<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>>& A) {
        int M = A.size(), N = A[0].size(), ans = 0;
        vector<int> h(N), nextSmaller(N);
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                h[j] = A[i][j] == '0' ? 0 : (h[j] + 1);
            }
            stack<int> s;
            for (int j = N - 1; j >= 0; --j) {
                while (s.size() && h[j] <= h[s.top()]) s.pop();
                nextSmaller[j] = s.size() ? s.top() : N;
                s.push(j);
            }
            s = {};
            for (int j = 0; j < N; ++j) {
                while (s.size() && h[j] <= h[s.top()]) s.pop();
                int prevSmaller = s.size() ? s.top() : -1;
                ans = max(ans, (nextSmaller[j] - prevSmaller - 1) * h[j]);
                s.push(j);
            }
        }
        return ans;
    }
};
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