<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 func(vector<int> &v){
        int n=v.size();
        stack<int> st;
        vector<int> l(n),r(n);
        for(int i=0;i<n;i++){
            while(!st.empty() && v[st.top()]>=v[i]) st.pop();
            l[i]=(st.empty()?i+1:i-st.top());
            st.push(i);
        }
        while(!st.empty()) st.pop();
        for(int i=n-1;i>=0;i--){
            while(!st.empty() && v[st.top()]>v[i]) st.pop();
            r[i]=(st.empty()?n-i:st.top()-i);
            st.push(i);
        }
        int maxi=INT_MIN;
        for(int i=0;i<n;i++){
            maxi=max(maxi,(l[i]+r[i]-1)*v[i]);
        }
        return maxi;
    }
    int maximalRectangle(vector<vector<char>>& matrix) {
        int n=matrix.size();
        int m=matrix[0].size();
        vector<vector<int>> grid(n,vector<int>(m,0));
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(i==0){
                    if(matrix[i][j]=='1') grid[i][j]=1;
                }
                else{
                    if(matrix[i][j]=='1') grid[i][j]=grid[i-1][j]+1;
                }
            }
        }
        int maxi=INT_MIN;
        for(int i=0;i<n;i++){
            int ans=func(grid[i]);
            maxi=max(maxi,ans);
        }
        return maxi;
    }
};
20556