Generate an N-length array having sum of each subarray divisible by K

Given two positive integers N and K, the task is to generate an array consisting of N distinct integers such that the sum of elements of each subarray of the constructed array is divisible by K.

Examples:

Input: N = 3, K = 3
Output: 3 6 9
Explanation:
The subarrays of the resultant array are {3}, {6}, {3, 6}, {9}, {6, 9}, {3, 6, 9}. Sum of all these subarrays are divisible by K.

Input: N = 5, K = 1
Output: 1 2 3 4 5

Approach: Follow the steps below to solve the problem:

1. Since the sum of elements of each subarray needs to be divisible by K, the most optimal approach would be to construct an array with each element being a multiple of K.
2. Therefore, iterate a loop from i = 1 to i = N and for each value of i, print K * i.

Below is the implementation of the above approach:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to construct an array
// with sum of each subarray
// divisible by K
void construct_Array(int N, int K)
{
    // Traverse a loop from 1 to N
    for (int i = 1; i <= N; i++) {

        // Print i-th multiple of K
        cout << K * i << " ";
    }
}

// Driver Code
int main()
{
    int N = 3, K = 3;
    construct_Array(N, K);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG
{

  // Function to construct an array
  // with sum of each subarray
  // divisible by K
  static void construct_Array(int N, int K)
  {

    // Traverse a loop from 1 to N
    for (int i = 1; i <= N; i++)
    {

      // Print i-th multiple of K
      System.out.print(K * i + " ");
    }
  }

  // Driver Code
  public static void main(String[] args)
  {
    int N = 3, K = 3;
    construct_Array(N, K);
  }
}

// This code is contributed by code hunt.

Python3

# Python program for the above approach

# Function to construct an array
# with sum of each subarray
# divisible by K
def construct_Array(N, K) :

    # Traverse a loop from 1 to N
    for i in range(1, N + 1):

        # Pri-th multiple of K
        print(K * i, end = " ")

# Driver Code
N = 3
K = 3
construct_Array(N, K)

# This code is contributed by splevel62.

C#

// C# program to implement
// the above approach
using System;
public class GFG
{

  // Function to construct an array
  // with sum of each subarray
  // divisible by K
  static void construct_Array(int N, int K)
  {

    // Traverse a loop from 1 to N
    for (int i = 1; i <= N; i++)
    {

      // Print i-th multiple of K
      Console.Write(K * i + " ");
    }
  }

  // Driver Code
  public static void Main(String[] args)
  {
    int N = 3, K = 3;
    construct_Array(N, K);
  }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript program for the above approach

// Function to construct an array
// with sum of each subarray
// divisible by K
function construct_Array(N, K)
{
    // Traverse a loop from 1 to N
    for (let i = 1; i <= N; i++) {

        // Print i-th multiple of K
       document.write( K * i + " ");
    }
}

// Driver Code
    let N = 3, K = 3;
    construct_Array(N, K);

// This code is contributed by todaysgaurav

</script>

Output:

3 6 9

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