<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 maxArea = 0;
        // Stack will store pairs of (index, height) for the histogram bars
        // This will be a monotonic stack, meaning the heights in the stack are 
        // in increasing order from bottom to top.
        //
        // The idea is that all the heights in the stack can extend upwards
        // (to the top element position, since the top element was the highest)
        // to form rectangles, and once we encounter a smaller height
        // so the extension to the right wouldn't be possible 
        // we calculate the maximum area for all larger heights
        stack<pair<int, int>> stack; 

        // Adding a 0-height bar at the end to ensure all bars are processed
        // so at the end stack is empty
        heights.push_back(0);

        for(int i = 0; i < heights.size(); ++i) {
            // 'start' will track the index where the current height can extend to the left (backwards)
            int start = i;
            while (!stack.empty() && stack.top().second > heights[i]) {
                auto top = stack.top();
                stack.pop();

                // Since the top element was taller than the current bar, we can extend
                // the rectangle with this bar's height backwards to the start index
                // of the popped rectangle.
                start = top.first;

                // before popping the element, check if it would yield the largest 
                // rectangle so far.
                maxArea = max(maxArea, (i - top.first) * top.second);
            }
            stack.emplace(start,heights[i]);
        }

        return maxArea;
        
    }
};
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