<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 {
public:
    int merge(vector<int>& nums, int left, int mid, int right) {
        vector<int> temp(right - left + 1);
        int count = 0;
        int i = left, j = mid + 1, k = 0;

        while (i <= mid && j <= right) {
            if (nums[i] > 2ll * nums[j]) {
                count += mid - i + 1;
                j++;
            } else {
                i++;
            }
        }
        inplace_merge(nums.begin() + left, nums.begin() + mid + 1, nums.begin() + right + 1);
        // while (i <= mid) {
        //     temp[k++] = nums[i++];
        // }
        // while (j <= right) {
        //     temp[k++] = nums[j++];
        // }
        // for (int p = 0; p < temp.size(); p++) {
        //     nums[left + p] = temp[p];
        // }
        return count;
    }

    int merge_sort(vector<int>& nums, int left, int right) {
        if (left >= right) return 0;
        int mid = left + (right - left) / 2;
        int count1 = merge_sort(nums, left, mid);
        int count2 = merge_sort(nums, mid + 1, right);
        int count3 = merge(nums, left, mid, right);
        return count1 + count2 + count3;
    }
    int reversePairs(vector<int>& nums) {
        int n = nums.size();

        return merge_sort(nums, 0, n - 1);
    }
};
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