Reverse Pairs
<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>
Binary Indexed Tree; Segment Tree; Array; Binary Search; Divide and Conquer; Ordered Set; Merge Sort
class Solution:
    def reversePairs(self, nums: List[int]) -> int:
        def merge_sort(l, r):
            if l >= r:
                return 0
            mid = (l + r) >> 1
            ans = merge_sort(l, mid) + merge_sort(mid + 1, r)
            t = []
            i, j = l, mid + 1
            while i <= mid and j <= r:
                if nums[i] <= 2 * nums[j]:
                    i += 1
                else:
                    ans += mid - i + 1
                    j += 1
            i, j = l, mid + 1
            while i <= mid and j <= r:
                if nums[i] <= nums[j]:
                    t.append(nums[i])
                    i += 1
                else:
                    t.append(nums[j])
                    j += 1
            t.extend(nums[i : mid + 1])
            t.extend(nums[j : r + 1])
            nums[l : r + 1] = t
            return ans

        return merge_sort(0, len(nums) - 1)
