<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) {
        int row = matrix.size();
        int col = matrix[0].size();

        // Step 1: Calculate column heights (sum of consecutive 1s)
        for (int j = 0; j < col; j++) {
            for (int i = 1; i < row; i++) {
                if (matrix[i][j] != 0) {
                    matrix[i][j] += matrix[i - 1][j];
                }
            }
        }

        int maxArea = 0;

        // Step 2: For each row, sort the heights and calculate the maximum area
        for (int i = 0; i < row; i++) {
            // Create a vector of column heights for this row
            vector<pair<int, int>> heightWithIndex;
            for (int j = 0; j < col; j++) {
                heightWithIndex.push_back({matrix[i][j], j});
            }

            // Sort by height in descending order
            sort(heightWithIndex.rbegin(), heightWithIndex.rend());

            // Calculate maximum area with this row as the base
            for (int j = 0; j < col; j++) {
                int height = heightWithIndex[j].first;
                int width = j + 1; // Number of columns considered so far
                maxArea = max(maxArea, height * width);
            }
        }

        return maxArea;
    }
};

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