<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 merge(vector<int> &a, vector<int> &b, vector<int> &res)
{   
    res.clear();
    int i = 0, j = 0;
    while (i < a.size() && j < b.size())
    {
        if (a[i] <= b[j])
            res.push_back(a[i++]);
        else
            res.push_back(b[j++]);
    }
    while(j < b.size()) res.push_back(b[j++]);
    while(i < a.size()) res.push_back(a[i++]);
    return;
}
int count_reverse_in2sorted(vector<int>& v1 , vector<int>& v2){
    int count = 0 , i = 0 , j = 0;
    while(i < v1.size() && j < v2.size()){
        if (v1[i] > 2*(long long)v2[j]){
            count += v1.size() - i;
            j++;
        }
        else i++;
    }
    return count;
}
int merge_sort_reverse_count(vector<int>&v){
    if(v.size() == 1)  return 0;
    vector<int> v_end , v_start;
    int n = v.size();
    for(int i = 0 ; i < n ; i++){
        if(i < n/2)v_start.push_back(v[i]);
        else v_end.push_back(v[i]);
    }
    int count_start = merge_sort_reverse_count(v_start);
    int count_end = merge_sort_reverse_count(v_end);

    int count = count_reverse_in2sorted(v_start,v_end) + count_start + count_end;

    merge(v_start , v_end , v);
    v_start.clear();
    v_end.clear();
    return count;
}
int reversePairs(vector<int>& nums) {
        return merge_sort_reverse_count(nums);
}
};
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