<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>st,st1;
        #define ll long long 

        int n=heights.size();
        vector<int>left(n,-1);
       for (int i = 0; i < n; i++) {
        while(st.size()>0 and heights[i]<=heights[st.top()])st.pop();
        if(st.size()==0) left[i]=-1; 
        else if (heights[i]>heights[st.top()])left[i]=st.top();
        st.push(i);
}
        vector<int>right(n,-1);
       for (int i= n-1;i>=0; i--) {
        while(st1.size()>0 and heights[i]<=heights[st1.top()])st1.pop();
        if(st1.size()==0) right[i]=-1; 
        else if (heights[i]>heights[st1.top()])right[i]=st1.top();
        st1.push(i);
}
        // printvec(left);
        // printvec(right);

        ll maxi=0;
        for(int i=0;i<n;i++)
        {
             int le=(left[i]==-1) ? 0: left[i]+1;
            int re=(right[i]==-1) ? n-1: right[i]-1;
            // maxi=max(maxi,heights[i]*(re-le+1));
            maxi=max(maxi,heights[i]*(re-le+1)*1LL);
        }
        return maxi;
    }
};
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