<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 m=matrix.size();
        int n=matrix[0].size();
        vector<int> temp(n,0);
        int area=0;
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(matrix[i][j]=='0')
                 temp[j]=0;
                else
                 temp[j]+=1;

            }
            area=max(area,MAH(temp));
        }
        return area;
    }

    int MAH(vector<int> &heights){
        int n=heights.size();
        /////make NSL array
        stack<pair<int,int>> nsl;
        vector<int> nsl1;
        for(int i=0; i<n; i++){
            if(nsl.empty()){
                nsl1.push_back(-1);
                nsl.push({heights[i],i});
            }
            else if(nsl.top().first>=heights[i]){
                while(!nsl.empty() && nsl.top().first>=heights[i])
                 nsl.pop();
                if(!nsl.empty())
                    nsl1.push_back(nsl.top().second);
                else
                    nsl1.push_back(-1);
                nsl.push({heights[i],i});
            }
            else{
            nsl1.push_back(nsl.top().second);
            nsl.push({heights[i],i});
            }
        }
        for(auto num:nsl1)
         cout<<num<<endl;

        /////make NSR array
        stack<pair<int,int>> nsr;
        vector<int> nsr1;
        for(int i=n-1; i>=0; i--){
            if(nsr.empty()){
                nsr1.push_back(n);
                nsr.push({heights[i],i});
            }

            else if(nsr.top().first>=heights[i]){
                while(!nsr.empty() && nsr.top().first>=heights[i])
                 nsr.pop();

                if(!nsr.empty())
                    nsr1.push_back(nsr.top().second);
                else
                  nsr1.push_back(n);
                nsr.push({heights[i],i});
            }
            else{
            nsr1.push_back(nsr.top().second);
            nsr.push({heights[i],i});
            }
        }
        reverse(nsr1.begin(),nsr1.end());

        for(auto num:nsr1)
         cout<<num<<endl;



        int max_area = 0;
        for (int i = 0; i < n; ++i) {
            int width = nsr1[i] - nsl1[i] - 1;
            int area = heights[i] * width;
            max_area = max(max_area, area);
        }

        return max_area;
    
    }
};
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