<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 {

// it is not a greedy approach, sorting does not solve it 
// brute force is getting each rectangle and evaluating the area 
// need to know nle and ple for each num in O(n)
// if we know the ple and nle with its respective positions, we can compare the max area of those two and store it in an array 
// called maxArea 
// at the end, we just iterate the maxArea array to see which is the highest
public:
    int largestRectangleArea(vector<int>& heights) {
        int n = heights.size();

        vector<int> nle(heights.size(),n);

        vector<int> ple(heights.size(),-1);

        int maxArea = INT_MIN;

        stack<int> s; 


        

        // get nle
        for(int i = 0; i < n; i++){
            int elem = heights[i];

            while(!s.empty() && elem < heights[s.top()]){
                nle[s.top()] = i; // NLE is found for the top of the stack
                s.pop();
            }

            s.push(i);
        }

        // clean 
        while(!s.empty()) 
            s.pop();

        // ple
        for(int i = 0; i < heights.size(); i++){
            while(!s.empty() && heights[s.top()] >= heights[i]){
                s.pop();
            }

            ple[i] = s.empty()? -1 : s.top();
            s.push(i);
        }


        for(int i = 0; i < n; i++){
            int b = nle[i]-ple[i]-1;
            int a = heights[i]*b;
            maxArea = max(a,maxArea);
        }


        return maxArea;
    }
};
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