<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 res=INT_MIN;
    vector<vector<int>> mat(matrix.size(), vector<int>(matrix[0].size(), 0));
    int n=matrix.size(),m=matrix[0].size();
    for(int i=0;i<n;i++)
    {  
        for(int j=0;j<m;j++)
        {
         if(matrix[i][j]=='1')
         mat[i][j]=(i==0)?1:(1+mat[i-1][j]);  
        }
        res=max(res,max_histogram(mat[i]));
    }
    return res;    
    }
   int max_histogram(vector<int>& arr)
       { int n=arr.size();
        stack<int>st;
        //int i
        int max_till_now=-1;
        for(int i=0;i<n;i++)
        {
            while(!st.empty()&&arr[st.top()]>arr[i])
            {
               int a=arr[st.top()];
               st.pop();
               int b;
               if(st.empty()==true)
                b=-1;
               else
                b=st.top();
               int curr_area=a*(i-b-1);
               max_till_now=max(max_till_now,curr_area);
               //st.pop();
            }
            st.push(i);
        }
        while(!st.empty())
        {
               int a=arr[st.top()];
               st.pop();
               int b;
               if(st.empty()==true)
                b=-1;
               else
                b=st.top();
               int curr_area=a*((n)-b-1);
               max_till_now=max(max_till_now,curr_area);  
        }
        cout<<max_till_now<<" ";
        return max_till_now;
    } 
};
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