<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>

from itertools import product

class Solution:
    def maxSideLength(self, mat: List[List[int]], threshold: int) -> int: 
        m, n = len(mat), len(mat[0])
        prefix = [[0]*(n+1) for _ in range(m+1)] 
        for i, j in product(range(m), range(n)):
            prefix[i+1][j+1] = prefix[i+1][j] + prefix[i][j+1] - prefix[i][j] + mat[i][j]
        
        def below(k): 
            for i, j in product(range(m+1-k), range(n+1-k)):
                if prefix[i+k][j+k] - prefix[i][j+k] - prefix[i+k][j] + prefix[i][j] <= threshold: 
                    return True
            return False 
                
        lo, hi = 1, min(m, n)
        while lo <= hi: 
            mid = (lo + hi)//2
            if below(mid): lo = mid + 1
            else: hi = mid - 1
            
        return hi
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