Maximum Side Length of a Square with Sum Less than or Equal to Threshold
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1292.Maximum%20Side%20Length%20of%20a%20Square%20with%20Sum%20Less%20than%20or%20Equal%20to%20Threshold/images/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>
Array; Binary Search; Matrix; Prefix Sum
func maxSideLength(mat [][]int, threshold int) int {
	m, n := len(mat), len(mat[0])
	s := make([][]int, m+1)
	for i := range s {
		s[i] = make([]int, n+1)
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + mat[i-1][j-1]
		}
	}
	check := func(k int) bool {
		for i := 0; i < m-k+1; i++ {
			for j := 0; j < n-k+1; j++ {
				if s[i+k][j+k]-s[i][j+k]-s[i+k][j]+s[i][j] <= threshold {
					return true
				}
			}
		}
		return false
	}
	l, r := 0, min(m, n)
	for l < r {
		mid := (l + r + 1) >> 1
		if check(mid) {
			l = mid
		} else {
			r = mid - 1
		}
	}
	return l
}