Find the good permutation of first N natural numbers

Given an integer N, the task is to print a good permutation of first N natural numbers. Let’s denote the ith element of the permutation be pi.
A good permutation is a permutation such that for all 1 ≤ i ≤ N the following equations hold true,

Basically above expressions mean, no value is equal to its position.

If no such good permutation exists then print -1.



Examples:

Input: N = 1
Output: -1
No good permutation exists.

Input: N = 2
Output: 2 1
Position of 2 is 1 and position of 1 is 2.

Approach: Consider permutation p such that pi = i. Actually, p is a sequence of numbers from 1 to N and ppi = i.
Now the only trick is to change the permutation to satisfy the second equation i.e. pi != i. Let’s swap every two consecutive elements. More formally, for each k: 2k ≤ n let's swap p2k – 1 and p2k. It’s easy to see that the obtained permutation satisfies both the equations for every n with the only exception: when n is odd, there is no answer and we should print -1.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to print the good permutation
// of first N natural numbers
int printPermutation(int n)
{
    // If n is odd
    if (n % 2 != 0)
        cout << -1;
  
    // Otherwise
    else
        for (int i = 1; i <= n / 2; i++)
            cout << 2 * i << " " << 2 * i - 1 << " ";
}
  
// Driver code
int main()
{
    int n = 4;
    printPermutation(n);
  
    return 0;
}
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// Java implementation of the approach
  
class GFG 
{
  
// Function to print the good permutation
// of first N natural numbers
static int printPermutation(int n)
{
    // If n is odd
    if (n % 2 != 0)
    {
        System.out.println("-1");
    }
  
    // Otherwise
    else
        for (int i = 1; i <= n / 2; i++)
        {
            System.out.print(2 * i + " " + ((2 * i) - 1) + " ");
        }
          
    return n;
      
}
  
// Driver code
public static void main(String []args)
{
    int n = 4;
    printPermutation(n);
}
}
  
// This code contributed by Rajput-Ji
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# Python3 implementation of the approach
  
# Function to print the good permutation
# of first N natural numbers
def printPermutation(n):
      
    # If n is odd
    if (n % 2 != 0):
        print(-1);
  
    # Otherwise
    else:
        for i in range(1, (n // 2) + 1):
            print((2 * i), (2 * i - 1), end = " ");
  
# Driver code
n = 4;
printPermutation(n);
  
# This code is contributed by mits
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// C# implementation of the approach
using System;
  
class GFG {
  
// Function to print the good permutation
// of first N natural numbers
static int printPermutation(int n)
{
    // If n is odd
    if (n % 2 != 0)
    {
        Console.WriteLine("-1");
    }
  
    // Otherwise
    else
        for (int i = 1; i <= n / 2; i++)
        {
            Console.WriteLine(2 * i + " " + ((2 * i) - 1) + " ");
        }
          
    return n;
      
}
  
// Driver code
public static void Main(String []args)
{
    int n = 4;
    printPermutation(n);
}
}
  
// This code is contributed by Kirti_Mangal
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<?phP
// PHP implementation of the approach
  
// Function to print the good permutation
// of first N natural numbers
function printPermutation($n)
{
    // If n is odd
    if ($n % 2 != 0)
    {
        echo("-1");
    }
  
    // Otherwise
    else
        for ($i = 1; $i <= $n / 2; $i++)
        {
            echo(2 * $i . " "
               ((2 * $i) - 1) . " ");
        }
          
    return $n;
}
  
// Driver code
$n = 4;
printPermutation($n);
  
// This code contributed
// by Code_Mech.
?>
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Output:
2 1 4 3




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