<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>

using ll = long long;
class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        ll n = heights.size();
        auto a = heights;
        stack<int> st;
        vector<ll> nx(n+5,n);
        for(int i = n-1;i>=0;i--){
            while(not empty(st) and a[st.top()]>=a[i]) st.pop();
            if(not empty(st)) nx[i] = st.top();
            st.emplace(i);
        }
        ll ans = 0;
        st = stack<int>();
        for(int i = 0;i<n;i++){
            while(not empty(st) and a[st.top()]>=a[i]) st.pop();
            ll prv = empty(st) ? -1:st.top();
            ll cur = nx[i] - i + (i - prv - 1);
            ans = max(ans,a[i] * cur);
            st.emplace(i);
        }
        return int(ans);
    }
};
87063