Count inversions in an array || Merge Sort

Problem link

Problem Statement: Given an array of N integers, count the inversion of the array.

What is an inversion of an array? Definition: for all i & j < size of array, if i < j then you have to find pair (A[i],A[j]) such that A[j] < A[i].

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>>Basically this problem required a direct implementation of merge sort. 

>>The number of inversion is the number of time it finds the array mismatched during a merge sort process. In a merge sort process we divide the array into two halves. until we have the unit of the array elements. Then just comparing the unit variables and put them in right places and merge back to the array.

Code

#include <bits/stdc++.h> 
long long Merge(long long *arr, long long *temp, long long left, long long mid, long long right){
    long long cnt = 0;
    long long i = left;
    long long j = mid;
    long long k = left;
    while((i<=mid-1) && (j<=right)){
        if(arr[i]<=arr[j]){
            temp[k++] = arr[i++];
        }
        else{
            temp[k++] = arr[j++];
            cnt = cnt + (mid-i);
        }
    }
    
    // leftovers
    while(i<=mid-1){
        temp[k++] = arr[i++];
    }
    while(j<=right){
        temp[k++] = arr[j++];
    }
    
    for( i = left; i<=right;i++){
        arr[i] = temp[i];
    }
    return cnt;
}

long long mergeSort(long long *arr, long long *temp, long long left,long long right){
    long long mid, cnt = 0;
    if(right>left){
        mid = (left+right)/2;
        cnt += mergeSort(arr,temp,left,mid);
        cnt += mergeSort(arr,temp,mid+1,right);
        cnt += Merge(arr,temp,left,mid+1,right);
    }
    return cnt;
}
long long getInversions(long long *arr, int n){
    long long temp[n];
    return mergeSort(arr,temp,0,n-1);
}