<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 <vector>
#include <stack>
#include <algorithm>

using namespace std;

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        stack<int> st;
        int n = heights.size();
        vector<int> lsmall(n, -1);
        vector<int> rsmall(n, n); // rsmall should be initialized to n (the end of the array)

        // Calculate left limits
        for (int i = 0; i < n; i++) {
            while (!st.empty() && heights[st.top()] >= heights[i]) {
                st.pop();
            }
            if (!st.empty()) lsmall[i] = st.top();
            st.push(i);
        }

        // Clear stack before calculating right limits
        while (!st.empty()) st.pop();

        // Calculate right limits
        for (int i = n - 1; i >= 0; i--) {
            while (!st.empty() && heights[st.top()] >= heights[i]) {
                st.pop();
            }
            if (!st.empty()) rsmall[i] = st.top();
            st.push(i);
        }

        // Calculate maximum area
        int maxArea = 0;
        for (int i = 0; i < n; i++) {
            maxArea = max(maxArea, (rsmall[i] - lsmall[i] - 1) * heights[i]);
        }
        return maxArea;
    }

    int maximalRectangle(vector<vector<char>>& matrix) {
        int m = matrix.size();
        if (m == 0) return 0;
        int n = matrix[0].size();
        vector<int> heights(n, 0);
        int ans = 0;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == '0') {
                    heights[j] = 0;
                } else {
                    heights[j] += (matrix[i][j] - '0');
                }
            }
            ans = max(ans, largestRectangleArea(heights));
        }
        return ans;
    }
};

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