<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 <iostream>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) 
    {
        stack<pair<int, int>> st;  // Stack to store pairs of (height, index)
        int maxArea = 0;
        int n = heights.size();
        int currentHeight=0;

        for (int i = 0; i <= n; i++) 
        {
            // Treat the end of the histogram as height 0 (sentinel value)
            if(i==n) 
            {
                currentHeight =  0;
            }
            else{
                currentHeight = heights[i];
            }
            int lastIndex = i;
            while (!st.empty() && st.top().first > currentHeight) 
            {
                int height = st.top().first;
                int index = st.top().second;
                st.pop();
                
                int width = i - index;
                maxArea = max(maxArea, height * width);
                lastIndex = index;  // Update the last index for the new pair
            }
            st.push({currentHeight, lastIndex});
        }
            
            // Push the current height and the last valid index
        
        
        return maxArea;
    }

    // Function to find the maximal rectangle in a binary matrix
    int maximalRectangle(vector<vector<char>>& matrix) {
        if (matrix.empty()) return 0;  // Handle edge case when matrix is empty
        
        int n = matrix.size();
        int m = matrix[0].size();
        int maxRectangle = 0;
        
        vector<int> heights(m, 0);  // Initialize a heights array to keep track of heights
        
        // Process each row of the matrix
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                // If we encounter '1', increase the height, else reset it to 0
                if (matrix[i][j] == '1') {
                    heights[j]++;
                } else {
                    heights[j] = 0;
                }
            }
            // Find the largest rectangle for the current row's histogram
            maxRectangle = max(maxRectangle, largestRectangleArea(heights));
        }
        
        return maxRectangle;
    }
};

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