<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:
    # the idea is that all the rectangles in the histogram must be able to extend to the end
    # if you are smaller than the previous value then the previous value cannot extend to the end
    # maintain a monotonically increasing stack
    # when you need to pop a value off calculate it's max possible value
        # do h * (curr - idx)
        # stretch back the current index to the idx and append it to the stack
    # process the values left in the stack
    def largestRectangleArea(self, heights: List[int]) -> int:
        stack = []
        _max = [0]
        for i, h in enumerate(heights):
            next_i = i
            while len(stack) != 0 and stack[-1][1] > h:
                pidx, ph = stack.pop()
                _max[0] = max(_max[0], (i - pidx) * ph)
                next_i = pidx
            stack.append((next_i, h))

        while len(stack) != 0:
            pidx, ph = stack.pop()
            _max[0] = max(_max[0], (len(heights) - pidx) * ph)
        
        return _max[0]
            
33100