<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:
        lookup = [[0 for _ in range(len(mat[0]) + 1)] for _ in range(len(mat) + 1)]

        curr = 0
        for i in range(len(mat)):
            curr2 = 0
            for j in range(len(mat[0])):
                curr2 += mat[i][j]
                lookup[i + 1][j + 1] = curr2 + lookup[i][j+1]
        

        out = 0
        for i in range(len(lookup)):
            for j in range(len(lookup[0])):
                lim = min(len(lookup) - i, len(lookup[0]) - j)
                for idx in range(1, lim):
                    curr = lookup[i+idx][j+idx] - lookup[i][j+idx] - \
                    lookup[i+idx][j] + lookup[i][j]
                    if curr <= threshold:
                        out = max(idx, out)

        return out
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