Generate a Permutation of 1 to N with no adjacent elements difference as 1

Given an integer N, the task is to construct a permutation from 1 to N where no adjacent elements have difference as 1. If there is no such permutation, print -1.

Permutation from 1 to N has all the numbers from 1 to N present exactly once.

Examples:

Input: N = 5 
Output: 4 2 5 3 1
Explanation: As for N = 5, [ 4 2 5 3 1 ] is the only permutation where the difference between the adjacent elements is not 1.

Input: N = 3
Output: -1

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

• Difference between any two even number is more than 1 and similarly, for all odd elements their difference is more than 1
• So, print all the even numbers first followed by the odd numbers.

Follow the steps mentioned below to implement the above approach:

• If N is less than or equal to 3, then no such permutation is possible.
• If N is even, print all even numbers from 2 to N firstly, and then all odds from 1 to N - 1 print all odd numbers.
• If the N is odd, then print all even numbers from 2 to N - 1 and then all odd numbers from 1 to N.

Below is the implementation of the above approach:

// C++ code for the above approach

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

// Function to find the permutation
// which satisfies the given condition
vector<int> permutation(int n)
{
    vector<int> ans;
    if (n <= 3) {
        ans.push_back(-1);
    }

    // If n is even
    if (n % 2 == 0) {
        for (int i = 2; i <= n; i += 2) {
            ans.push_back(i);
        }
        for (int i = 1; i < n; i += 2) {
            ans.push_back(i);
        }
    }

    // If n is odd
    else {
        for (int i = 2; i <= n - 1; i += 2) {
            ans.push_back(i);
        }
        for (int j = 1; j <= n; j += 2) {
            ans.push_back(j);
        }
    }
    return ans;
}

// Driver Code
int main()
{
    int N = 5;
    vector<int> ans = permutation(N);
    for (int x : ans)
        cout << x << " ";
    return 0;
}

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

class GFG {

  // Function to find the permutation
  // which satisfies the given condition
  static List<Integer> permutation(int n)
  {
    List<Integer> ans = new ArrayList<>();
    if (n <= 3) {
      ans.add(-1);
    }

    // If n is even
    if (n % 2 == 0) {
      for (int i = 2; i <= n; i += 2) {
        ans.add(i);
      }
      for (int i = 1; i < n; i += 2) {
        ans.add(i);
      }
    }

    // If n is odd
    else {
      for (int i = 2; i <= n - 1; i += 2) {
        ans.add(i);
      }
      for (int j = 1; j <= n; j += 2) {
        ans.add(j);
      }
    }
    return ans;
  }

  // Driver Code
  public static void main (String[] args) {
    int N = 5;
    List<Integer> ans = permutation(N);
    for (Integer x : ans)
      System.out.print(x + " ");
  }
}

// This code is contributed by hrithikgarg03188.

# Python program to generate permutation of 1 to n 
# with no adjacent element difference as 1
def permutation(n):

    # for storing the resultant permutations
    ans = []

    if n <= 3:
        ans.append(-1)

    # if n is even
    if n % 2 == 0:
        i = 0
        while i <= n:
            ans.append(i)
            i += 2
        i = 1
        while i < n:
            ans.append(i)
            i += 2

    # if n is odd
    else:
        i = 2
        while i <= n-1:
            ans.append(i)
            i += 2
        j = 1
        while j <= n:
            ans.append(j)
            j += 2
    return ans

# Driver Code
if __name__ == '__main__':

    n = 5
    ans = permutation(n)
    for i in ans:
        print(i, end=" ")

        # This code is contributed by Amnindersingh1414.

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

public class GFG {

  // Function to find the permutation
  // which satisfies the given condition
  static List<int> permutation(int n) {
    List<int> ans = new List<int>();
    if (n <= 3) {
      ans.Add(-1);
    }

    // If n is even
    if (n % 2 == 0) {
      for (int i = 2; i <= n; i += 2) {
        ans.Add(i);
      }
      for (int i = 1; i < n; i += 2) {
        ans.Add(i);
      }
    }

    // If n is odd
    else {
      for (int i = 2; i <= n - 1; i += 2) {
        ans.Add(i);
      }
      for (int j = 1; j <= n; j += 2) {
        ans.Add(j);
      }
    }
    return ans;
  }

  // Driver Code
  public static void Main(String[] args) {
    int N = 5;
    List<int> ans = permutation(N);
    foreach (int x in ans)
      Console.Write(x + " ");
  }
}

// This code is contributed by gauravrajput1

  <script>
        // JavaScript code for the above approach

// Function to find the permutation
// which satisfies the given condition
function permutation(n)
{
    let ans=[];
    if (n <= 3) {
        ans.push(-1);
    }

    // If n is even
    if (n % 2 == 0) {
        for (let i = 2; i <= n; i += 2) {
            ans.push(i);
        }
        for (let i = 1; i < n; i += 2) {
            ans.push(i);
        }
    }

    // If n is odd
    else {
        for (let i = 2; i <= n - 1; i += 2) {
            ans.push(i);
        }
        for (let j = 1; j <= n; j += 2) {
            ans.push(j);
        }
    }
    return ans;
}

// Driver Code

    let N = 5;
    let ans = permutation(N);
    for (let x of ans)
        document.write( x + " ");
   
      // This code is contributed by Potta Lokesh

    </script>

2 4 1 3 5 

Time complexity: O(N)
Auxiliary Space: O(N) because it is using extra space for vector ans.

mayank007rawa

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Article Tags :

• Misc
• Pattern Searching
• DSA
• permutation

Practice Tags :

• Misc
• Pattern Searching
• permutation