<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:
    vector<int> pse(vector<int> arr){
        int n=arr.size();
        stack<int> st;
        vector<int> v;
        for(int i=0;i<n;i++){
            while(!st.empty() && arr[st.top()]>=arr[i]){
                st.pop();
            }
            if(st.empty()){
                v.push_back(-1);
            }
            else{
                v.push_back(st.top());
            }
            st.push(i);
        }
        return v;
    }
    vector<int> nse(vector<int> arr){
        int n=arr.size();
        int z=n;
        stack<int> st;
        vector<int> v(z);
        n=n-1;
        while(n>=0){
            while(!st.empty() && arr[st.top()]>=arr[n]){
                st.pop();
            }
            if(st.empty()){
                v[n]=z;
            }
            else{
                v[n]=st.top();
            }
            st.push(n);
            n--;
        }
        return v;
    }
    int histogram(vector<int> arr){
        vector<int> p=pse(arr);
        vector<int> n=nse(arr);
        int maxarea=0;
        for(int i=0;i<arr.size();i++){
            int psee=p[i];
            int nsee=n[i];
            maxarea=max(maxarea,(arr[i]*(nsee-psee-1)));
        }
        return maxarea;
    }
    int maximalRectangle(vector<vector<char>>& matrix) {
        if (matrix.empty()) return 0;

        int n = matrix.size();
        int m = matrix[0].size();
        vector<int> heights(m, 0);  // This will store the heights of histograms.

        int maxArea = 0;

        for (int i = 0; i < n; i++) {  // Iterate over each row.
            for (int j = 0; j < m; j++) {  // Update the histogram heights.
                if (matrix[i][j] == '1') {
                    heights[j] += 1;
                } else {
                    heights[j] = 0;
                }
            }
            maxArea = max(maxArea, histogram(heights));  // Calculate the max area for the current histogram.
        }

        return maxArea;
    }
};
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