<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://assets.leetcode.com/uploads/2021/10/27/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://assets.leetcode.com/uploads/2021/10/27/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>

class Solution {
public:
    int numSubmat(vector<vector<int>>& mat) 
    {
        int m = mat.size();
    int n = mat[0].size();
    vector<vector<int>> height(m, vector<int>(n, 0));
    
    for (int i = 0; i < m; ++i) 
    {
     for (int j = 0; j < n; ++j) 
     {
            if (mat[i][j] == 1)
             {
                height[i][j] = (i ==0) ? 1 : height[i-1][j] + 1;
            }
        }
    }
    int count = 0;
    for (int i = 0; i <m; ++i) 
    {
        vector<int> dp(n, 0);

        for (int j = 0; j <n; ++j)
         {
            int min_height = height[i][j];
            for (int k = j; k >=0 && min_height > 0; --k) 
            {
                min_height = min(min_height, height[i][k]);
                dp[j] += min_height;
            }
            count += dp[j];
        }
    }
    return count;
    }
};
18900