<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 ans=0;
 void find(vector<int> v,int n)
 {
      vector<int> left(n,-1);
       vector<int> right(n,n);
       stack<int> st;
       st.push(0);
       for(int i=1;i<n;i++)
       {
          while(!st.empty() && v[st.top()]>=v[i])
          {
            st.pop();
          }
          if(!st.empty())
          {
            left[i]=st.top();
          }
          st.push(i);
       }
       while(!st.empty())
       {
        st.pop();
       }
       st.push(n-1);
       for(int i=n-2;i>=0;i--)
       {
          while(!st.empty() && v[st.top()]>=v[i])
          {
            st.pop();
          }
          if(!st.empty())
          {
            right[i]=st.top();
          }
          st.push(i);
       }
       for(int i=0;i<n;i++)
       {
         ans=max(ans,(right[i]-left[i]-1)*v[i]);
       }


 }

    int maximalRectangle(vector<vector<char>>& mat) 
    {
        int n=mat.size();
        int m=mat[0].size();
        vector<vector<int>> v(n,vector<int>(m,0));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(mat[i][j]=='1')
                {
                    if(i>0)
                    {
                        v[i][j]=v[i-1][j]+1;
                    }
                    else
                    {
                        v[i][j]=1;
                    }
                }
            }
        } 
        for(int i=0;i<n;i++)
        {
            find(v[i],m);
        } 
        return ans;  
    }
};
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