Solution:
    def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
        def check(x):
            for i in range(n):
                j = i + mid - 1
                if j < n and f[j + 1] - f[i] <= maxCost:
                    return True
            return False

        n = len(s)
        f = list(accumulate((abs(ord(a) - ord(b)) for a, b in zip(s, t)), initial=0))
        l, r = 0, n
        while l < r:
            mid = (l + r + 1) >> 1
            if check(mid):
                l = mid
            else:
                r = mid - 1
        return l
