Count Inversion Sort

//count the inversion in a given array while sorting
#include<iostream>
using namespace std;
long long  merge (int arr[],int l, int mid,int r){
    long long inv =0;
    int n1=mid-l+1;
    int n2=r-mid;
    int a[n1];
    int b[n2]; //we are making two temporary array as we have to compare two values and then inputing the lower value
    for(int i=0;i<n1;i++){
        a[i]=arr[l+i];  //as our indexing start with leftmost value of original array
    }
    for(int i=0;i<n2;i++){
        b[i]=arr[mid+1+i];  //as our indexing start with mid+1 value of original array
    }
    int i=0;
    int j=0;
    int k=l;//our indexinging will start with leftmost value
    while(i<n1 && j<n2){
        if(a[i]<b[j]){
            arr[k]=a[i];
            k++;
            i++; //no inversion as a[i]<b[j]
        }
        else{
            arr[k]=b[j];
            inv = inv + n1-i;  //n1-i as a[i+1],a[i+2]... and so on are greater than b[j]   //major catch
            k++;  
            j++;  //inversion as a[i]>b[j]
        }
    }
    while(i<n1){
        arr[k]=a[i];
        k++;
        i++;
    }
    while(j<n2){
        arr[k]=b[j];
        k++;
        j++;
    }
    return inv;
}
long long mergeSort(int arr[],int l,int r){
    long long inv =0;
    if(r>l){
        int mid=(l+r)/2;
        inv += mergeSort(arr,l,mid);  //inversion in left side 
        inv += mergeSort(arr,mid+1,r);  //inversion in right side
        inv += merge(arr,l,mid,r);  //inversion while merging 
    }
    return inv;
}
int main(){
int n ;
cout<<"Enter size of array ";
cin>>n;
int arr[n];
cout<<"Enter values of array ";
for(int i=0;i<n;i++){
    cin>>arr[i];
}
//brut force approach
int inv=0;
for(int i=0;i<n;i++){
    for(int j=i+1;j<n;j++){
        if(arr[i]>arr[j]){
            inv++;
        }
    }
}
cout<<inv<<endl;//time complexity of order o(n^2)
//optimise approach
cout<<(mergeSort(arr,0,n));
return 0;
}