<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 squareSum(self, mat, x, y, r):
        return sum(sum(mat[x+i][y:y+r]) for i in range(r))

    def sq_contains(self,sq1, sq2):
        return (sq1[0] <= sq2[0]
            and sq1[1] <= sq2[1]
            and (sq2[0]-sq1[0]+sq2[2])<sq1[2]
            and (sq2[1]-sq1[1]+sq2[2])<sq1[2])


    def maxSideLength(self, mat: List[List[int]], threshold: int) -> int:
        m = len(mat)
        n = len(mat[0])

        biggest = 0

        too_big = set()

        for x in (range(m)):
            for y in reversed(range(n)):
                curr_too_big = False
                for r in range(biggest+1,1+min(m-x,n-y)):
                    for sq in too_big:
                        if self.sq_contains(sq,(x,y,r)):
                            curr_too_big = True
                    if self.squareSum(mat, x, y, r) <= threshold or curr_too_big:
                        biggest = r
                        too_big.add((x,y,r))
                    else:
                        break
        return biggest
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