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 {
public:
    int reversePairs(vector<int>& nums) {
        int n = nums.size();
        int t[n];
        function<int(int, int)> mergeSort = [&](int l, int r) -> int {
            if (l >= r) {
                return 0;
            }
            int mid = (l + r) >> 1;
            int ans = mergeSort(l, mid) + mergeSort(mid + 1, r);
            int i = l, j = mid + 1, k = 0;
            while (i <= mid && j <= r) {
                if (nums[i] <= nums[j] * 2LL) {
                    ++i;
                } else {
                    ans += mid - i + 1;
                    ++j;
                }
            }
            i = l;
            j = mid + 1;
            while (i <= mid && j <= r) {
                if (nums[i] <= nums[j]) {
                    t[k++] = nums[i++];
                } else {
                    t[k++] = nums[j++];
                }
            }
            while (i <= mid) {
                t[k++] = nums[i++];
            }
            while (j <= r) {
                t[k++] = nums[j++];
            }
            for (i = l; i <= r; ++i) {
                nums[i] = t[i - l];
            }
            return ans;
        };
        return mergeSort(0, n - 1);
    }
};