<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>& nums) {

        int n = nums.size();
        vector<int> prev(n, -1), next(n, n);
        stack<int> sk;

        auto print = [](auto nums){
            for(auto a: nums)
                cout<<a<<"\t";
            cout<<endl;
        };

        // index           -1. 0. 1. 2. 3. 4. 5  6. 
        // nums                2. 1. 5. 6. 2. 3
        // next                1  6. 4. 4. 6. 6
        // prev.              -1 -1  1  2  1  4
        // Diff                2. 7. 3. 2. 5. 2
        // final               1. 6. 2. 1. 4. 1

    
        for(int i=0 ; i<n ; i++){
            while(sk.size() and nums[sk.top()] > nums[i]){
                next[sk.top()] = i;
                sk.pop();
            }
            sk.push(i);
        }

        while(sk.size())    sk.pop();

        for(int i=n-1; i>=0; i--){
            while(sk.size() and nums[sk.top()] > nums[i]){
                prev[sk.top()] = i;
                sk.pop();
            }
            sk.push(i);
        }

        // print(nums);
        // print(next);
        // print(prev);

        int ans = 0;
        for(int i=0 ; i<n; i++){
            ans = max(ans , (next[i] - prev[i] - 1) * nums[i]);
        }

        return ans;

        
        
    }
};
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