<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<long> &nums, int l, int m, int r) {
        int count = 0;
        int leftN = m - l + 1;
        int rightN = r - m;

        vector<long> left(leftN);
        vector<long> right(rightN);

        for(int i = l; i <= m; i++) {
            left[i - l] = nums[i];
        }
        for(int i = m + 1; i <= r; i++) {
            right[i - m - 1] = nums[i];
        }

        // update current count as both left and right array are sorted
        for(int i = 0; i < leftN; i++) {
            count += lower_bound(right.begin(), right.end(), left[i]/2) - right.begin();
        }

        // merge
        int i = l, lc = 0, rc = 0;
        while(lc < leftN && rc < rightN) {
            if(left[lc] <= right[rc]) {
                nums[i++] = left[lc++];
            } else {
                nums[i++] = right[rc++];
            }
        }
        while(lc < leftN) {
            nums[i++] = left[lc++];
        }
        while(rc < rightN) {
            nums[i++] = right[rc++];
        }

        return count;
    }

    int mergeSort(vector<long> &nums, int l, int r) {
        if(l >= r) {
            return 0;
        }

        int mid = (l + r)/2;

        int leftCount = mergeSort(nums, l, mid);
        int rightCount = mergeSort(nums, mid + 1, r);
        int currentCount = merge(nums, l, mid, r);
        
        return leftCount + rightCount + currentCount;
    }

    int reversePairs(vector<int>& nums) {
        int n = nums.size();
        vector<long> copy(n);

        for(int i = 0; i < n; i++) {
            copy[i] = (long)(2 * (long)nums[i]);
        }

        return mergeSort(copy, 0, n-1);
    }
};
173524