<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>

import math
class Solution:
    # TIME n*(logn)*(logn)
    def floor(self, li, key):
        i,j = 0, len(li)-1
        pos = -1
        while i <= j:
            mid = (i+j)//2
            if li[mid] < key:
                pos = mid
                i = mid+1
            else:
                j = mid-1
        return pos
    
    def merge(self, li, l, m, r):
        L, R,n1,n2 = [],[], m-l+1, r-m
        for i in range(n1):
            L.append(li[l+i])
        for i in range(n2):
            R.append(li[m+1+i])
        
		######ANSWER LOGIC#######
        for i in range(n1):
            key = math.ceil(L[i]/2)
            j = self.floor(R, key)
            if j!=-1:
                self.count += j+1
                
        ######MERGE LOGIC########
        i,j,k=0,0,l        
        while i<n1 and j<n2:
            if L[i] <= R[j]:
                li[k] = L[i]
                i+=1
            else:
                li[k] = R[j]
                j+=1
            k+=1
        while i<n1:
            li[k] = L[i]
            i+=1
            k+=1
        while j<n2:
            li[k] = R[j]
            j+=1
            k+=1
    
    def merge_sort(self, li, i, j):
        if i < j:
            m = (i+j)//2
            self.merge_sort(li,i, m)
            self.merge_sort(li,m+1,j)
            self.merge(li, i, m, j)
              
    
    def reversePairs(self, nums: List[int]) -> int:
        #couting inversion logic
        self.count = 0
        self.merge_sort(nums, 0, len(nums)-1)
        return self.count
22700