<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 m=matrix.size(),n=matrix[0].size();
        int rs=0;
        vector<int>dp(n,0);
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(matrix[i][j]=='0') dp[j]=0;
                else dp[j]++;
            }

            vector<int>l(n,0);
            stack<int>st;
            for(int j=0;j<n;j++){
                while(!st.empty()&&dp[j]<=dp[st.top()]){
                    st.pop();
                }
                l[j]=st.empty()?0:(st.top()+1); 
                st.push(j);
            }
 
            stack<int>st2;
            for(int j=n-1;j>=0;j--){
                while(!st2.empty()&&dp[j]<=dp[st2.top()]){
                    st2.pop();
                }
                int start=st2.empty()?(n-1):(st2.top()-1);
                rs=max(rs, dp[j]*(start-l[j]+1));
                st2.push(j);
            }
             
        }

        return rs;
    }
};
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