<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 largestRectangleArea(vector<int>& arr) {
            int n = arr.size();
            int prevsmaller[n];
            stack<int> st;

            for (int i = 0 ; i < n ; i++) {
                while(st.size()!=0 and arr[st.top()]>=arr[i])
                    st.pop();
                if(st.size()==0){
                    prevsmaller[i]=0;
                }else{
                    prevsmaller[i]=st.top()+1;
                }
                st.push(i);
            }
            while (!st.empty())
                st.pop();

            int rigthsmaller[n];
            for (int i = n - 1 ; i >= 0 ; i--) {
                while(st.size()!=0 and arr[st.top()]>=arr[i])
                    st.pop();
                if(st.size()==0){
                    rigthsmaller[i]=n-1;
                }else{
                    rigthsmaller[i]=st.top()-1;
                }
                st.push(i);
            }
            int maxi = -100000;
            for (int i = 0 ; i < n ; i++) {
                maxi = max(maxi, ((rigthsmaller[i] - prevsmaller[i] + 1 ) * arr[i] ));
            }
            return maxi;
    }
    int maximalRectangle(vector<vector<char>>& arr) {
        int n = arr.size(),m = arr[0].size();
        vector<vector<int>> dp(n,vector<int>(m,0));
        int area = 0 ;
        for(int i = 0 ; i< n; i++){
            for(int j =0 ; j < m ; j ++){
                if(i==0){
                    dp[i][j]=arr[i][j]-'0';
                }else if(i!=0 and arr[i][j]=='1'){
                    dp[i][j]+=dp[i-1][j]+1;
                }
            }
            area = max(area, largestRectangleArea(dp[i]));
        }
        return area ;
    }
};
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