<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:
     vector<int> nsr(vector<int>& arr){
        int n=arr.size();
          vector<int> right;
        stack<pair<int,int>> s;
        
        for(int i=n-1;i>=0;i--){
            
            if(s.size()==0){
                right.push_back(n);
            }
            
            else if(s.size()>0 && s.top().first <arr[i]){
                right.push_back(s.top().second);
            }
            
            else if(s.size()>0 && s.top().first >=arr[i]){
                
            while(s.size()>0 && s.top().first >=arr[i]){
                s.pop();
            }
            
                if(s.size()==0){
                    right.push_back(n);
                }
                else{
                    right.push_back(s.top().second);
                }
            }
            s.push({arr[i],i});
        }
        
        reverse(right.begin(),right.end());
        
        return right;
        
        
    }
    
    //next smallest to left
    vector<int> nsl(vector<int>& arr){
        int n=arr.size();
          vector<int> left;
        stack<pair<int,int>> s;
        
        for(int i=0;i<n;i++){
            
            if(s.size()==0){
                left.push_back(-1);
            }
            
            else if(s.size()>0 && s.top().first <arr[i]){
                left.push_back(s.top().second);
            }
            
            else if(s.size()>0 && s.top().first >=arr[i]){
                
            while(s.size()>0 && s.top().first >=arr[i]){
                s.pop();
            }
            
                if(s.size()==0){
                    left.push_back(-1);
                }
                else{
                    left.push_back(s.top().second);
                }
            }
            s.push({arr[i],i});
        }
        
        
        
        return left;
        
        
    }
    int MAH(vector<int>& arr) {
        int n=arr.size();
        
        
        vector<int> right=nsr(arr);
        vector<int> left=nsl(arr);
        int ans=INT_MIN;
        
        
        for(int i=0;i<n;i++){
            int a=(right[i]-left[i]-1)*arr[i];
            ans=max(a,ans);            
        }
        
        return ans;
        
    }
    int maximalRectangle(vector<vector<char>>& matrix) {
        int row=matrix.size();
        int col=matrix[0].size();
        
        vector<int> v(col,0);
        int mx=0;
        
        
        
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
            
                if(matrix[i][j]=='0'){
                    v[j]=0;
                }
                else{
                    v[j]=v[j]+1;
                }
            
            }
            mx=max(mx,MAH(v));
            
        }
        
        return mx;
        
        
        
    }
};
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