<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> &height) {
            
            int ret = 0;
            height.push_back(0);
            vector<int> index;
            
            for(int i = 0; i < height.size(); i++)
            {
                while(index.size() > 0 && height[index.back()] >= height[i])
                {
                    int h = height[index.back()];
                    index.pop_back();
                    
                    int sidx = index.size() > 0 ? index.back() : -1;
                    if(h * (i-sidx-1) > ret)
                        ret = h * (i-sidx-1);
                }
                index.push_back(i);
            }
            
            return ret;
        }
    };
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