<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(), maxarea=0;
        stack<pair<int, int>> st1, st2;
        vector<int> nsl, nsr;
        for(int i=0; i<n; i++){
            if(st1.empty()) nsl.push_back(-1);
            if(!st1.empty() && st1.top().first<heights[i])
                nsl.push_back(st1.top().second);
            else if(!st1.empty() && st1.top().first>=heights[i]){
                while(!st1.empty() && st1.top().first>=heights[i])
                    st1.pop();
                if(st1.empty()) nsl.push_back(-1);
                else nsl.push_back(st1.top().second);
            }
            st1.push({heights[i], i});
        }
        for(int i=n-1; i>=0; i--){
            if(st2.empty()) nsr.push_back(n);
            if(!st2.empty() && st2.top().first<heights[i])
                nsr.push_back(st2.top().second);
            else if(!st2.empty() && st2.top().first>=heights[i]){
                while(!st2.empty() && st2.top().first>=heights[i])
                    st2.pop();
                if(st2.empty()) nsr.push_back(n);
                else nsr.push_back(st2.top().second);
            }
            st2.push({heights[i], i});
        }
        reverse(nsr.begin(), nsr.end());
        for(int i=0; i<n; i++)
            maxarea=max(maxarea, (nsr[i]-nsl[i]-1)*heights[i]);
        
        return maxarea;
    }
};
94823