<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(), ans=0;
        stack<int> st; //stores indices

        for(int i=0; i<n; i++){
            while(!st.empty() && heights[i]<heights[st.top()]){
                int topIdx=st.top();
                st.pop();
                int h=heights[topIdx];
                int w=st.empty()? i:(i-st.top()-1);
                ans=max(ans, h*w);
            }
            st.push(i);
        }
        while(!st.empty()){
            int topIdx=st.top();
            st.pop();
            int h=heights[topIdx];
            int w=st.empty()? n:(n-st.top()-1);
            ans=max(ans, h*w);
        }
        return ans;        
    }
};
79546