HackerRank Absolute Permutation challenge

absolutePermutation.py

# We define  to be a permutation of the first  natural numbers in the range . Let  denote the value at position  in permutation  using -based indexing.

#  is considered to be an absolute permutation if  holds true for every .

# Given  and , print the lexicographically smallest absolute permutation . If no absolute permutation exists, print -1.

# Example


# Create an array of elements from  to , . Using  based indexing, create a permutation where every . It can be rearranged to  so that all of the absolute differences equal :

# pos[i]  i   |pos[i] - i|
#   3     1        2
#   4     2        2
#   1     3        2
#   2     4        2
# Function Description

# Complete the absolutePermutation function in the editor below.

# absolutePermutation has the following parameter(s):

# int n: the upper bound of natural numbers to consider, inclusive
# int k: the absolute difference between each element's value and its index
# Returns

# int[n]: the lexicographically smallest permutation, or  if there is none
# Input Format

# The first line contains an integer , the number of queries.
# Each of the next  lines contains  space-separated integers,  and .

# Constraints

# Sample Input

# STDIN   Function
# -----   --------
# 3       t = 3 (number of queries)
# 2 1     n = 2, k = 1
# 3 0     n = 3, k = 0
# 3 2     n = 3, k = 2
# Sample Output

# 2 1
# 1 2 3
# -1
# Explanation

# Test Case 0:

def absolutePermutation(n, k):
    # Write your code here
    output = []
    for i in range(1,n+1):
        for j in range(n,0,-1):
            if abs(i-j) == k and j not in output:
                output.append(j)
    if len(output) == n: 
        print(*output)
    else: print(-1)

absolutePermutation(4,1)

N**2 complexity.