<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:
  struct Pair {
    int index;
    int height;
  };

  int largestRectangleArea(vector<int>& heights) {
    int max_area = 0;
    std::vector<Pair> st;

    for (int i = 0; i < heights.size(); ++i) {
      int start = i;
      while (!st.empty() && st.back().height > heights[i]) {
        auto [cur_ind, cur_h] = st.back();
        st.pop_back();
        // area that can be formed with popped height
        int cur_area = cur_h * (i - cur_ind);
        max_area = std::max(max_area, cur_area);
        start = cur_ind;
      }
      st.push_back({start, heights[i]});
    }

    for (auto [index, height] : st) {
      int cur = (heights.size() - index) * height;
      max_area = std::max(max_area, cur);
    }
    return max_area;
  }
};
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