# Arrange first N natural numbers such that absolute difference between all adjacent elements > 1

Given an integer N. The task is to find the permutation of first N natural numbers such that the absolute difference between any two consecutive numbers > 1. If no such permutation is possible then print -1.

Examples:

Input: N = 5
Output: 5 3 1 4 2

Input: N = 3
Output: -1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: There may be many such arrangements possible but one of the most common and greedy approach is to arrange all odd numbers in decreasing (or increasing) order and after that arrange all even numbers in decreasing (or increasing) order. Note that if N = 3 or N = 2 then there will be no such arrangement possible and if N = 1 then the sequence will consist of a single element i.e. 1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to print the required permutation ` `void` `arrange(``int` `N) ` `{ ` ` `  `    ``if` `(N == 1) { ` `        ``cout << ``"1"``; ` `        ``return``; ` `    ``} ` ` `  `    ``// No permutation is possible ` `    ``// satisfying the given condition ` `    ``if` `(N == 2 || N == 3) { ` `        ``cout << ``"-1"``; ` `        ``return``; ` `    ``} ` ` `  `    ``// Maximum even and odd elements < N ` `    ``int` `even = -1, odd = -1; ` `    ``if` `(N % 2 == 0) { ` `        ``even = N; ` `        ``odd = N - 1; ` `    ``} ` `    ``else` `{ ` `        ``odd = N; ` `        ``even = N - 1; ` `    ``} ` ` `  `    ``// Print all odd elements in decreasing order ` `    ``while` `(odd >= 1) { ` `        ``cout << odd << ``" "``; ` ` `  `        ``// Next element must be odd ` `        ``odd = odd - 2; ` `    ``} ` ` `  `    ``// Print all even elements in decreasing order ` `    ``while` `(even >= 2) { ` `        ``cout << even << ``" "``; ` ` `  `        ``// Next element must be even ` `        ``even = even - 2; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 5; ` `    ``arrange(N); ` ` `  `    ``return` `0; ` `} `

## Java

// Java implementation of the approach
class GFG
{

// Function to print the required
// permutation
static void arrange(int N)
{
if (N == 1)
{
System.out.println(“1”);
return;
}

// No permutation is possible
// satisfying the given condition
if (N == 2 || N == 3)
{
System.out.println(“-1”);
return;
}

// Maximum even and odd elements < N int even = -1, odd = -1; if (N % 2 == 0) { even = N; odd = N - 1; } else { odd = N; even = N - 1; } // Print all odd elements in // decreasing order while (odd >= 1)
{
System.out.print(odd);
System.out.print(” “);

// Next element must be odd
odd = odd – 2;
}

// Print all even elements in
// decreasing order
while (even >= 2)
{
System.out.print(even);
System.out.print(” “);

// Next element must be even
even = even – 2;
}
}

// Driver code
public static void main(String[] args)
{
int N = 5;
arrange(N);
}
}

// This code is contributed
// by Akanksha Rai

## Python 3

 `# Python 3 implementation of the approach ` ` `  `# Function to print the required permutation ` `def` `arrange(N): ` ` `  `    ``if` `(N ``=``=` `1``) : ` `        ``print``(``"1"``) ` `        ``return` ` `  `    ``# No permutation is possible ` `    ``# satisfying the given condition ` `    ``if` `(N ``=``=` `2` `or` `N ``=``=` `3``) : ` `        ``print``(``"-1"``) ` `        ``return` ` `  `    ``# Maximum even and odd elements < N ` `    ``even ``=` `-``1` `    ``odd ``=` `-``1` `    ``if` `(N ``%` `2` `=``=` `0``): ` `        ``even ``=` `N ` `        ``odd ``=` `N ``-` `1` `    ``else` `: ` `        ``odd ``=` `N ` `        ``even ``=` `N ``-` `1` ` `  `    ``# Print all odd elements in  ` `    ``# decreasing order ` `    ``while` `(odd >``=` `1``): ` `        ``print``(odd, end ``=` `" "``) ` ` `  `        ``# Next element must be odd ` `        ``odd ``=` `odd ``-` `2` ` `  `    ``# Print all even elements in  ` `    ``# decreasing order ` `    ``while` `(even >``=` `2``): ` `        ``print``(even, end ``=` `" "``) ` ` `  `        ``# Next element must be even ` `        ``even ``=` `even ``-` `2` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``N ``=` `5` `    ``arrange(N) ` ` `  `# This code is contributed by ita_c `

## C#

 `// C# implementation of the approach ` `using` `System;  ` ` `  `class` `GFG  ` `{  ` ` `  `// Function to print the required  ` `// permutation ` `static` `void` `arrange(``int` `N) ` `{ ` `    ``if` `(N == 1)  ` `    ``{ ` `        ``Console.WriteLine(``"1"``); ` `        ``return``; ` `    ``} ` ` `  `    ``// No permutation is possible ` `    ``// satisfying the given condition ` `    ``if` `(N == 2 || N == 3)  ` `    ``{ ` `        ``Console.WriteLine(``"-1"``); ` `        ``return``; ` `    ``} ` ` `  `    ``// Maximum even and odd elements < N ` `    ``int` `even = -1, odd = -1; ` `    ``if` `(N % 2 == 0)  ` `    ``{ ` `        ``even = N; ` `        ``odd = N - 1; ` `    ``} ` `    ``else` `    ``{ ` `        ``odd = N; ` `        ``even = N - 1; ` `    ``} ` ` `  `    ``// Print all odd elements in ` `    ``// decreasing order ` `    ``while` `(odd >= 1)  ` `    ``{ ` `        ``Console.Write(odd); ` `        ``Console.Write(``" "``); ` `     `  `        ``// Next element must be odd ` `        ``odd = odd - 2; ` `    ``} ` ` `  `    ``// Print all even elements in ` `    ``// decreasing order ` `    ``while` `(even >= 2)  ` `    ``{ ` `        ``Console.Write(even); ` `        ``Console.Write(``" "``); ` ` `  `        ``// Next element must be even ` `        ``even = even - 2; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main()  ` `{  ` `    ``int` `N = 5; ` `    ``arrange(N); ` `} ` `} ` ` `  `// This code is contributed  ` `// by Shivi_Aggarwal `

## PHP

 `= 1)  ` `    ``{  ` `        ``echo` `\$odd``, ``" "``;  ` ` `  `        ``// Next element must be odd  ` `        ``\$odd` `= ``\$odd` `- 2;  ` `    ``}  ` ` `  `    ``// Print all even elements in  ` `    ``// decreasing order  ` `    ``while` `(``\$even` `>= 2)  ` `    ``{  ` `        ``echo` `\$even``, ``" "``;  ` ` `  `        ``// Next element must be even  ` `        ``\$even` `= ``\$even` `- 2;  ` `    ``}  ` `}  ` ` `  `// Driver code  ` `\$N` `= 5;  ` `arrange(``\$N``); ` ` `  `// This code is contributed by Ryuga ` `?> `

Output:

```5 3 1 4 2
```

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