<p>Given a <code>rows x cols</code>&nbsp;binary <code>matrix</code> filled with <code>0</code>&#39;s and <code>1</code>&#39;s, find the largest rectangle containing only <code>1</code>&#39;s and return <em>its area</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/14/maximal.jpg" style="width: 402px; height: 322px;" />
<pre>
<strong>Input:</strong> matrix = [[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The maximal rectangle is shown in the above picture.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[&quot;0&quot;]]
<strong>Output:</strong> 0
</pre>

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

<pre>
<strong>Input:</strong> matrix = [[&quot;1&quot;]]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>rows == matrix.length</code></li>
	<li><code>cols == matrix[i].length</code></li>
	<li><code>1 &lt;= row, cols &lt;= 200</code></li>
	<li><code>matrix[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class Solution {
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        int n = matrix.size(), m = matrix[0].size(),i,j;
        vector<vector<int>> max_to_left(n, vector<int> (m,0));
        int curr = 0;
        for(i = 0; i < n ; i ++  ){

            max_to_left[i][0] = (matrix[i][0] == '1' );
            curr = max_to_left[i][0];
            for(j = 1; j < m ; j ++ ){
                if (matrix[i][j] == '1'){
                    curr ++;
                    
                }
                else {
                    curr = 0;
                }
                max_to_left[i][j] = curr;
            }

        }
        int cur_max = 0,k;
        int ans = 0;
        for (i = 0; i < n ; i ++){
            for (j = 0; j < m ; j ++){
                cur_max = max_to_left[i][j];
                for(k = i; k >=0; k -- ){
                    cur_max = min(cur_max,max_to_left[k][j]);
                    ans = max(ans, cur_max * (i - k + 1));
                }
            }
        }
        return ans;

        
    }
};
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