<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 const m = mat.size();
        int const n = mat[0].size();
        int count = 0;
        
          // number of one to the right of (i,j) along i till first zero  
        vector<vector<int>> prefix(m, vector<int>(n,0));
            
        for(int i=0; i<m; ++i){
            for(int j=n-1; j>=0; --j){
                if(mat[i][j] == 1){
                    if(j!=n-1){
                        prefix[i][j]+=prefix[i][j+1];
                    }
                    prefix[i][j] += mat[i][j];
                }
            }
        }
        
        for(int j=0; j<n; ++j){
            int i = m-1;
            
            stack<pair<int, int>> st;
            int sum =0;
            while(i>=0){
                int c = 0;
                while(!st.empty() && st.top().first > prefix[i][j]){
                    sum -= (st.top().second + 1) * (st.top().first - prefix[i][j]);
                    c+= st.top().second + 1;
                    st.pop();
                }
                sum += prefix[i][j];
                count +=sum;
                
                st.push({prefix[i][j], c});
                --i;
            }
        }
        
        return count;
        
    }
    

};
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