<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:
    //next smaller element
    vector<int> findNSE(vector<int> &arr)
    {
        int n=arr.size();
        stack<int>st;
        vector<int>nse(n,-1);
        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;
    }
    //previous smaller element
    vector<int> findPSE(vector<int> &arr)
    {
        int n=arr.size();
        stack<int>st;
        vector<int>pse(n,-1);
        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;
    }
    //BRUTE FORCE
    int largestRectangleArea(vector<int>& heights) {
        int n=heights.size();
        vector<int>area(n,0);
        vector<int>nse=findNSE(heights);
        vector<int>pse=findPSE(heights);
        int maxi=0;
        for(int i=0;i<n;i++)
        {
            int currHeight=(nse[i]-pse[i]-1)*heights[i];
            maxi=max(maxi,currHeight);
        }
        return maxi;
    }
    int maximalRectangle(vector<vector<char>>& matrix) {
        int n=matrix.size();
        int m=matrix[0].size();
        int maxArea=0;
        vector<vector<int>>prefixSum(n,vector<int>(m,0));
        //using prefix sum concept
        for(int j=0;j<m;j++)
        {
            int sum=0;
            for(int i=0;i<n;i++)
            {
                sum+=1;
                if(matrix[i][j]=='0')
                    sum=0;
                prefixSum[i][j]=sum;
            }
        }
        for(int i=0;i<n;i++)
        {
            int currMax=largestRectangleArea(prefixSum[i]);
            maxArea=max(maxArea,currMax);
        }
        return maxArea;
    }
};
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