<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();
        int mA = 0;
        vector<int> dp1(n,0);
        stack<pair<int,int>> s;
        s.push({-1,-1});
        for(int i=0;i<n;i++){
            while(s.top().first >= heights[i]) s.pop();
            dp1[i] = heights[i] * (i - s.top().second);
            s.push({heights[i], i});
        }
        while(!s.empty())   s.pop();
        s.push({-1,n});
        int tmp;
        for(int i=n-1;i>-1;i--){
            while(s.top().first >= heights[i]) s.pop();
            tmp = heights[i] * (s.top().second - i);
            s.push({heights[i], i});
            mA = max(mA, dp1[i]+tmp-heights[i]);
        }
        return mA;
    }
};
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