<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();
        vector<long long>left(n);
        vector<long long>right(n);
        stack<pair <long long , long long >>st;
        stack<pair <long long , long long >>st1;
        for (int i = n - 1 ; i >= 0 ; --i) {
            while (!st.empty() && st.top().first >= heights[i])
                st.pop();
            if (st.empty())
                right[i] = n;
            else
                right[i] = st.top().second;
            st.push({heights[i] , i});
        }
        for (int i = 0; i < n ; ++i) {
            while (!st1.empty() && st1.top().first >= heights[i])
                st1.pop();
            if (st1.empty())
                left[i] = -1;
            else
                left[i] = st1.top().second;
            st1.push({heights[i] , i});
        }
        long long max = -1;
        for(int i = 0 ; i < n ; i++) {
            long long ans = (right[i] - left[i] - 1) * heights[i];
            if(ans > max) max = ans;
        }
        return max;      
    }
};
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