<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) {
        vector<vector<int>> sums(matrix.size(), vector<int>(matrix[0].size(), 0));
        for (int j = 0; j < matrix[0].size(); j++) {
            sums[0][j] = matrix[0][j];
        }
        for (int i = 1; i < matrix.size(); i++) {
            for (int j = 0; j < matrix[i].size(); j++) {
                sums[i][j] = matrix[i][j] == 0 ? 0 : sums[i-1][j] + 1;
            }
        }
        int max_rect = 0;
        for (int i = 0; i < sums.size(); i++) {
            sort(sums[i].begin(), sums[i].end());
            for (int j = 0; j < sums[i].size(); j++) {
                int curr_rect = sums[i][j] * (sums[i].size() - j);
                max_rect = max(curr_rect, max_rect);
            }
        }
        return max_rect;
    }
};
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