<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> st; // Stack to store indices
    int max_area = 0;
    int i = 0;
    
    while (i < n) {
        // If the stack is empty or the current height is greater than the height at the top of the stack
        if (st.empty() || heights[i] >= heights[st.top()]) {
            st.push(i++);
        } else {
            // Pop the top
            int h = heights[st.top()];
            st.pop();
            // Calculate the width of the rectangle with height h
            int width = st.empty() ? i : i - st.top() - 1;
            // Calculate the area with height h
            max_area = max(max_area, h * width);
        }
    }
    
    // Process remaining bars in the stack
    while (!st.empty()) {
        int h = heights[st.top()];
        st.pop();
        int width = st.empty() ? i : i - st.top() - 1;
        max_area = max(max_area, h * width);
    }
    
    return max_area;
    }
};
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