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

Absolute Permutation

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  • 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

  • gcysne
     + 0 comments

    Simple approach in python:

        P = [0]*(n+1)
    # Try to create P that fulfills |P[i] - i| = k
    for i in range(1,n+1):
        if P[i] == 0:
            # P[i] can be either i+k or i-k.
            # Greedy approach, try to fit the lexicographically smallest first
            if i-k > 0 and P[i-k] == 0:
                P[i], P[i-k] = i-k, i
            elif i+k <= n and P[i+k] == 0:
                P[i], P[i+k] = i+k, i
            else:
                # Not possible
                return [-1]
    # Skips the first element, added just for better code readability
    return P[1:]