<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) {
        stack <pair<int, int>> s;
        int n = heights.size();
        vector<int> min_right(n, -1);
        vector<int> min_left(n, -1);
        s.push({heights[0], 0});
        for(int i = 1; i < n; i++){
            if(heights[i] >= s.top().first){
                s.push({heights[i], i});
            }else{
                while( !s.empty() && heights[i] < s.top().first){
                    min_right[s.top().second] = i -1; 
                    s.pop();
                }
                s.push({heights[i], i});
            }
        }
        while(!s.empty()){
            s.pop();
        }
        s.push({heights[n-1], n-1});
        for(int i = n-1; i > -1; i--){
            if(heights[i] >= s.top().first){
                s.push({heights[i], i});
            }else{
                while( !s.empty() && heights[i] < s.top().first){
                    min_left[s.top().second] = i;
                    s.pop();
                }
                s.push({heights[i], i});
            }
        }
        int ans = 0;
        for(int i =0; i < n; i++){
            int temp, l, r;
            l = min_left[i];
            r = min_right[i];
            if(min_right[i] == -1)
                r = n - 1;       
            temp = heights[i]*(r- l ) ;
            if(temp > ans)
                ans = temp;
        }
        return ans;


    }
};

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