<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.size();
        int m = matrix[0].size();
        vector<int> left(m, INT_MAX), right(m, INT_MAX), height(m, 0);
        int ans = 0;

        for (int i = 0; i < n; i++) {
            int curLeft = 0, curRight = 0;
            for (int j = 0; j < m; j++) {
                if (matrix[i][j] == '1') {
                    height[j]++;
                    curLeft++;
                } else {
                    height[j] = 0;
                    curLeft = 0;
                }

                left[j] = height[j] <= 1 ? curLeft : min(left[j], curLeft);
            }

            for (int j = m - 1; j >= 0; j--) {
                if (matrix[i][j] == '1') {
                    curRight++;
                } else {
                    curRight = 0;
                }

                right[j] = height[j] <= 1 ? curRight : min(right[j], curRight);
            }

            for (int j = 0; j < m; j++) {
                cout << "(" << height[j] << ' ' << left[j] << ' ' << right[j] << "), ";
                ans = max(ans, height[j] * ( left[j] + right[j] - 1 ));
            }
            cout << '\n';
        }

        return ans;
    }
};
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