<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://assets.leetcode.com/uploads/2019/12/05/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>

class Solution:
    def maxSideLength(self, mat: List[List[int]], threshold: int) -> int:
        m, n = len(mat), len(mat[0])
        prefix_sum = [[0] * (n + 1) for _ in range(m + 1)]
        
        for i in range(m):
            for j in range(n):
                prefix_sum[i + 1][j + 1] = mat[i][j] + prefix_sum[i + 1][j] + prefix_sum[i][j + 1] - prefix_sum[i][j]
        
        def get_sum(x1, y1, x2, y2):
            return (prefix_sum[x2 + 1][y2 + 1] - prefix_sum[x1][y2 + 1] - prefix_sum[x2 + 1][y1] + prefix_sum[x1][y1])

        low, high = 0, min(m, n)
        
        while low < high:
            mid = (low + high + 1) // 2
            found = False
            for i in range(m - mid + 1):
                for j in range(n - mid + 1):
                    if get_sum(i, j, i + mid - 1, j + mid - 1) <= threshold:
                        found = True
                        break
                if found:
                    break
            if found:
                low = mid
            else:
                high = mid - 1
        return low
22100