<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:
    int largestRectangleArea(vector<int>& a) {
       stack<int>s;
       vector<int>pre(a.size());
       for(int i=0;i<a.size();i++){
             while(!s.empty()&&a[s.top()]>=a[i]){
                s.pop();
             }
             if(s.empty()){
                pre[i]=i+1;
             }
             else{
                pre[i]=i-(s.top());
             }
             //cout<<pre[i]<<" ";
             s.push(i);
       }
       while(!s.empty()){
        s.pop();
       }
       int maxi=0;
       int n=a.size();
       for(int i=a.size()-1;i>=0;i--){
        while(!s.empty()&&a[s.top()]>=a[i]){
            s.pop();
        }
        int p;
        if(s.empty()){
           p=n-i-1;
        }
        else{
            p=s.top()-i-1;
        }
        maxi=max(maxi,(p+pre[i])*a[i]);
        s.push(i);

       }
       return maxi;
    }
};
81001