<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>

class Solution {
public:
    int reversePairs(vector<int>& nums) {
        int count=0;

        auto mergeSort = [&](auto&& self, int l, int r)->void{
            if(l == r)return;
            int m = (l+r)/2;
            self(self, l, m);
            self(self, m+1, r);

            vector<int> temp(r-l+1);
            int j = m+1;
            int first = m+1;
            for(int i=l, k = 0;i<=m;++i,++k)
            {
                while(first <= r && nums[i]/2.0> nums[first])first++;
                while(j<=r && nums[j]<nums[i])temp[k++]=nums[j++];
                count+=first-m-1;
                temp[k] = nums[i];
            }
            for(int i=0;i<j-l;i++)nums[i+l]=temp[i];
        };
        mergeSort(mergeSort, 0, nums.size()-1);
        return count;
    }
};
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