<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 count = 0;

    void single(const vector<long long>& x, const vector<long long>& y) {
        long long i = 0;
        long long j = 0;
        while (i < x.size() && j < y.size()) {
            if (x[i] > 2 * y[j]) {
                count += x.size() - i; // Count pairs where x[i] > 2 * y[j]
                j++;
            } else {
                i++;
            }
        }
    }

    void merge(vector<long long>& a, vector<long long>& b, vector<long long>& ans) {
        long long i = 0;
        long long j = 0;
        long long k = 0;
        while (i < a.size() && j < b.size()) {
            if (a[i] < b[j]) {
                ans[k++] = a[i++];
            } else {
                ans[k++] = b[j++];
            }
        }
        while (i < a.size()) {
            ans[k++] = a[i++];
        }
        while (j < b.size()) {
            ans[k++] = b[j++];
        }
    }

    void mergeSort(vector<long long>& nums) {
        long long size = nums.size();
        if (size <= 1) return;

        long long mid = size / 2;
        vector<long long> left(mid);
        vector<long long> right(size - mid);

        for (long long i = 0; i < mid; ++i) {
            left[i] = nums[i];
        }
        for (long long i = mid; i < size; ++i) {
            right[i - mid] = nums[i];
        }

        mergeSort(left);
        mergeSort(right);

        single(left, right);
        merge(left, right, nums);
    }

    int reversePairs(vector<int>& nums) {
        vector<long long> nums_ll(nums.begin(), nums.end());
        mergeSort(nums_ll);
        return count;
    }
};

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