<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 maximalRectangle(vector<vector<char>>& matrix) {
        if (matrix.empty()) return 0;  // Edge case: empty matrix

        int n = matrix.size();  // Number of rows
        int c = matrix[0].size();  // Number of columns
        stack<int> stc;
        vector<int> next_right(c, c);
        vector<int> next_left(c, -1);
        vector<int> arr(c, 0);  // Heights array
        int max_sum = 0;

        for (int i = 0; i < n; i++) {
            // Update the heights array for the current row
            for (int j = 0; j < c; j++) {
                if (matrix[i][j] == '1') {
                    arr[j]++;  // Increment height if '1'
                } else {
                    arr[j] = 0;  // Reset height if '0'
                }
            }

            // Clear the stack for fresh calculations
            while (!stc.empty()) stc.pop();

            // Compute next left for each column
            for (int j = 0; j < c; j++) {
                while (!stc.empty() && arr[stc.top()] >= arr[j]) {
                    stc.pop();
                }
                next_left[j] = stc.empty() ? -1 : stc.top();
                stc.push(j);
            }

            // Clear the stack again for next right calculations
            while (!stc.empty()) stc.pop();

            // Compute next right for each column
            for (int j = c - 1; j >= 0; j--) {
                while (!stc.empty() && arr[stc.top()] >= arr[j]) {
                    stc.pop();
                }
                next_right[j] = stc.empty() ? c : stc.top();
                stc.push(j);
            }

            // Calculate the area for each column
            for (int j = 0; j < c; j++) {
                int width = next_right[j] - next_left[j] - 1;
                int area = width * arr[j];
                max_sum = max(max_sum, area);
            }
        }

        return max_sum;
    }
};

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