<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=[0]
        def merge(left,right):
            i,j=0,0
            res=[]
            while i<len(left) and j<len(right):
                if left[i]>right[j]:
                    res.append(right[j])
                    j+=1
                else:
                    res.append(left[i])
                    i+=1
            while i<len(left):
                res.append(left[i])
                i+=1
            while j<len(right):
                res.append(right[j])
                j+=1
            
            i,j=0,0
            while i<len(left) and j<len(right):
                if left[i]>2*right[j]:
                    c[0]+=len(left)-i
                    j+=1
                else:
                    i+=1




            return res


        def partition(nums):
            if len(nums)<=1:
                return nums
            mid=len(nums)//2
            left=partition(nums[:mid])
            right=partition(nums[mid:])
            return merge(left,right)
        partition(nums)
        return c[0]
24300