<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 h, w;
    vector<vector<int>> subsequentOnes;
    vector<vector<char>> m;

    void solveSubsequent() {
        for(int i = h - 1; i >= 0; i--) {
            for(int j = w - 1; j >= 0; j--) {
                if(m[i][j] == '0') subsequentOnes[i][j] = 0;
                else {
                    if(j == w - 1) subsequentOnes[i][j] = 1;
                    else subsequentOnes[i][j] = 1 + subsequentOnes[i][j+1];
                }
            }
        }
    }

    int solveRectangles() {
        int maxArea = 0;
        for(int i = 0; i < h; i++) {
            for(int j = 0; j < w; j++) {
                if(m[i][j] == '0') continue;
                int minOnes = INT_MAX;
                for(int k = i; k < h; k++) {
                    if(m[k][j] == 0) break;
                    minOnes = min(minOnes, subsequentOnes[k][j]);
                    int tempArea = (k - i + 1) * minOnes;
                    maxArea = max(maxArea, tempArea);
                }
            }
        }
        return maxArea;
    }

    int maximalRectangle(vector<vector<char>>& matrix) {
        h = matrix.size(); w = matrix[0].size();
        subsequentOnes.resize(h, vector<int>(w, -1));
        m = matrix;
        solveSubsequent();
        return solveRectangles();
    }
};
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