Constructing Palindromic Arrays with First Element as X

Given two integers N and X (1 <= X <= N). The task is to find a permutation P of [1,2,3...N] with the first element as X, such that the difference array Diff[], where Diff_i = P_(i+1) - P_i for all i (1 <= i < N), forms a palindrome. If no such permutation exists, output -1.

Examples:

Input: N = 4, X = 3
Output: P[] = {3, 1, 4, 2}
Explanation: If P[] = {3, 1, 4, 2}, Then:

• First element of P[] is equal to X = 3, which is true.
• Calculating Diff[] for all i (1 <= i < N) gives = {(P₂-P₁), (P₃-P₂), (P₄-P₃)} = {(1-3), (4-1), (2-4)} = {-2, 3, -2}, which is palindrome.

Thus, P[] follows all the conditions, having length N and all elements are from the range [1, N].

Input: N = 3, X = 2
Output: -1
Explanation: It can be verified that no such P[] exists for the given values of N and X.

Approach:

The problem is observation based. The main idea is:

• If N is odd and X is equal to (N + 1)/2, then no such P[] exists.
• Otherwise, a possible permutation P[] will always be there. Which can be obtained by following below sequence:
  • First, we have to print the X.
  • Then print all numbers from 1 to N except X and (N - X + 1).
  • Finally print (N - X + 1).

Step-by-step approach:

• Check Special Case:
  • If N is odd and X is equal to (N + 1) / 2, output -1.
• Generate Permutation:
  • Calculate Y = N - X + 1.
  • Print X.
  • Print all numbers from 1 to N except X and Y.
  • Print Y.

Below is the implementation of the above approach:

4 2 3 1 

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