<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) {
        if(matrix.empty()){
            return 0;
        }
        int n = matrix.size();
        int c = matrix[0].size();
        stack<int>stc;
        vector<int>next_right(c,c);
        vector<int>next_left(c,-1);
        vector<int>arr(c,0);
        int max_sum = 0;

        for(int i = 0 ; i<n ; i++){
            
            for(int j = 0 ; j<c ;j++){
                if(matrix[i][j]=='1'){
                    arr[j]++;
                }
                else{
                    arr[j] = 0 ;
                }
            }
            while(!stc.empty()){
                stc.pop();
            }
            for(int j = 0 ; j<c ; j++){
                while(!stc.empty() && arr[stc.top()]>=arr[j]){
                    stc.pop();
                }
                next_left[j] = stc.empty() ? -1 : stc.top();
                stc.push(j);
             
            }
            while(!stc.empty()){
                stc.pop();
            }
            for(int j = c-1 ; j>=0 ; j--){
                while(!stc.empty() && arr[stc.top()]>=arr[j]){
                    stc.pop();
                }
                next_right[j] = stc.empty() ? c : stc.top();
                stc.push(j);
             
            }         
            for(int j = 0 ; j<c ; j++){
                int k = next_right[j]-next_left[j]-1;
                int area = k*arr[j];
                max_sum = max(max_sum , area);
            }          
        }
        return max_sum;
        
    }
};
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