<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 {
public:
    int largestSubmatrix(vector<vector<int>>& matrix) {
        // for each i j where el==1
        // count how many 1 it has from j to j-1;
        // then iterate over each row;
        // for each row, take a pq and push the count of 1s for each el;
        // then iterate over pq, and increment count(count= noOfRows) for each pq.pop() where subMatrixmatrixArea= count*pq.front();
        // now ans=max(for each row max(max El of that row, max subMatrixmatrixArea of that row))
        for (int j = 0; j < matrix[0].size(); j++) {
            for (int i = matrix.size() - 2; i >= 0; i--) {
                if (matrix[i][j] == 1) {
                    matrix[i][j] += matrix[i + 1][j];
                }
            }
        }
        int ans = INT_MIN;
        for (int i = 0; i < matrix.size(); i++) {
            int rowAns = *max_element(matrix[i].begin(), matrix[i].end());
            priority_queue<int> pq;
            for (int j = 0; j < matrix[0].size(); j++) {
                pq.push(matrix[i][j]);
            }
            int row = 1;
            while (!pq.empty()) {
                int subMatrixArea = row * pq.top();
                pq.pop();
                rowAns = max(rowAns, subMatrixArea);
                row++;
            }
            ans = max(ans, rowAns);
        }
        return ans;
    }
};
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