<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 solve(vector<int> arr)
    {
        int maxi=0;

        stack<int> st;
        

        for(int i=0;i<arr.size();i++)
        {
            while(!st.empty() && arr[st.top()]>arr[i])
            {
                int num=arr[st.top()];
                st.pop();

                int nse=i;
                int pse = st.empty() ? -1 : st.top();  // Previous Smaller Element index

                maxi=max(maxi,num*(nse-pse-1));
            }

            st.push(i);
        }

        while(!st.empty())
        {
             int num=arr[st.top()];
                st.pop();

                int nse=arr.size();
                int pse = st.empty() ? -1 : st.top();  // Previous Smaller Element index

                maxi=max(maxi,num*(nse-pse-1));
        }

        return maxi;
    }

    int maximalRectangle(vector<vector<char>>& matrix) {

        int prefix=0;
        vector<vector<int>> matrix1(matrix.size(),vector<int>(matrix[0].size()));

        for(int i=0;i<matrix[0].size();i++)
        {
            for(int j=0;j<matrix.size();j++)
            {
                if(matrix[j][i]=='0')
                prefix=0;

                prefix=prefix+(matrix[j][i]-'0');
                matrix1[j][i]=prefix;
            }
            prefix=0;
        }


        
        




        int maxi=0;

        for(int i=0;i<matrix1.size();i++)
        {
            maxi=max(maxi,solve(matrix1[i]));
        }
        return maxi;
        
    }
};
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