<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>& a = heights;

        int pse[n], nse[n];
        fill(pse, pse + n, -1);
        fill(nse, nse + n, n);

        stack<int> s;
        for(int i = 0; i < n; i++)
        {
            while(!s.empty() && a[i] < a[s.top()])
            {
                nse[s.top()] = i; //need to store the index instead of the element value
                s.pop();
            }
            if(!s.empty())
                pse[i] = s.top();
            s.push(i);
        }

        int maxArea = INT_MIN;
        for(int i = 0; i < n; i++)
        {
            int area = (nse[i] - pse[i] - 1) * a[i];
            maxArea = max(maxArea, area);
        }
        return maxArea;
    }
};
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