<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:
        n = len(nums)

        def mergesort(start, end):
            if start >= end:
                return

            mid = start + (end - start) // 2
            mergesort(start, mid)
            mergesort(mid + 1, end)

            merge(start, mid, end)

        def merge(start, mid, end):
            # Count qualifying pairs using range
            nonlocal res
            j = mid + 1  # Do not reset this for each i
            for i in range(start, mid + 1):
                while j <= end and nums[i] > nums[j] * 2:
                    j += 1
                res += j - (mid + 1)

            # Common merging logic
            left, right = nums[start : mid + 1], nums[mid + 1 : end + 1]
            i, j, k = 0, 0, start

            while i < len(left) and j < len(right):
                if left[i] <= right[j]:
                    nums[k] = left[i]
                    i += 1
                else:
                    nums[k] = right[j]
                    j += 1
                k += 1

            nums[k:end + 1] = left[i:] + right[j:]

        res = 0
        mergesort(0, n - 1)

        return res

29600