Count Submatrices With All Ones
<p>Given an <code>m x n</code> binary matrix <code>mat</code>, <em>return the number of <strong>submatrices</strong> that have all ones</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/1500-1599/1504.Count%20Submatrices%20With%20All%20Ones/images/ones1-grid.jpg" style="width: 244px; height: 245px;" />
<pre>
<strong>Input:</strong> mat = [[1,0,1],[1,1,0],[1,1,0]]
<strong>Output:</strong> 13
<strong>Explanation:</strong> 
There are 6 rectangles of side 1x1.
There are 2 rectangles of side 1x2.
There are 3 rectangles of side 2x1.
There is 1 rectangle of side 2x2. 
There is 1 rectangle of side 3x1.
Total number of rectangles = 6 + 2 + 3 + 1 + 1 = 13.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1504.Count%20Submatrices%20With%20All%20Ones/images/ones2-grid.jpg" style="width: 324px; height: 245px;" />
<pre>
<strong>Input:</strong> mat = [[0,1,1,0],[0,1,1,1],[1,1,1,0]]
<strong>Output:</strong> 24
<strong>Explanation:</strong> 
There are 8 rectangles of side 1x1.
There are 5 rectangles of side 1x2.
There are 2 rectangles of side 1x3. 
There are 4 rectangles of side 2x1.
There are 2 rectangles of side 2x2. 
There are 2 rectangles of side 3x1. 
There is 1 rectangle of side 3x2. 
Total number of rectangles = 8 + 5 + 2 + 4 + 2 + 2 + 1 = 24.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= m, n &lt;= 150</code></li>
	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>
Stack; Array; Dynamic Programming; Matrix; Monotonic Stack
impl Solution {
    pub fn num_submat(mat: Vec<Vec<i32>>) -> i32 {
        let m = mat.len();
        let n = mat[0].len();
        let mut g = vec![vec![0; n]; m];

        for i in 0..m {
            for j in 0..n {
                if mat[i][j] == 1 {
                    if j == 0 {
                        g[i][j] = 1;
                    } else {
                        g[i][j] = 1 + g[i][j - 1];
                    }
                }
            }
        }

        let mut ans = 0;
        for i in 0..m {
            for j in 0..n {
                let mut col = i32::MAX;
                let mut k = i as i32;
                while k >= 0 && col > 0 {
                    col = col.min(g[k as usize][j]);
                    ans += col;
                    k -= 1;
                }
            }
        }
        ans
    }
}
