<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>

using namespace std;

class Solution {
private:
    vector<int> nextsmallele(vector<int> arr, int n) {
        stack<int> st;
        vector<int> ans(n);
        st.push(-1);
        for (int i = n - 1; i >= 0; i--) {
            int curr = arr[i];
            while ((st.top() != -1) && (arr[st.top()] >= curr)) {
                st.pop();
            }
            ans[i] = st.top();
            st.push(i);
        }
        return ans;
    }
    
    vector<int> prevsmallele(vector<int> arr, int n) {
        stack<int> st;
        vector<int> ans(n);
        st.push(-1);
        for (int i = 0; i < n; i++) {
            int curr = arr[i];
            while ((st.top() != -1) && (arr[st.top()] >= curr)) {
                st.pop();
            }
            ans[i] = st.top();
            st.push(i);
        }
        return ans;
    }

    int largestRectangleArea(vector<int> heights) {
        int n = heights.size();

        vector<int> next = nextsmallele(heights, n);
        vector<int> prev = prevsmallele(heights, n);

        int area = 0;  // Area cannot be negative
        for (int i = 0; i < n; i++) {
            int height = heights[i];
            if (next[i] == -1) {
                next[i] = n;
            }
            int width = next[i] - prev[i] - 1;
            int newArea = height * width;
            area = max(area, newArea);
        }
        return area;
    }
    
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        if (matrix.empty() || matrix[0].empty()) return 0;  // Check for empty matrix
        
        int rows = matrix.size();
        int cols = matrix[0].size();
        
        vector<int> heights(cols, 0);
        int maxArea = 0;

        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                // Update the height of the histogram bars
                heights[j] = matrix[i][j] == '0' ? 0 : heights[j] + 1;
            }
            // Find the largest rectangle area in the updated histogram
            maxArea = max(maxArea, largestRectangleArea(heights));
        }
        return maxArea;
    }
};

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