<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:
    vector<int> nextSmallerRight(vector<int>& heights) {
        // we need to iterate from the left hand side of the array
        int n = heights.size();
        stack<pair<int, int>> s;
        vector<int> v(n,n);
        for (int i = n - 1; i >= 0; i--) {
            while (!s.empty() and s.top().first >= heights[i]) {
                s.pop();
            }
            if (s.empty()) {
                v[i]=n;
            } else {
                v[i]=(s.top().second);
            }
            s.push(make_pair(heights[i], i));
        }
        return v;
    }

    vector<int> nextSmallerLeft(vector<int>& heights) {
        // we need to iterate from the left hand side of the array
        int n = heights.size();
        stack<pair<int, int>> s;
        vector<int> v;
        for (int i = 0; i <= n-1; i++) {
            while (!s.empty() and s.top().first >= heights[i]) {
                s.pop();
            }
            if (s.empty()) {
                v.push_back(-1);
            } else {
                v.push_back(s.top().second);
            }
            s.push(make_pair(heights[i], i));
        }
        return v;
    }

    int largestRectangleArea(vector<int>& heights) {
        vector<int> nsr=nextSmallerRight(heights);
        vector<int> nsl=nextSmallerLeft(heights);
        int maxi=INT_MIN;
        for(int i=0;i<heights.size();i++){
            maxi=max(maxi,heights[i]*((nsr[i]-nsl[i])-1));
        }
        return maxi;

    }
};
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