Permutation of first N elements with absolute adjacent difference in increasing order

Given a positive integer N, the task is to construct a permutation from 1 to N such that the absolute difference of elements is in strictly increasing order. 

Note: N cannot be 0 or 1.

Examples:

Input: N = 10
Output: 6 5 7 4 8 3 9 2 10 1
Explanation: abs(6 - 5) i.e., 1 < abs(5 - 7) i.e., 2 < abs(7 - 4) i.e., 3 .... < abs(2 - 10) i.e., 8 < abs(10 - 1) i.e., 9

Input: 3
Output: 2 3 1 
Explanation: abs(2 - 3) = 1 and abs(3 - 1) = 2, 1 < 2 hence it is in strictly increasing order.

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

Observation:

Let's say, you have the i =1 and j = N, the largest absolute difference made is by subtracting 1 and N = (N - 1)

Next Time, i increment by 1, i = 2 and j remains same i.e., N, So, the absolute difference is = (N - 2).
Next Time, i remains same i.e., 2 and j decrement by 1, j = N-1, So, the absolute difference is = (N - 1 - 2) = (N - 3).
Next Time, i increment by 1, i = 3 and j remains same i.e., N-1, So, the absolute difference is = (N - 1 - 3) = (N - 4).
Next Time, i remains same i.e., 3 and j decrement by 1, j = N-2, So, the absolute difference is = (N - 2 - 3) = (N - 5)......

Now, this way the series go, and at last two condition possible,

• When i = j + 1, [If N is odd], absolute difference = 1
• Or, j = i + 1, [If N is even], absolute difference = 1

So, this way the series become for given N, series = (N - 1), (N - 2), (N - 3), .... 3, 2, 1.

Follow the below steps to solve the problem:

• Initialize a pointer i = 1 and j = N.
• Declare an array of size N.
• Run a loop (using iterator x) from 0 to N - 1.
  • If x is even then set, arr[x] = i and increment i by 1.
  • Else then set, arr[x] = j and decrement j by 1.
• After executing the loop, print the array in reverse order.

Below is the implementation of the above approach:

6 5 7 4 8 3 9 2 10 1 

Time Complexity: O(N)
Auxiliary Space: O(N)