<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:
        # prefix sum
        new_mat = [[0 for i in range(len(mat[0]) + 1)] for j in range(len(mat) + 1)]
        for r in range(len(mat)):
            for c in range(len(mat[0])):
                new_mat[r+1][c+1] = mat[r][c] + new_mat[r][c+1] + new_mat[r+1][c] - new_mat[r][c]
        def get_square_area(r, c, n):
            return new_mat[r+1][c+1] - new_mat[r-n+1][c+1] - new_mat[r+1][c-n+1] + new_mat[r-n+1][c-n+1]
        ret = 0
        for r in range(len(mat)):
            for c in range(len(mat[0])):
                low = ret
                high = min(r,c) + 1
                while low <= high:
                    mid = low + (high - low)//2
                    if get_square_area(r,c, mid) <= threshold:
                        ret = max(ret, mid) 
                        low = mid + 1
                    else:
                        high = mid - 1
        return ret

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