Create Array of distinct parts the place odd listed parts are a number of of left neighbour

Given an integer N, the duty is to generate an array A[] of size N such that it satisfies the next situations for all 1 ≤ i ≤ N−1:

• A_i is a number of of A_i-1 when i is odd
• A_i is just not a number of of A_i-1 when i is even
• All A_i are pairwise distinct
• 1 ≤ A_i ≤ 2⋅N

Observe: If there are a number of solutions print any of them.

    Examples:

Enter: N = 4
Output: 3 6 4 8 

Clarification:  [3, 6, 4, 8] is a sound array as a result of:
A1 = 3 is a number of of A2 = 6
A2 = 6 is just not a number of of A3 = 4
A3 = 4 is a number of of A4 = 8.

Enter: N = 6
Output: 4 8 5 10 6 12

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

Observations:

Let x = N − ⌈N / 2⌉ + 1. Then, the next sequence is legitimate: [x, 2⋅x, x + 1, 2⋅(x + 1), x + 2, …]

• It’s straightforward to see that the weather at odd indices are in growing order from x→N. Equally, the weather at even indices are in growing order from 2⋅x→2⋅N (from 2⋅x→2⋅(N − 1) when N is odd).
• Then, 2⋅x = 2⋅(N − ⌈N / 2⌉ + 1) > N implies the units {x, x + 1, …, N} and {2⋅x, 2⋅(x + 1), …, 2⋅N} are disjoint. Due to this fact, all parts of the above sequence are distinctive.
• A_i is a number of of A_i-1 could be trivially verified to be true for all odd
  A_i is just not a number of of A_i-1 holds true for all even i.

Thus, the supplied sequence fulfills all necessities of the issue, and is due to this fact legitimate!

Observe the beneath steps to resolve the issue:

• Initialize a variable oddElement = (N / 2) + 1 for odd listed parts.
• Initialize a variable evenElement = oddElement * 2 for even listed parts.
• Traverse a loop from 1 until N on i:
  • If i is odd print oddElement.
    • Assign evenElement = oddElement * 2.
    • Increment the evenElement.   
  • Else print the evenElement.
     

Under is the implementation of the above method.

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