Find permutation such that adjacent elements has difference of at least 2

Given a number N, the task is to find a permutation A[] of first N integers such that the absolute difference of adjacent elements is at least 2 i.e., | A_i+1 ? A_i | ? 2 for all 0 ? i < N?1.If no such permutation exists, print -1.

Examples:

Input: N = 4
Output: 3 1 4 2
?Explanation: Here A[] = {3, 1, 4, 2} satisfies the given condition. 
Since  | A_i+1 ? A_i | ? 2 for all 0 ? i < N?1

Input: N = 2
Output: -1
Explanation: No such permutation is possible that satisfies the given condition

Approach: The problem can be solved based on the following observation:

• If N = 2 or N = 3, then no array exists that satisfy the above condition.
• Otherwise, array exists that satisfy the above condition such as:
  • First print all odd numbers from N to 1 in decreasing order and after that print all even numbers in decreasing order. 

Follow the steps mentioned below to implement the idea:

• If N = 2 or N = 3, print -1.
• Otherwise, check whether N is odd or even:
  • If N is odd, iterate a loop from N to 1 to print all odd numbers after that iterate another loop from N – 1 to 2 to print even numbers.
  • If N is even, iterate a loop from N – 1 to 1 to print all odd numbers after that iterate another loop from N to 2 to print even numbers.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the array
void findArray(int n)
{
  if (n == 2 || n == 3)
    cout << -1;
  else {

    // If n is odd
    if ((n % 2) == 1) {

      // Loops to print the permutation
      for (int i = n; i >= 1; i -= 2) {
        cout << i << " ";
      }
      for (int i = n - 1; i >= 2; i -= 2) {
        cout << i << " ";
      }
    }
    // If n is even
    else {

      // Loops to print the permutation
      for (int i = n - 1; i >= 1; i -= 2) {
        cout << i << " ";
      }
      for (int i = n; i >= 2; i -= 2) {
        cout << i << " ";
      }
    }
  }
}

// Driver Code
int main()
{
  int N = 4;

  // Function call
  findArray(N);
  return 0;
}

// This code is contributed by aarohirai2616.

Java

// Java code to implement the approach

import java.io.*;
import java.util.*;

public class GFG {

    // Function to find the array
    public static void findArray(int n)
    {
        if (n == 2 || n == 3)
            System.out.println(-1);
        else {

            // If n is odd
            if ((n % 2) == 1) {

                // Loops to print the permutation
                for (int i = n; i >= 1; i -= 2) {
                    System.out.print(i + " ");
                }
                for (int i = n - 1; i >= 2; i -= 2) {
                    System.out.print(i + " ");
                }
            }

            // If n is even
            else {

                // Loop to print the permutation
                for (int i = n - 1; i >= 1; i -= 2) {
                    System.out.print(i + " ");
                }
                for (int i = n; i >= 2; i -= 2) {
                    System.out.print(i + " ");
                }
            }
            System.out.println();
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        int N = 4;

        // Function call
        findArray(N);
    }
}

Python3

# Python3 code to implement the approach

# Function to find the array
def findArray(n):

  if (n == 2 or n == 3) :
    print(-1);

  else :

    # If n is odd
    if ((n % 2) == 1) :

      # Loops to print the permutation
      for i in range(n, 0, -2) :
        print(i,end=" ");

      for i in range(n - 1, 1, -2) :
        print(i,end = " ");

    # If n is even
    else :

      # Loops to print the permutation
      for i in range( n - 1, 0, -2) :
        print(i,end=" ");

      for i in range(n, 1, -2) :
        print(i, end=" ");

# Driver Code
if __name__ == "__main__" :
    N = 4;

    # Function call
    findArray(N);

    #  This code is contributed by AnkThon

C#

// C# code to implement the approach
using System;

public class GFG {

  // Function to find the array
  static void findArray(int n)
  {
    if (n == 2 || n == 3)
      Console.WriteLine(-1);
    else {

      // If n is odd
      if ((n % 2) == 1) {

        // Loops to print the permutation
        for (int i = n; i >= 1; i -= 2) {
          Console.Write(i + " ");
        }
        for (int i = n - 1; i >= 2; i -= 2) {
          Console.Write(i + " ");
        }
      }

      // If n is even
      else {

        // Loop to print the permutation
        for (int i = n - 1; i >= 1; i -= 2) {
          Console.Write(i + " ");
        }
        for (int i = n; i >= 2; i -= 2) {
          Console.Write(i + " ");
        }
      }
      Console.WriteLine();
    }
  }

  static public void Main()
  {

    // Code
    int N = 4;

    // Function call
    findArray(N);
  }
}

// This code is contributed by lokeshmvs21.

Javascript

// Javascript code to implement the approach

    // Function to find the array
    function findArray(n)
    {
        if (n == 2 || n == 3)
            process.stdout.write(-1);
        else {

            // If n is odd
            if ((n % 2) == 1) {

                // Loops to print the permutation
                for (let i = n; i >= 1; i -= 2) {
                   process.stdout.write(i + " ");
                }
                for (let i = n - 1; i >= 2; i -= 2) {
                    process.stdout.write(i + " ");
                }
            }

            // If n is even
            else {

                // Loop to print the permutation
                for (let i = n - 1; i >= 1; i -= 2) {
                    process.stdout.write(i + " ");
                }
                for (let i = n; i >= 2; i -= 2) {
                    process.stdout.write(i + " ");
                }
            }

        }
    }

    // Driver code

        let N = 4;

        // Function call
        findArray(N);

    // This code is contributed by aarohirai2616.

Output

3 1 4 2 

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