<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 {
private:
    int count = 0;
    void merge(vector<int>& nums, int low, int high, int mid) {
        vector<int> temp;

        int left = low, right = mid + 1;
        while (left <= mid && right <= high) {
            if (nums[left] <= nums[right]) {
                temp.push_back(nums[left]);
                left++;
            }

            if (nums[right] <= nums[left]) {
                temp.push_back(nums[right]);
                right++;
            }
        }
        while (left <= mid) {
            temp.push_back(nums[left]);
            left++;
        }

        while (right <= high) {
            temp.push_back(nums[right]);
            right++;
        }
        for (int i = low; i <= high; i++) {
            nums[i] = temp[i - low];
        }
    }
    void countPairs(vector<int>& nums, int low, int high, int mid) {
        int left = low;
        int right = mid + 1;
        for (int i = low; i <= mid; i++) {
            while (right <= high && nums[i] > (long long)2 * nums[right]) {
                right++;
            }

            count += right - (mid + 1);
        }
    }
    void mergeSort(vector<int>& nums, int low, int high) {
        if (low >= high)
            return;

        int mid = (low + high) / 2;

        mergeSort(nums, low, mid);
        mergeSort(nums, mid + 1, high);
        countPairs(nums, low, high, mid);
        merge(nums, low, high, mid);
    }

public:
    int reversePairs(vector<int>& nums) {
        int n = nums.size();
        mergeSort(nums, 0, n - 1);
        return count;
    }
};
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