<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 {
    private:
    vector<int> nse_finder(vector<int>arr){
        vector<int>nse(arr.size());
        stack<int>st;
        for(int i=arr.size()-1;i>=0;i--){
            while(!st.empty() && arr[i] <= arr[st.top()]){
                st.pop();
            }
            if(st.empty()){
                nse[i]=arr.size();
            }
            else{
                nse[i]=st.top();
            }
            st.push(i);
        }
        return nse;
    }
    vector<int> pse_finder(vector<int>arr){
        vector<int>pse(arr.size());
        stack<int>st;
        for(int i=0;i<arr.size();i++){
            while(!st.empty() && arr[i] <= arr[st.top()]){
                st.pop();
            }
            if(st.empty()){
                pse[i]=-1;
            }
            else{
                pse[i]=st.top();
            }
            st.push(i);
        }
        return pse;
    }
    int LargestRectangleinHistogram(vector<int>arr){
        vector<int>nse=nse_finder(arr);
        vector<int>pse=pse_finder(arr);
        int largest=INT_MIN;
        for(int i=0;i<arr.size();i++){
            largest=max(largest,arr[i]*(nse[i]-pse[i]-1));
        }
        return largest;
    }
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
     vector<vector<int>> arr(matrix.size(), vector<int>(matrix[0].size(), 0));

    for(int i = 0; i < matrix.size(); i++) {
        for(int j = 0; j < matrix[0].size(); j++) {
            if(i == 0) {
                arr[i][j] = matrix[i][j] - '0';
            } else {
                if(matrix[i][j] == '1') {
                    arr[i][j] = arr[i-1][j] + 1; 
                } else {
                    arr[i][j] = 0; 
                }
            }
        }
    }
     int maxArea=INT_MIN;
    for(int i = 0; i < arr.size(); i++) {
        int Area=LargestRectangleinHistogram(arr[i]);
       maxArea =max(maxArea,Area);
    }
    return maxArea;
    }
};
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