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 muliple 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++

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// C++ implementation of the
// above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Funtion 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);
}

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Java

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// Java program for the above approach
class GFG{ 
  
// Funtion 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

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Python3

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# 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);

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C#

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

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Output:

6 12 18 24 30 36


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

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Improved By : dewantipandeydp, Code_Mech