<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:
        
        def merge(i,mid,j,arr):
            
            temp=[]
            count=0
            y=mid+1
            for x in range(i,mid+1):
                while y<=j and arr[x]>arr[y]*2:
                    y+=1
                count+=(y-(mid+1))
        
            x,y=i,mid+1
            while x<=mid and y<=j:
                if arr[x]<arr[y]:
                    temp.append(arr[x])
                    x+=1
                else:
                    temp.append(arr[y])
                    y+=1
            while x<=mid:
                temp.append(arr[x])
                x+=1
            while y<=j:
                temp.append(arr[y])
                y+=1
            for m in range(i,j+1):
                arr[m]=temp[m-i]
            return count
        def mergeSort(i,j,arr):
            count=0
            if i>=j:
                return 0
            mid=(i+j)//2
            count+=mergeSort(i,mid,arr)
            count+=mergeSort(mid+1,j,arr)
            count+=merge(i,mid,j,arr)
            return count
        
        
        n=len(nums)
        res=mergeSort(0,n-1,nums)
        return res

        
        
23900