<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<int> ns;
        stack<int> ps;
        int len = heights.size();

        vector<int> nxtsml(len, 0);
        vector<int> presml;

        ns.push(-1);

        for(int i=0;i<len;i++){

            while(ns.top()!= -1 && heights[ns.top()] >= heights[i]){
                ns.pop();
            }

            presml.push_back(ns.top());
            ns.push(i);
        }
    
        ps.push(len);

        for(int i=len-1;i>=0;i--){

            while(ps.top()!= len && heights[ps.top()] >= heights[i]){
                ps.pop();
            }

            nxtsml[i] = (ps.top());
            ps.push(i);
        }
        int ans = 0;
        for(int i=0;i<len;i++){
            ans = max(ans,abs(nxtsml[i] - presml[i] -1) * heights[i]);
        }

        return ans;
    }
};
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