<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:
    vector<int> solve1(vector<int>& heights){
        stack<int>s;
        s.push(-1);
        vector<int>next(heights.size());
        for(int i=heights.size()-1;i>=0;i--){
            while(s.top()!=-1 && heights[s.top()]>=heights[i]){
                s.pop();
            }
            next[i]=s.top();
            s.push(i);
        }
        return next;
    }
    vector<int> solve2(vector<int>& heights){
        stack<int>s;
        s.push(-1);
        vector<int>prev(heights.size());
        for(int i=0;i<heights.size();i++){
            while(s.top()!=-1 && heights[s.top()]>=heights[i]){
                s.pop();
            }
            prev[i]=s.top();
            s.push(i);
        }
        return prev;
    }
    int largestRectangleArea(vector<int>& heights) {
        vector<int>next= solve1(heights);
        vector<int>prev = solve2(heights);
        int ans =0;
        for(int i=0;i<heights.size();i++){
            if(next[i]==-1){
                next[i]=heights.size();
            }
            int len = next[i]-prev[i]-1;
            int area = len*heights[i];
            ans = max(ans,area);
        }
        return ans;

    }
    int maximalRectangle(vector<vector<char>>& matrix) {
        vector<vector<int>>v;
        for(int i=0;i<matrix.size();i++){
            vector<int>p;
            for(int j=0;j<matrix[0].size();j++){
                p.push_back(matrix[i][j]-'0');
            }
            v.push_back(p);
        }
        int area = largestRectangleArea(v[0]);
        for(int i=1;i<v.size();i++){
            for(int j=0;j<v[0].size();j++){
                if(v[i][j]==0)continue;
                v[i][j]+=v[i-1][j];
            }
            area = max(area,largestRectangleArea(v[i]));
        }
        return area;
    }
};
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