<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>& arr) {
        int n=arr.size();
        stack<pair<int,int>> st;

        vector<int> prefix(n),suffix(n);

        for(int i=0;i<n;i++)
        {
            int curr=arr[i];
            while(st.size()>0&&st.top().first>=curr) st.pop();
            if(st.size()==0) prefix[i]=0;
            else prefix[i]=st.top().second+1;
            st.push({curr,i});
        }
        stack<pair<int,int>> dup;
        swap(dup,st);

        for(int i=n-1;i>=0;i--)
        {
            int curr=arr[i];

            while(st.size()>0&&st.top().first>=curr) st.pop();
            if(st.size()==0) suffix[i]=n-1;
            else suffix[i]=st.top().second-1;
            st.push({curr,i});
        }
        vector<int> len(n);
        for(int i=0;i<n;i++) len[i]=(suffix[i]-prefix[i]+1);
        vector<long long> ans(n);
        for(int i=0;i<n;i++) ans[i]=arr[i]*len[i];
        return *max_element(ans.begin(),ans.end());
        // return 0;
    }
};
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