<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 {
    private:
    vector<int>nextSmaller(vector<int>arr){
        int n=arr.size();
        stack<int>st;
        vector<int>nse(n);
        for(int i=n-1;i>=0;i--){
            while(!st.empty()&&arr[st.top()]>arr[i])st.pop();
            if(st.empty())nse[i]=n;
            else{
                nse[i]=st.top();
            }
            st.push(i);
        }
        return nse;
    }
    vector<int>prevSmaller(vector<int>arr){
        int n=arr.size();
        stack<int>st;
        vector<int>pse(n);
        for(int i=0;i<n;i++){
            while(!st.empty()&&arr[st.top()]>=arr[i])st.pop();
            if(st.empty())pse[i]=-1;
            else{
                pse[i]=st.top();
            }
            st.push(i);
        }
        return pse;
    }
public:
    int maximalRectangle(vector<vector<char>>& mat) {
        int n=mat.size();
        int m=mat[0].size();
        vector<int>height(m,0);
        int maxi=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                height[j]=mat[i][j]=='1'?height[j]+1:0;
            }

            vector<int>nse=nextSmaller(height);
            vector<int>pse=prevSmaller(height);
            for(int i=0;i<m;i++){
            int width=nse[i]-pse[i]-1;
            maxi=max(maxi,width*height[i]);
        }

        }
        return maxi;
    }
};
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