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
class Solution {
    public int numSubmat(int[][] mat) {
        int m = mat.length, n = mat[0].length;
        int[][] g = new int[m][n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (mat[i][j] == 1) {
                    g[i][j] = j == 0 ? 1 : 1 + g[i][j - 1];
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int col = 1 << 30;
                for (int k = i; k >= 0 && col > 0; --k) {
                    col = Math.min(col, g[k][j]);
                    ans += col;
                }
            }
        }
        return ans;
    }
}