<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 n = heights.size();

        stack<int> height;height.push(-1);
        int max_area = 0;
        
        int a = 0;
        int b = 0;

        for( int i = 0 ; i < n ; i ++ )
        {
           while( height.top() != -1 and heights[height.top()] >= heights[i])
           {
                 int h = heights[height.top()]; height.pop();
                 int w = abs(i - height.top() - 1);
                 max_area = max( max_area , h * w);
           }
           height.push(i);

        }

        while(height.top() != -1)
       {
         int h = heights[height.top()]; height.pop();
          int w = abs(n - height.top() - 1);
          max_area = max(max_area, h*w);
       }

        return max_area;
                
    }
};
77848