Count inversions

/*
T.C O(nlogn) S.C O(n)
logic:
*We want pair of elements where first element should be greater and second should be smaller.
*We applied merge sort but while merging two sorted arrays,if left array ie. arr1 ith
 element which is greater than jth element of right array ie. arr2 then all the elements
 after ith element including ith element will be greater than jth element of arr2 because
 both array is sorted so that satisfies our condition so we will add that in answer.
*/

long long int res=0;
class Solution{
  public:
    void merge(int l,int mid,int r,vector <long long int> &arr){
    vector <long long int> arr1,arr2;
    for(long long int i=l;i<=mid;i++)
        arr1.push_back(arr[i]);
    for(long long int i=mid+1;i<=r;i++)
        arr2.push_back(arr[i]);

    long long int i=0,j=0;
    long long int n1=arr1.size();
    long long int n2=arr2.size();
    long long int k=l;

    while(i<n1 && j<n2){
          if(arr1[i]>arr2[j]){
              res=res+(n1-i);  //all elements of left array after ith element and including
              arr[k++]=arr2[j++];  //ith element will be greater than jth element
          }
          else{
              arr[k++]=arr1[i++];
          }
    }
    if(i<n1){
        while(i<n1){
            arr[k++]=arr1[i++];
        }
    }
    if(j<n2){
        while(j<n2)
           arr[k++]=arr2[j++];
    }
}
void count_inversions(long long int l,long long int r,vector <long long int> &arr){
    if(l<r){
        long long int mid=(l+r)/2;
        count_inversions(l,mid,arr);
        count_inversions(mid+1,r,arr);
        merge(l,mid,r,arr);  //l to mid is sorted and mid+1 to r is sorted.
    }
}
long long int inversionCount(long long ar[], long long N)
    {
    res=0;
    vector <long long int> arr;
    for(long long int i=0;i<N;i++)
       arr.push_back(ar[i]);

    count_inversions(0,N-1,arr);
    return res;
}
};