<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) {
        int n = matrix[0].size();
        vector<int> heights(n);
        int ans = 0;
        for (auto& row : matrix) {
            for (int j = 0; j < n; ++j) {
                if (row[j] == '1')
                    ++heights[j];
                else
                    heights[j] = 0;
            }
            ans = max(ans, largestRectangleArea(heights));
        }
        return ans;
    }

    int largestRectangleArea(vector<int>& heights) {
        int res = 0, n = heights.size();
        stack<int> stk;
        vector<int> left(n, -1);
        vector<int> right(n, n);
        for (int i = 0; i < n; ++i) {
            while (!stk.empty() && heights[stk.top()] >= heights[i]) {
                right[stk.top()] = i;
                stk.pop();
            }
            if (!stk.empty()) left[i] = stk.top();
            stk.push(i);
        }
        for (int i = 0; i < n; ++i)
            res = max(res, heights[i] * (right[i] - left[i] - 1));
        return res;
    }
};
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