Discover permutation such that adjoining components has distinction of no less than 2

Given a quantity N, the duty is to discover a permutation A[] of first N integers such that absolutely the distinction of adjoining components is no less than 2 i.e., | A_i+1 − A_i | ≥ 2 for all 0 ≤ i < N−1.If no such permutation exists, print -1.

Examples:

Enter: N = 4
Output: 3 1 4 2

Clarification: Right here A[] = {3, 1, 4, 2} satisfies the given situation. 
Since  | A_i+1 − A_i | ≥ 2 for all 0 ≤ i < N−1

Enter: N = 2
Output: -1
Clarification: No such permutation is feasible that satisfies the given situation

Method: The issue could be solved based mostly on the next commentary:

• If N = 2 or N = 3, then no array exists that fulfill the above situation.
• In any other case, array exists that fulfill the above situation resembling:
  • First print all odd numbers from N to 1 in lowering order and after that print all even numbers in lowering order. 

Comply with the steps talked about beneath to implement the concept:

• If N = 2 or N = 3, print -1.
• In any other case, verify whether or not N is odd and even:
  • If N is odd, iterate a loop from N to 1 to print all odd numbers after that iterate one other loop from N – 1 to 2 to print even numbers.
  • If N is even, iterate a loop from N – 1 to 1 to print all odd numbers after that iterate one other loop from N to 2 to print even numbers.

Beneath is the implementation of the above method.

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