Generate a unique Array of length N with sum of all subarrays divisible by N

Given an integer N, the task is to make an array of unique elements of length N such that all subarrays sum modulo N equals to zero.

Examples:

Input: N = 6 
Output: 6 12 18 24 30 36 
Explanation:
Since all elements are a multiple of 6. Hence all subarrays add up to a sum divisible by 6.
Input: N = 4 
Output: 4 8 12 16 

Approach:
We can observe that for all subarrays to be divisible by N, the elements of the array need to be a multiple of N.
Illustration:

For N = 4, if we consider the array elements to {4, 8, 12, 16}, All possible subarrays are: 
{4}, {8}, {12}, {16}, {4, 8}, {8, 12}, {12, 16}, {4, 8, 12}, {8, 12, 16}, {4, 8, 12, 16} 
Hence, all subarrays have a sum divisible by N. 

Hence, to solve the problem, we just need to print {N, 2*N, 3*N, ….., N*N} to get the desired array. 
Below is the implementation of the above approach:

C++

// C++ implementation of the
// above approach

#include <bits/stdc++.h>
using namespace std;

// Function to print the required
// array
void makeArray(int a[], int n)
{
    // Print the array
    for (int i = 1; i <= n; i++)
        cout << i * n << " ";
}

// Driver Program
int main()
{
    int N = 6;
    int arr[N];
    makeArray(arr, N);
}

Java

// Java program for the above approach
class GFG{

// Function to print the required
// array
static void makeArray(int a[], int n)
{

    // Print the array
    for(int i = 1; i <= n; i++)
       System.out.print(i * n + " ");
}

// Driver code
public static void main(String[] args)
{
    int N = 6;
    int arr[] = new int[N];

    makeArray(arr, N);
}
}

// This code is contributed by Pratima Pandey

Python3

# Python3 implementation of the
# above approach

# Function to print the
# required array
def makeArray(n):

    # Print Array
    for i in range(n):
        print((i + 1) * n, end =" ")


# Driver code
n = 6;
makeArray(n);

C#

// C# program for the above approach
using System;
class GFG{

// Function to print the required
// array
static void makeArray(int []a, int n)
{

    // Print the array
    for(int i = 1; i <= n; i++)
    Console.Write(i * n + " ");
}

// Driver code
public static void Main()
{
    int N = 6;
    int []arr = new int[N];

    makeArray(arr, N);
}
}

// This code is contributed by Code_Mech

Javascript

<script>
    // javascript program for the above approach

// Function to print the required
// array

function makeArray(n)
{

    // Print the array
    for(var i = 1; i <= n; i++)
    document.write(i * n + " ");
}

// Driver code

    var N = 6;

    makeArray(N);



</script>

Output:

6 12 18 24 30 36

Time Complexity: O(N) 
Auxiliary Space: O(1)