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

  • Last Updated : 08 Sep, 2021

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 tha 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)

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