<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>

#include<bits/stdc++.h>
class Solution {
public:

    int largesthist(vector<int> &A){
        stack<int> st;
        int maxarea=0;
        for(int i=0;i<A.size();i++){
            while(!st.empty() && A[st.top()]>A[i]){
                int ele= st.top();
                st.pop();
                int nse= i;
                int pse= st.empty()?-1 : st.top();
                maxarea= max(maxarea, (nse-pse-1)*A[ele]);
            }

            st.push(i);
        }

        while(!st.empty()){
                int ele= st.top();
                st.pop();
                int nse= A.size();
                int pse= st.empty()?-1 : st.top();
                maxarea= max(maxarea, (nse-pse-1)*A[ele]);
        }
        return maxarea;
    }


    int maximalRectangle(vector<vector<char>>& matrix) {
        int n=matrix.size();
        int m= matrix[0].size();
        vector<vector<int>> psum(n, vector<int>(m));
        
        for(int j=0;j<m;j++){
            int sum=0;
            for(int i=0;i<n;i++){
                if (matrix[i][j] == '1'){
                    sum+=1;
                }
                else {
                    sum=0;
                }
                psum[i][j]= sum;
            }
        }

        //now finding larget area, row wise
        int maxi=0;
        for(int i=0;i<n;i++){
            maxi= max(maxi, largesthist(psum[i]));
        }

        return maxi;
    }
};
18696