<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> left;
        vector<int> right;
        stack<int>  q,q2;
        q.push(0);
        left.push_back(-1);
        for(int i=1;i<n;i++){
            while(!q.empty() && heights[q.top()]>heights[i] ){
                    q.pop();
            }
            if(q.empty()) left.push_back(-1);
            else{
                left.push_back(q.top());
            }
            q.push(i);
        }

        right.push_back(n);
        q2.push(n-1);
        for(int i=n-2;i>=0;i--){
            while(!q2.empty() && heights[q2.top()]>=heights[i]){
                q2.pop();
            }
            if(q2.empty()) right.push_back(n);
            else{
                right.push_back(q2.top());
            }
            q2.push(i);
        }

        int ans = INT_MIN;
        reverse(right.begin(),right.end());
        for(int i=0;i<n;i++){
            cout<<left[i]<<" "<<right[i]<<endl;
            ans = max(ans,((right[i]-left[i]-1)*heights[i]));
        }
    return ans;
    }
};
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