<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>

const auto init = [](){
       std::cin.tie(nullptr);
       std::cout.tie(nullptr);
       std::ios_base::sync_with_stdio(false);
       return nullptr;
}();

class Solution {
    vector<int> swag;
    void inc(int i) {
        for(; i < swag.size(); i += i&-i) swag[i]++;
    }
    int sum(int i) {
        int o = 0;
        for(; i; i -= i&-i) o += swag[i];
        return o;
    }
    
public:
    int reversePairs(vector<int>& nums) {
        int n = nums.size(), count = 0;
        swag.resize( n + 1 );
        
        vector<pair<int,int>> sorted;
        for( int i = 0; i < n; i++ ) sorted.push_back( {nums[i], n - i} );
        sort( sorted.begin(), sorted.end(), []( auto &x, auto &y ) { return x.first < y.first; } );
        
        for( int i = 0, j = 0; i < n; i++ )
        {
            while( j < n && 2L*sorted[j].first < sorted[i].first ) inc( sorted[j++].second );
            count += sum( sorted[i].second - 1 );
        }
        return count;
    }
};
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