<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 solve(vector<int> &dp){
        int n = dp.size();
        stack<int> nse;
        stack<int> pse;

        vector<int> l(n, -1);
        vector<int> r(n, n);

        for(int i=0;i<n;i++){
            while(!pse.empty() && dp[pse.top()]>=dp[i]){
                pse.pop();
            }
            if(!pse.empty()) l[i] = pse.top();
            pse.push(i);
        }

        for(int i=n-1;i>=0;i--){
            while(!nse.empty() && dp[nse.top()]>=dp[i]){
                nse.pop();
            }
            if(!nse.empty()) r[i] = nse.top();
            nse.push(i);
        }

        int maxi = 0;
        int area = 0;
        for(int i=0;i<n;i++){
            area = dp[i]*(r[i]-l[i]-1);
            maxi = max(area, maxi);
        }

        return maxi;
    }
    int maximalRectangle(vector<vector<char>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        vector<vector<int>> dp(m, vector<int>(n));

        for(int j=0;j<n;j++){
            dp[0][j] = (matrix[0][j]=='1') ? 1 : 0;
        }
        int maxi = solve(dp[0]);

        for(int i=1;i<m;i++){
            for(int j=0;j<n;j++){
                dp[i][j] = (matrix[i][j]=='1') ? 1 + dp[i-1][j] : 0;
            }
            int area = solve(dp[i]);
            maxi = max(maxi, area);
        }

        return maxi;
    }
};
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