<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();
        vector<int> prevSmaller, nextSmaller(n);

        vector<pair<int,int>> st, st2;
        st.push_back({-1,-1});
        st2.push_back({-1,n});

        for(int i=0; i<n; i++)
        {
            while(!st.empty() && st.back().first >= heights[i])
                st.pop_back();
            
            prevSmaller.push_back(st.back().second);
            st.push_back({heights[i], i});

            while(!st2.empty() && st2.back().first >= heights[n-1-i])
                st2.pop_back();
            
            nextSmaller[n-1-i] = st2.back().second;
            st2.push_back({heights[n-1-i], n-1-i});
        }

        int ans = 0;
        for(int i=0; i<n; i++)
        {
            int temp = heights[i] * (nextSmaller[i]-prevSmaller[i]-1);
            ans = max(ans, temp);
        }

        return ans;
    }
};
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