Maximum Side Length of a Square with Sum Less than or Equal to Threshold
<p>Given a <code>m x n</code> matrix <code>mat</code> and an integer <code>threshold</code>, return <em>the maximum side-length of a square with a sum less than or equal to </em><code>threshold</code><em> or return </em><code>0</code><em> if there is no such square</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1292.Maximum%20Side%20Length%20of%20a%20Square%20with%20Sum%20Less%20than%20or%20Equal%20to%20Threshold/images/e1.png" style="width: 335px; height: 186px;" />
<pre>
<strong>Input:</strong> mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,3,2]], threshold = 4
<strong>Output:</strong> 2
<strong>Explanation:</strong> The maximum side length of square with sum less than 4 is 2 as shown.
</pre>

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

<pre>
<strong>Input:</strong> mat = [[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2]], threshold = 1
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 300</code></li>
	<li><code>0 &lt;= mat[i][j] &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= threshold &lt;= 10<sup>5</sup></code></li>
</ul>
Array; Binary Search; Matrix; Prefix Sum
class Solution:
    def maxSideLength(self, mat: List[List[int]], threshold: int) -> int:
        def check(k: int) -> bool:
            for i in range(m - k + 1):
                for j in range(n - k + 1):
                    v = s[i + k][j + k] - s[i][j + k] - s[i + k][j] + s[i][j]
                    if v <= threshold:
                        return True
            return False

        m, n = len(mat), len(mat[0])
        s = [[0] * (n + 1) for _ in range(m + 1)]
        for i, row in enumerate(mat, 1):
            for j, x in enumerate(row, 1):
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + x
        l, r = 0, min(m, n)
        while l < r:
            mid = (l + r + 1) >> 1
            if check(mid):
                l = mid
            else:
                r = mid - 1
        return l
