<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>

#define fst(t) (get<0>(t))
#define snd(t) (get<1>(t))
using pii = tuple<int,int>;

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        stack<pii> s;
        vector<int> ans = heights;

        for(int i=0; i<heights.size(); i++) {
            while(!s.empty() and fst(s.top()) >= heights[i]) s.pop();
            int j = -1;
            if(!s.empty()) j = snd(s.top());
            s.push({ heights[i],i });
            int area = (i-j-1)*heights[i];
            ans[i] += area;
        }

        while(!s.empty()) s.pop();

        int a=0;
        for(int i=heights.size()-1; i>=0; i--) {
            while(!s.empty() and fst(s.top()) >= heights[i]) s.pop();
            int j = heights.size();
            if(!s.empty()) j = snd(s.top());
            s.push({ heights[i],i });
            int area = (j-i-1)*heights[i];
            ans[i] += area;
            a = max(a, ans[i]);
        }

        return a;
        // return *max_element(begin(ans), end(ans));
    }
};
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