<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:
    def largestRectangleArea(self, heights: List[int]) -> int:
        largest = 0
        stack = []

        for idx, curr_height in enumerate(heights):
            
            # The new starting index for the height with value v. Defaults to its original index.
            curr_width_idx = idx
            

            # if prev height > curr height
            while stack and stack[-1][1] > curr_height:
                old_idx, prev_height = stack.pop()
                
                # previous is larger, so see if we can make a larger rectangle using our now larger width (update curr width idx to use prev heights idx)
                new_width = idx - old_idx
                curr_width_idx = old_idx

                largest = max(largest, new_width * prev_height)



            stack.append((curr_width_idx, curr_height))

        # check all heights remaining in the stack (An item will remain in the stack if a smaller height hasn't been encountered) Since these rectangles can now extend to the end of the histogram.

        for (old_idx, curr_height) in stack:
            width = len(heights) - old_idx
            largest = max(largest, width * curr_height)

        return largest
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