<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:
    vector<int> presin(vector<int>& heights)
    {
        stack<int> st;
        int n = heights.size();
        vector<int> pse(n);
        for(int i = 0; i < n; i++)
        {
            while(!st.empty() && heights[i] <= heights[st.top()])
                st.pop();
            
            if(st.empty())
                pse[i] = -1;
            
            else
                pse[i] = st.top();

            st.push(i);
        }
        return pse;
    }
    vector<int> nextsin(vector<int>& heights)
    {
        stack<int> st;
        int n = heights.size();
        vector<int> nse(n);
        for(int i = n - 1; i >= 0; i--)
        {
            while(!st.empty() && heights[i] <= heights[st.top()])
                st.pop();
            
            if(st.empty())
                nse[i] = n;
            
            else
                nse[i] = st.top();

            st.push(i);
        }
        return nse;
    }


    int largestRectangleArea(vector<int>& heights) {
        vector<int> nse = nextsin(heights);
        vector<int> pse = presin(heights);
        int n = heights.size();

        int maxi = 0;
        
        for(int i = 0; i < n; i++)
        {
            int tot = nse[i] - pse[i] - 1;
            maxi = max(maxi , tot * heights[i]);
        }
        return maxi;
        
    }
    int maximalRectangle(vector<vector<char>>& matrix) {
        int n = matrix.size();
        int m = matrix[0].size();

        vector<vector<int>> temp(n , vector<int> (m));
        for(int i = 0; i < m; i++)
        {
            int cnt = 0;
            for(int j = 0; j < n; j++)
            {
                if(matrix[j][i] == '1')
                {
                    cnt++;
                    temp[j][i] = cnt;
                }
                else
                {
                    cnt = 0;
                    temp[j][i] = 0;
                }
            }
        }

        int maxi = 0;
        for(int i = 0; i < n; i++)
        {
            maxi = max(maxi , largestRectangleArea(temp[i]));
        }
        return maxi;
    }
};
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