<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:
    def reversePairs(self, nums: List[int]) -> int:
        c=self.mergesort(nums,0,len(nums)-1)
        return c

    def mergesort(self,arr,l,h):
        c=0 #count
        if l<h:
            mid=(l+h)//2
            c+=self.mergesort(arr,l,mid)
            c+=self.mergesort(arr,mid+1,h)
            c+=self.merge(arr,l,mid,h)
        return c
    
    def merge(self,a,l,mid,h):
        j=mid+1
        temp=[]
        c=0
        for i in range(l,mid+1):
            while j<=h and a[i]>2*a[j]:
                j+=1
            c+=(j-(mid+1))
        j=mid+1
        i=l
        while i<=mid and j<=h:
            if a[i]<=a[j]:
                temp.append(a[i])
                i+=1
            else:
                temp.append(a[j])
                j+=1
        while i<=mid:
            temp.append(a[i])
            i+=1
        while j<=h:
            temp.append(a[j])
            j+=1
        for i in range(l,h+1):
            a[i]=temp[i-l]
        return c
23800