<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 = len(mat)
        n = len(mat[0])
        left = 0
        right = min(m,n)
        prefix_sum = [[0] * (n + 1) for _ in range(m + 1)]
    
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                prefix_sum[i][j] = mat[i-1][j-1] + prefix_sum[i-1][j] + prefix_sum[i][j-1] - prefix_sum[i-1][j-1]

        def helper(mid,threshold,mat):
            for i in range(mid, m + 1):
                for j in range(mid, n + 1):
                    # 使用前缀和计算 (i-mid, j-mid) 到 (i, j) 子矩阵的和
                    total_sum = prefix_sum[i][j] - prefix_sum[i-mid][j] - prefix_sum[i][j-mid] + prefix_sum[i-mid][j-mid]
                    if total_sum <= threshold:
                        return True
            return False
        
        while left < right:
            mid = (right + left + 1) // 2
            if helper(mid,threshold,mat):
                left = mid
            else:
                right = mid-1
        
        return left
22600