We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Implementation
  4. Absolute Permutation
  5. Discussions

Absolute Permutation

Sort by

recency

|

538 Discussions

|

  • kpprajwal54
     + 0 comments

    solution in javascript:

    function absolutePermutation(n, k) {
   const useSet = new Set();
   const res = [];
   
   for(let i = 1; i<=n; i++){
    if((i-k)>=1 && !useSet.has(i-k)){
        res.push(i-k);
        useSet.add(i-k);
    } else if((i+k) <= n && !useSet.has(i+k)){
        res.push(i+k);
        useSet.add(i+k);
    } else{
        return [-1];
    }
   }
   
   return res;
}

  • 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