<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 largest = 0;
        vector<pair<int, int>> mins;
        for (int i = 0; i < heights.size(); ++i)
        {
            while (!mins.empty() && mins.back().first >= heights[i])
                mins.pop_back();
            if (mins.empty() || mins.back().first <= heights[i])
                mins.push_back(make_pair(heights[i], i));
            if (largest < mins[0].first * (i + 1))
                largest = mins[0].first * (i + 1);
            for (int j = 1; j < mins.size(); ++j)
                if (largest < mins[j].first * (i - mins[j - 1].second))
                    largest = mins[j].first * (i - mins[j - 1].second);
        }
        return largest;
    }
};
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