<p>Given an array of integers <code>heights</code> representing the histogram&#39;s bar height where the width of each bar is <code>1</code>, return <em>the area of the largest rectangle in the histogram</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/04/histogram.jpg" style="width: 522px; height: 242px;" />
<pre>
<strong>Input:</strong> heights = [2,1,5,6,2,3]
<strong>Output:</strong> 10
<strong>Explanation:</strong> The above is a histogram where width of each bar is 1.
The largest rectangle is shown in the red area, which has an area = 10 units.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/04/histogram-1.jpg" style="width: 202px; height: 362px;" />
<pre>
<strong>Input:</strong> heights = [2,4]
<strong>Output:</strong> 4
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= heights.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= heights[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
unsigned long long maximum(unsigned long long a, unsigned long long b)
      {
          if(a>b)return a;
          else return b;
      }
    int largestRectangleArea(vector<int>& arr) {
      int  n=arr.size();
         long long left[n],right[n]; //will note the index of the left and right boundary
        stack< long long>p; 
        for(int i=0;i<n;i++) //for the left boundary
          {
              while(!p.empty() && arr[p.top()]>=arr[i])p.pop();
              if(p.empty())left[i]=0;
              else left[i]=p.top()+1;
              p.push(i);
                
          }
          while(!p.empty())p.pop();
          for(int i=n-1;i>=0;i--)
           {
               while(!p.empty() && arr[p.top()]>=arr[i])p.pop();
               if(p.empty())right[i]=n-1;
               else right[i]=p.top()-1;
               p.push(i);
           }
           //calculation and storing the max area
            long long maxm=0;
           for(int i=0;i<n;i++)
             {
                 maxm=maximum(maxm,(right[i]-left[i]+1)*arr[i]);
             }
             return maxm;
    }
        
    
};
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