<p>Given an integer array <code>nums</code>, return <em>the number of <strong>reverse pairs</strong> in the array</em>.</p>

<p>A <strong>reverse pair</strong> is a pair <code>(i, j)</code> where:</p>

<ul>
	<li><code>0 &lt;= i &lt; j &lt; nums.length</code> and</li>
	<li><code>nums[i] &gt; 2 * nums[j]</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,2,3,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The reverse pairs are:
(1, 4) --&gt; nums[1] = 3, nums[4] = 1, 3 &gt; 2 * 1
(3, 4) --&gt; nums[3] = 3, nums[4] = 1, 3 &gt; 2 * 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,4,3,5,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The reverse pairs are:
(1, 4) --&gt; nums[1] = 4, nums[4] = 1, 4 &gt; 2 * 1
(2, 4) --&gt; nums[2] = 3, nums[4] = 1, 3 &gt; 2 * 1
(3, 4) --&gt; nums[3] = 5, nums[4] = 1, 5 &gt; 2 * 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>

using namespace __gnu_pbds;

class Solution {
public:
    int reversePairs(vector<int>& nums) {
        /*
        Rewrite the condition in a different manner
        nums[i] / 2 > nums[j]
        
        for each element, we want to know how many previous elements/2 are larger
        For index i, if there are k larger elements before it, we get k distinct pairs
        
        How to find number of larger elements? Store the previous elements / 2 in a sorted
        array
        Search for an element that is greater than nums[j]
        if no such element then no pair
        if you find such element then len(arr) - index is your k value
        
        
        So how to keep a sorted structure? Using rb_tree in C++ is a good idea
        
        */
        
        tree<pair<long long,long long>, null_type, greater<pair<long long,long long>>, rb_tree_tag, tree_order_statistics_node_update> t;
        long long ans = 0;
        /*
        This is a great piece of code for logn searche. What would we be without c++
        
        order_of_key returns the elements that are greater than the query, which is why we use greater<pair> above
        for negative values the % gets muddy, so use ceil() and typecast it back to long long
        */
        for(long long i = 0; i < nums.size(); i++){
            ans += t.order_of_key({nums[i], INT_MAX});
            t.insert({(long long) ceil((float) nums[i]/2), i});
        }
        return ans;
        
        
    }
};
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