Generate an N-length permutation such that absolute difference between adjacent elements are present in the range [2, 4]

Given a positive integer N, the task is to construct a permutation of first N natural numbers such that the absolute difference between adjacent elements is either 2, 3, or 4. If it is not possible to construct such a permutation, then print “-1”.

Examples:

Input: N = 4
Output: 3 1 4 2
Explanation:
Consider a permutation {3, 1, 4, 2}. Now, the absolute difference between adjacent elements are {2, 3, 2}.

Input: N = 9
Output: 9 7 5 3 1 4 2 6 8 

Approach: The given problem can be solved by grouping up consecutive even and odd elements together to construct the permutation. Follow the steps below to solve the problem:

• If the value of N is less than 4 then print -1 as it is impossible to construct a permutation according to the given conditions for N less than 4.
• Initialize a variable, say i as N, and perform the following steps below:
  • If the value of i is even, then decrement the value of i by 1.
  • Iterate until the value of i is at least 1 and print the value of i and decrement the value of i by 2.
  • Print 4 and 2 and update the value of i to 6.
  • Iterate in the range [i, N] and print the value of i and increment the value of i by 2.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

// Function to print the permutation of
// size N with absolute difference of
// adjacent elements in range [2, 4]
void getPermutation(int N)
{
    // If N is less than 4
    if (N <= 3) {
        cout << -1;
        return;
    }

    int i = N;

    // Check if N is even
    if (N % 2 == 0)
        i--;

    // Traverse through odd integers
    while (i >= 1) {
        cout << i << " ";
        i -= 2;
    }

    cout << 4 << " " << 2 << " ";

    // Update the value of i
    i = 6;

    // Traverse through even integers
    while (i <= N) {
        cout << i << " ";
        i += 2;
    }
}

// Driver Code
int main()
{
    int N = 9;
    getPermutation(N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to print the permutation of
// size N with absolute difference of
// adjacent elements in range [2, 4]
static void getPermutation(int N)
{

    // If N is less than 4
    if (N <= 3)
    {
        System.out.print(-1);
        return;
    }

    int i = N;

    // Check if N is even
    if (N % 2 == 0)
        i--;

    // Traverse through odd integers
    while (i >= 1)
    {
        System.out.print(i + " ");
        i -= 2;
    }

    System.out.print(4 + " " + 2 +" ");

    // Update the value of i
    i = 6;

    // Traverse through even integers
    while (i <= N)
    {
        System.out.print(i + " ");
        i += 2;
    }
}

// Driver Code
public static void main(String[] args)
{
    int N = 9;

    getPermutation(N);
}
}

// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach

# Function to print permutation of
# size N with absolute difference of
# adjacent elements in range [2, 4]
def getPermutation(N):

    # If N is less than 4
    if (N <= 3):
        print(-1)
        return

    i = N

    # Check if N is even
    if (N % 2 == 0):
        i -= 1

    # Traverse through odd integers
    while (i >= 1):
        print(i, end = " ")
        i -= 2

    print(4, 2, end = " ")

    # Update the value of i
    i = 6

    # Traverse through even integers
    while (i <= N):
        print( i, end = " ")
        i += 2

# Driver Code
if __name__ == '__main__':
    N = 9
    getPermutation(N)

    # This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to print the permutation of
// size N with absolute difference of
// adjacent elements in range [2, 4]
static void getPermutation(int N)
{

    // If N is less than 4
    if (N <= 3)
    {

        Console.Write(-1);
        return;
    }

    int i = N;

    // Check if N is even
    if (N % 2 == 0)
        i--;

    // Traverse through odd integers
    while (i >= 1)
    {
        Console.Write(i + " ");
        i -= 2;
    }

    Console.Write(4 + " " + 2 +" ");

    // Update the value of i
    i = 6;

    // Traverse through even integers
    while (i <= N)
    {
        Console.Write(i +" ");
        i += 2;
    }
}

// Driver Code
public static void Main()
{
    int N = 9;

    getPermutation(N);
}
}

// This code is contributed by SURENDRA_GANGWAR

Javascript

<script>

// JavaScript program for the above approach

// Function to print the permutation of
// size N with absolute difference of
// adjacent elements in range [2, 4]
function getPermutation(N)
{

    // If N is less than 4
    if (N <= 3)
    {
        document.write(-1);
        return;
    }

    let i = N;

    // Check if N is even
    if (N % 2 == 0)
        i--;

    // Traverse through odd integers
    while (i >= 1)
    {
        document.write(i + " ");
        i -= 2;
    }

    document.write(4 + " " + 2 + " ");

    // Update the value of i
    i = 6;

    // Traverse through even integers
    while (i <= N)
    {
        document.write(i + " ");
        i += 2;
    }
}

// Driver Code
let N = 9;

getPermutation(N);

// This code is contributed by Potta Lokesh

</script>

Output:

9 7 5 3 1 4 2 6 8

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