Generate an Array such that the sum of any Subarray is not divisible by its Length

Last Updated : 10 Jan, 2024

Given an integer N. Then the task is to output an array let's say A[] of length N such that:

The sum S, of any subarray, let's say A[L, R] where (L != R) must not leave the remainder as 0 when dividing by the length of A[L, R]. Formally, Sum(A[L, R])%(length of A[L, R]) != 0.
A[] must contain all unique elements.

Note: Subarrays must be chosen at least of length equal to 2.

Examples:

Input: N = 3

Output: A[] = {7, 2, 5}

Explanation: There can be three possible subarrays of length at least equal to 2:

A[1, 2]: The length and sum of subarray is 2 and (7+2) = 9 respectively. We can see that (9%2) != 0
A[2, 3]: The length and sum of subarray is 2 and (2+5) = 7 respectively. We can see that (7%2) != 0
A[1, 3]: The length and sum of subarray is 3 and (7+2+5) = 14 respectively. We can see that (14%3) != 0
There is no subarray such that its sum is perfectly divisible by its length. A[] also contains N distinct elements. Therefore, output array A[] is valid.

Input: N = 2

Output: A[] = {2, 1}

Explanation: It can be verified that both conditions are followed by the output array A[].

Approach: Implement the idea below to solve the problem

The problem is observation based and can be solved using those observations. Let us see some observations:

When N is an even number, we can output a permutation.
When n is an odd number, we can replace N with N + 1 and again we get another array consisting of N distinct integers.
Examples:
N = 7, A[] = {2, 1, 4, 3, 6, 5, 8}
N = 8, A[] = {2, 1, 4, 3, 6, 5, 8, 7}
General formula:

N is Odd: A[] = {2, 1, 3, 4, ... N-1, N-2, N+1}
N is Even: A[] = {2, 1, 3, 4, ... N, N-1}

Steps were taken to solve the problem:

If (N is Odd)
- Output {2, 1, 3, 4, ... N-1, N-2, N+1} using loop.
If (N is Even)
- Output {2, 1, 3, 4, ... N, N-1} using loop.

Code to implement the approach:

C++

#include <iostream>  void GFG(int N) {     // Check if N is odd or even     if (N % 2 == 1) {         // Print elements for the odd N         for (int i = 1; i < N; i += 2)             std::cout << i + 1 << " " << i << " ";         std::cout << N + 1 << std::endl;     } else {         // Print elements for even N         for (int i = 1; i <= N; i += 2)             std::cout << i + 1 << " " << i << " ";     } } int main() {     // Input     int N = 8;     // Function call     GFG(N);     return 0; }

Java

// Java code to implement the approach  // Driver Class public class Main {      // Driver Function     public static void main(String[] args)     {         // Input         int N = 8;          // Function call         Print_array(N);     }      // Method to print A[]     public static void Print_array(int N)     {         if (N % 2 == 1) {             for (int i = 1; i < N; i += 2)                 System.out.print(i + 1 + " " + i + " ");             System.out.println(N + 1);         }         else {             for (int i = 1; i <= N; i += 2)                 System.out.print(i + 1 + " " + i + " ");         }     } }

Python3

# Python code to implement the approach  # Method to print A[] def Print_array(N):     if N % 2 == 1:         for i in range(1, N, 2):             print(i + 1, i, end=" ")         print(N + 1)     else:         for i in range(1, N+1, 2):             print(i + 1, i, end=" ")  # Driver function def main():     # Input     N = 8      # Function call     Print_array(N)  if __name__ == '__main__':     main()  # This code is contributed by ragul21

using System;  class GFG {     static void Main(string[] args)     {         // Input         int N = 8;         // Function call         PrintArray(N);     }      // Method to print elements based on N     static void PrintArray(int N)     {         // Check if N is odd or even         if (N % 2 == 1)         {             // Print elements for the odd N             for (int i = 1; i < N; i += 2)                 Console.Write(i + 1 + " " + i + " ");             Console.WriteLine(N + 1);         }         else         {             // Print elements for even N             for (int i = 1; i <= N; i += 2)                 Console.Write(i + 1 + " " + i + " ");         }     } }

JavaScript

function GFG(N) {      // Check if N is odd or even      if (N % 2 === 1) {          // Print elements for the odd N          for (let i = 1; i < N; i += 2)              console.log(i + 1, i);          console.log(N + 1);      } else {          // Print elements for even N          for (let i = 1; i <= N; i += 2)              console.log(i + 1, i);      }  }  // Input  let N = 8;  // Function call  GFG(N);

Output

2 1 4 3 6 5 8 7

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

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

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Generate an Array such that the sum of any Subarray is not divisible by its Length

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