Absolute Permutation in Java

 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:

Test Case 1:

Test Case 2:
No absolute permutation exists, so we print -1 on a new line.

SOLUTION:

import java.io.*;

import java.util.*;

import java.text.*;

import java.math.*;

import java.util.regex.*;

public class Solution {

    static Scanner in = new Scanner(System.in);

    public static void main(String[] args) {

        int t = in.nextInt();

        for (int a0 = 0; a0 < t; a0++) {

            int n = in.nextInt();

            int k = in.nextInt();

            if (k == 0) {

                for (int i = 1; i <= n; i++) {

                    System.out.print(i + " ");

                }

                System.out.println();

            } else if (n % 2 == 0 && n % (2 * k) == 0) {

                int blocks = n / k;

                int currentNumber = k;

                for (int i = 0; i < blocks; i++) {

                    for (int j = 0; j < k; j++) {

                        currentNumber++;

                        System.out.print(currentNumber + " ");

                    }

                    if (i % 2 != 0) {

                        currentNumber = currentNumber + (2 * k);

                    } else {

                        currentNumber = currentNumber - (2 * k);

                    }

                }

                System.out.println();

            } else {

                System.out.println(-1);

            }

        }

    }                

}