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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Absolute Permutation
  5. Discussions

Absolute Permutation

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529 Discussions

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  • lherrada
     + 0 comments

    This a is bit more intuitive approach. 

    def condition(i: int, k: int, n: int, a: set) -> bool:
    return 0 < i + k <= n and (i + k) not in a


def absolutePermutation(n: int, k: int) -> int | list[int]:
    ans = [-1 for _ in range(n + 1)]
    a = set()
    for i in range(1, n + 1):
        if condition(i, -k, n, a) is True:
            ans[i] = i - k
            a.add(i - k)
        elif condition(i, k, n, a) is True:
            ans[i] = i + k
            a.add(i + k)
        else:
            return [-1]
    return ans[1:]

  • vuhuutuan0
     + 0 comments

    can't understand question

  • dorirgob
     + 0 comments

    Java solution:

    public static List<Integer> absolutePermutation(int n, int k) {
  // Write your code here

  List<Integer> result = new ArrayList<>();

  if (k == 0) {
    for (int i = 1; i <= n; i++) {
      result.add(i);
    }
    return result;
  }
  if (n % (2 * k) != 0 || k > n || n == 0) {
    return Collections.singletonList(-1);
  }

  boolean add = true; // add or substract
  for (int i = 1; i <= n; i++) {
    if (add) {
      result.add(i + k);

    } else {
      result.add(i - k);
    }

    if (i % k == 0) {
      add = !add;
    }
  }

  return result;
}
}

  • alex_k5
     + 1 comment

    Python

    def absolutePermutation(n, k):

    if k !=0 and n% (2*k) !=0:
    return [-1]
if k == 0:
    return list(range(1, n+1))
return [(u + k if (u-1) % (2*k) + 1 <= k else u - k) for u in range(1, n+1)]

  • dhihanlaksanaaj1
     + 0 comments

    swift

    func absolutePermutation(n: Int, k: Int) -> [Int] {
    var results: [Int] = [Int]()
    var isUpper: Bool = false
    
    if k == 0 { return Array(1 ... n) }
    
    for number in 0 ... n - 1 {        
        if number % k == 0 { isUpper = !isUpper }
        
        if isUpper { results.append((number + 1) + k) }
        else { results.append((number + 1) - k) }
    
        if results.last! > n { return [-1] }
    }
    return results
}