<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>

#pragma GCC optimize("O3")
class Solution {
public:  
    int largestSubmatrix(vector<vector<int>>& matrix) {
        int m=matrix.size(), n=matrix[0].size();
        int area=count(matrix[0].begin(), matrix[0].end(), 1);
        if (m==1) return area;
        if (n==1){
            #pragma unroll
            for(int i=1; i<m; i++){
                if (matrix[i][0]!=0)
                    matrix[i][0]+=matrix[i-1][0];
                area=max(area, matrix[i][0]);
            }
            return area;
        }
        
        #pragma unroll
        for(int i=1; i<m; i++){
            #pragma unroll
            for(int j=0;  j<n; j++){
                if (matrix[i][j]!=0)
                    matrix[i][j]+=matrix[i-1][j];
            }
       
            auto row=matrix[i];
            int minH=i+1, maxH=0;
            #pragma unroll
            for(int x: row){
                minH=min(minH, x);
                maxH=max(maxH, x);
            }
            vector<int> freq(maxH-minH+1, 0);
            #pragma unroll
            for(int x: row){
                freq[x-minH]++;
            }
            int acc=0;
            #pragma unroll
            for(int x=maxH-minH; acc<n; x--){
                if (freq[x]>0){
                    acc+=freq[x];
                    area=max(area, acc*(x+minH));
                }
            }  
        }
        return area;
    }
};
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