<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>& h) {
        vector<pair<int, int> > s;
        int n = h.size();
        int l[n],r[n];
        for(int i=n-1;i>-1;i--) {
            int count = 0;
            while(!s.empty() && s[s.size()-1].first>=h[i]) {
                count+=r[s[s.size()-1].second];
                s.pop_back(); 
            }
            s.push_back({h[i],i});
            r[i] = count+1; 
        }
        s.clear();
        for(int i=0;i<n;i++) {
            int count = 0;
            while(!s.empty() && s[s.size()-1].first>=h[i]) {
                count+=l[s[s.size()-1].second];
                s.pop_back(); 
            }
            s.push_back({h[i],i});
            l[i] = count+1;
        }
        int mx = INT_MIN;
        for(int i=0;i<n;i++) {
            mx = max(mx, h[i]*(l[i]+r[i]-1));
        }
        return mx;
    }
};
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