<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 m(vector<int>& arr,int s,int e,int mid){
        int n1=mid-s+1;
        int n2=e-mid;
        int k=s;
        int j=mid+1;
        int i;
        for(int i=s; i<=mid; i++){
            while(j<=e && arr[i] > 2*(long long)arr[j]){
                j++;
            }
            count += j-(mid+1);
        }
        int *arr1=new int[n1];
        int *arr2=new int[n2];
        for(int i=0;i<n1;i++){
            arr1[i]=arr[k++];
        }
         for(int i=0;i<n2;i++){
            arr2[i]=arr[k++];
        }
         i=0,j=0;
        k=s;
      

        while(i<n1&&j<n2){
            if(arr1[i]<=arr2[j]){
                arr[k++]=arr1[i++];
            

            }
            else{
                arr[k++]=arr2[j++];
            
            }
        }
         while(i<n1){
            arr[k++]=arr1[i++];
          
            
        }
        while(j<n2){
            arr[k++]=arr2[j++];
          
            
        }

        delete []arr1;
        delete []arr2;
    }
    void ms(vector<int>& arr,int s, int e){
        if(s>=e){
            return ;
        }
        int mid=s+(e-s)/2;
        ms(arr,s,mid);
        ms(arr,mid+1,e);
        m(arr,s,e,mid);

    }
    int reversePairs(vector<int>& nums) {
        int n=nums.size();
       ms(nums,0,n-1);

        
        return count;
        
    }
};
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