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

Absolute Permutation

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537 Discussions

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  • peteratcod
     + 0 comments

    So, I initially used list.contains which creates a problem regarding time complexity(O(N2)). Then, chatGPT suggested to use another means of checking if it's used or not..But the logic works and is originally mine.

    public static List<Integer> absolutePermutation(int n, int k) {
    List<Integer> result = new ArrayList<>();
    boolean[] used = new boolean[n + 1];

    for (int i = 1; i <= n; i++) {
        if (i - k > 0 && !used[i - k]) {
            result.add(i - k);
            used[i - k] = true;
        } else if (i + k <= n && !used[i + k]) {
            result.add(i + k);
            used[i + k] = true;
        } else {
            return Arrays.asList(-1);
        }
    }
    return result;
}

  • yashdeora98294
     + 0 comments

    Here is problem solution in python java c++ c and Javascript - https://programmingoneonone.com/hackerrank-absolute-permutation-problem-solution.html

  • raihanbilal711
     + 1 comment

    Hi everybody, With n =10 and k = 1, i obtain 2 3 4 5 6 7 8 9 10 9 but the expected output is 2 1 4 3 6 5 8 7 10 9

    Does somebody could explain why my output is not considered as a good answer ?

    Thanks

  • mootezceam
     + 0 comments

    A simple python solution that passes all the test cases consists of doing +k -k operations then give privilege to - k operation if applicable else + k operation if applicable to get the smallest P. If - k and + k can not be possible return -1. NOTE that using a set to test if the element has been already used is mandatory, otherwise test cases from 9 to 12 will exceed allowed time:

    def absolutePermutation(n, k):
    res = []
    used = set()
    for index in range(1, n + 1):
        lower_option = index - k
        upper_option = index + k
        if lower_option > 0 and lower_option not in used:
            res.append(lower_option)
            used.add(lower_option)
        elif upper_option <= n and upper_option not in used:
            res.append(upper_option)
            used.add(upper_option)
        else:
            return [-1]
    return res

  • alex_cel_lemon
     + 0 comments

    Easy python

    def absolutePermutation(n, k):
    if k == 0: return list(range(1,n+1))
    if (n/k)%2 != 0: return [-1]
    res = []
    up = True
    for i in range(n//k):
        for j in range(k):
            res.append(i*k+1+j + (k if up else -k))
        up = not up
        
        
    return  res