<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>

from sortedcontainers import SortedList
class Solution:
    def reversePairs(self, nums: List[int]) -> int:
        n = len(nums)
        result = 0

        def mergesort(i,m,j):
            x = i
            y = m+1
            temp = [0]*(j-i+1)
            v = 0
            while(x<=m and y<=j):
                temp[v] = min(nums[x], nums[y])
                v+=1
                if nums[x]<=nums[y]:
                    x+=1
                else:
                    y+=1
            while(x<=m):
                temp[v] = nums[x]
                v+=1
                x+=1
            while(y<=j):
                temp[v] = nums[y]
                y+=1
                v+=1
            v = 0
            for x in range(i,j+1):
                nums[x] = temp[v]
                v+=1

        def merge(i,j):
            nonlocal result
            if i>=j:
                return
            m = (i+j)//2
            merge(i,m)
            merge(m+1,j)
            # for x in range(i,m+1):
                # p = bisect_left(nums, (nums[x]/2), m+1, j+1) - 1
                # result += p-m           
            l,r = i, m+1
            while(l<=m and r<=j):
                if nums[l]>2*nums[r]:
                    result += (m+1 - l)
                    r+=1
                else:
                    l+=1

            mergesort(i,m,j)

        merge(0, n-1)
        return result
26800