<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:
    void my_sort(vector<int>& nums, int start, int mid, int end) {
        vector<int> temp;
        std::copy(nums.begin() + start, nums.begin() + end + 1, back_inserter(temp));
        mid -= start;
        end -= start;

        int i = 0, j = mid + 1;
        int ind = start;
        while (i <= mid || j <= end) {
            if (j > end || (i <= mid && temp[i] < temp[j])) {
                nums[ind++] = temp[i++];
            } else {
                nums[ind++] = temp[j++];
            }
        }

       // std::copy(temp.begin(), temp.end(), nums.begin() + start);
    }

    int merge(vector<int>& nums, int start, int end) {
        if (start >= end) return 0;

        int mid = start + (end - start) / 2;

        int cnt = merge(nums, start, mid) + merge(nums, mid + 1, end);

        for (int i = start, j = mid + 1; i <= mid; i++) {
            while (j <= end && nums[i] > (long)nums[j] * 2) {
                j++;
            }

            cnt += j - (mid + 1);
        }

        // sort(nums.begin() + start, nums.begin() + end + 1);
        my_sort(nums, start, mid, end);

        return cnt;
    }
    int reversePairs(vector<int>& nums) {
        return merge(nums, 0, ssize(nums) - 1);
    }
};
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