<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 {
    vector<int> presmaller(vector<int> &row) {
        int n = row.size();
        vector<int> ans;
        stack<int> s;
        s.push(-1);
        for (int i = 0; i < n; i++) {
            while (s.size() > 1 && row[s.top()] >= row[i]) {
                s.pop();
            }
            ans.push_back(s.top());
            s.push(i);
        }
        return ans;
    }
    
    vector<int> nextsmaller(vector<int> &row) {
        int n = row.size();
        vector<int> ans;
        stack<int> s;
        s.push(-1);

        for (int i = n - 1; i >= 0; i--) {
            while (s.size() > 1 && row[s.top()] >= row[i]) {
                s.pop();
            }
            if (s.top() != -1)
                ans.push_back(s.top());
            else
                ans.push_back(n);
            s.push(i);
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
    
    void histogram(vector<int> &row, int &area) {
        int newArea = 0;
        vector<int> pre = presmaller(row);
        vector<int> next = nextsmaller(row);
        int n = row.size();
        for (int i = 0; i < n; i++) {
            int previous = pre[i];
            int nxt = next[i];
            newArea = (nxt - previous - 1) * row[i];
            area = max(area, newArea);
        }
    }

public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        int n = matrix.size();
        int m = matrix[0].size();
        int area = 0;
        vector<int> row;
        
        for (int i = 0; i < m; i++) {
            if (matrix[0][i] == '0') {
                row.push_back(0);
            } else {
                row.push_back(1);
            }
        }
        
        for (int i = 1; i < n; i++) {
            histogram(row, area);
            for (int j = 0; j < m; j++) {
                if (matrix[i][j] == '0') {
                    row[j] = 0;
                } else {
                    row[j]++;
                }
            }
        }
        
        histogram(row, area);
        
        return area;
    }
};
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