<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:
    long long int mod=1e9+7;
    int solve(vector<int>& arr) {
        vector<long long>v(arr.size(),0);
        stack<int>st;
        long long ans=0;
        int n=arr.size();
        v[n-1]=n;
        st.push(n-1);
        for(int i=n-2;i>=0;i--){
            while(!st.empty() && arr[st.top()]>=arr[i])st.pop();
            if(st.empty())v[i]=n;
            else v[i]=st.top();
            st.push(i);
            
        }
        vector<long long>v2(arr.size(),0);
        while(!st.empty())st.pop();
        v2[0]=-1;
        st.push(0);
        for(int i=1;i<n;i++){
            while(!st.empty() && arr[st.top()]>=arr[i])st.pop();
            if(st.empty())v2[i]=-1;
            else v2[i]=st.top();st.push(i);
            
        }
        // for(auto it:v)cout<<it<< " ";
        // cout<<endl;
        // for(auto it:v2)cout<<it<< " ";
        for(int i=0;i<n;i++){
            ans=max(ans,(((long long)(i-v2[i])+(v[i]-i)-1)*arr[i]));
        }
        return ans;
    }
    int largestRectangleArea(vector<int>& arr) {
        return solve(arr);
        
    }
};
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