<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:
#define pii pair<int, int>
    int largestRectangleArea(vector<int>& h) {
        int n = h.size();
        int res = 0;
        stack<pii> st;
        for (int i = 0; i < n; i++) {
            int idx = i;
            while (!st.empty() && h[i] <= st.top().second) {
                res = max(res, (i-st.top().first) * st.top().second);
                idx = st.top().first;
                st.pop();
            }
            st.push({idx, h[i]});
        }
        while (!st.empty()) {
            res = max(res, (n-st.top().first) * st.top().second);
            st.pop();
        }
        return res;
    }
};
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