<p>You are given a binary matrix <code>matrix</code> of size <code>m x n</code>, and you are allowed to rearrange the <strong>columns</strong> of the <code>matrix</code> in any order.</p>

<p>Return <em>the area of the largest submatrix within </em><code>matrix</code><em> where <strong>every</strong> element of the submatrix is </em><code>1</code><em> after reordering the columns optimally.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/29/screenshot-2020-12-30-at-40536-pm.png" style="width: 500px; height: 240px;" />
<pre>
<strong>Input:</strong> matrix = [[0,0,1],[1,1,1],[1,0,1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> You can rearrange the columns as shown above.
The largest submatrix of 1s, in bold, has an area of 4.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/29/screenshot-2020-12-30-at-40852-pm.png" style="width: 500px; height: 62px;" />
<pre>
<strong>Input:</strong> matrix = [[1,0,1,0,1]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> You can rearrange the columns as shown above.
The largest submatrix of 1s, in bold, has an area of 3.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[1,1,0],[1,0,1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Notice that you must rearrange entire columns, and there is no way to make a submatrix of 1s larger than an area of 2.
</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>matrix[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
private:
    void computeColumnArea(vector< vector<int> > &A, int col, vector< priority_queue<int> > &rows_height)
    {
        int n = A.size();
        int m = A[0].size();
        
        vector<int> column_area(n, 0);
        column_area[0] = A[0][col];
        if(column_area[0] != 0)
            rows_height[0].push(column_area[0]);
        for(int row = 1; row < n; row++)
        {
            if(A[row][col] != 0)
            {
                column_area[row] = column_area[row-1] + A[row][col];
                rows_height[row].push(column_area[row]);
            }
        }
        return;
    }

    int findMaxArea(vector< priority_queue<int> > &rows_height)
    {
        int n = rows_height.size();
        
        int max_area = 0;
        for(int i = 0; i < n; i++)
        {
            int width = 0;
            int height = 0;
            while(!rows_height[i].empty())
            {
                height = rows_height[i].top();
                rows_height[i].pop();
                width += 1;
                
                max_area = max(max_area, width * height);
            }
        }
        
        return max_area;
    }

public:
    int largestSubmatrix(vector<vector<int>>& A) {
        int n = A.size();
        int m = A[0].size();
        
        vector< priority_queue<int> > rows_height(n);
        for(int j = 0; j < m; j++)
            computeColumnArea(A, j, rows_height);
            
        int max_area = findMaxArea(rows_height);
        return max_area;
    }
};
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