<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>& heights) {
        int n  = heights.size();
        vector<int>arr = heights;
         vector<int> pse(n, -1);  
        vector<int> nse(n, n);   
        stack<pair<int, int>> s1; 
        stack<pair<int, int>> s2; 
        
        for (int i = 0; i < n; ++i) {
            while (!s1.empty() && s1.top().first >= arr[i]) {
                s1.pop();
            }
            if (!s1.empty()) {
                pse[i] = s1.top().second;
            }
            s1.push({arr[i], i});
        }
        
        for (int i = n - 1; i >= 0; --i) {
            while (!s2.empty() && s2.top().first > arr[i]) {
                s2.pop();
            }
            if (!s2.empty()) {
                nse[i] = s2.top().second;
            }
            s2.push({arr[i], i});
        }
        int maxi = 0;
        for(int i=0; i<n; i++){
            maxi = max(maxi, arr[i]*(nse[i]-pse[i]-1));
        }
        return maxi;
    }
};
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