<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 l,int h)
    {
        vector<int>temp;
        int count=0;
        int mid=(l+h)/2;
        int i=l,j=mid+1;
        while(i<=mid && j<=h)
        {

          if(static_cast<long long int>(nums[i])>2*(static_cast<long long int>(nums[j]))) 
          {
                count+=h-j+1;
                temp.push_back(nums[i]);
                i++;
          }
          else
          {
            temp.push_back(nums[j]);
            j++;
          }
        }
          if(i<=mid)
        {
            while(i<=mid)
            {
                temp.push_back(nums[i]);
                i++;
                
            }
        }
        if(j<=h)
        {
            while(j<=h)
            {
                temp.push_back(nums[j]);
                j++;
            }
        }
        sort(temp.begin(),temp.end(),greater<int>());

        for(int k=l;k<=h;k++)
        {
            nums[k]=temp[k-l];
        }
        return count;
    }

    int divide(vector<int>& nums,int l,int h)
    {
        if(l<h)
        {
            return divide(nums,l,(l+h)/2)+divide(nums,1+(l+h)/2,h)+ merge(nums,l,h);
        }
        return 0;
    }

    int reversePairs(vector<int>& nums) 
    {
       return divide(nums,0,nums.size()-1);
    }
};
232114