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

Absolute Permutation

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  • agentkaumal88
     + 0 comments

    Same code logic using python and C# but this may not be the optimal way. If you use a list instead of hashset 4 test cases will be failed due to time limit exceed.

    Python

    def absolutePermutation(n, k):
    permutation = []
    #if you use a list instead of a hashset 4 tests are failed.
    rn = set(range(1, n + 1))
    
    if k == 0:
        return list(rn)
    
    for i in range(1, n + 1):
        low = i - k
        up = i + k
        if low in rn:
            rn.discard(low)
            permutation.append(low)
        elif up in rn:
            rn.discard(up)
            permutation.append(up)
        else:
            return [-1]
            
    return permutation

    C#

    public static List<int> absolutePermutation(int n, int k)
{
	var permutation = new List<int>();
	var rn = Enumerable.Range(1, n);
	/*
	if you use a list
	instead of a hashset
	4 tests are failed.
	*/
	var hs = new HashSet<int>(rn);
	
	if (k == 0)
		return rn.ToList();
	
	for (int i = 1; i <= n; ++i)
	{
		var low = i - k;
		var up = i + k;
		
		if (hs.Contains(low))
		{
			hs.Remove(low);
			permutation.Add(low);
		}
		else if (hs.Contains(up))
		{
			hs.Remove(up);
			permutation.Add(up);
		}
		else
		{
			return new List<int>(){ - 1};
		}
	}
	
	return permutation;
}

  • renushankarjha
     + 0 comments
    def absolutePermutation(n, k):
    # Write your code here
    
    if ((n % 2 != 0) & (k != 0)) | (k > n // 2):
        return [-1]
    arr = [i+1 for i in range(n)]
    if k == 0:
        return arr
    changed = [0 for _ in range(n)]
    for i in range(n - k):
        if (not changed[i]) & (not changed[i+k]):
            arr[i], arr[i+k] = arr[i+k], arr[i]
            changed[i], changed[i+k] = 1, 1
    if 0 in changed[n-k:]:
        return [-1]
    return arr

  • G_Ruiz
     + 1 comment

    Python 3 Solution

        def absolutePermutation(n, k):

            if k == 0:
                    return [x for x in range(1, n+1)]

            elif n % (2*k) != 0:
                    return [-1]

            else:
                    abs_permutation = list()

                    for i in range(0, int(n/(2*k))):

                            for j in range(1, (2*k)+1):

                                    if j <= k:
                                            abs_permutation.append(((i*(k*2))+j)+k)
                                    else:
                                            abs_permutation.append(((i*(k*2))+j)-k)

                    return abs_permutation

  • claravivian75
     + 1 comment

    Can someone help to find why my code in Python 3 would have some cases to have exceeded the time limit?

    def absolutePermutation(n, k):

    pos = [i for i in range(1, n+1)]

    length = len(pos)

    item = pos

    new = []

    if k != 0:

    for i in item:

        for j in range(length):

            if abs(pos[j] - i) == k and pos[j] not in new:
                    new.append(pos[j])
                    break
            continue

    else: new = pos if len(new) != n: new = [-1] return new

  • mohamedcherrou21
     + 0 comments

    java 8 :

    public static List<Integer> absolutePermutation(int n, int k) {
    // Write your code here
        List<Integer> permutation = new ArrayList<>();
        
        if (k == 0) {
            for (int i = 1; i <= n; i++) {
                permutation.add(i);
            }
            return permutation;
        }

        if (n % (2 * k) != 0) {
            return Arrays.asList(-1); // Not possible to form absolute permutation
        }
        
        int sign = 1;
        for (int i = 1; i <= n; i++) {
            permutation.add(i + sign * k);
            if(i%k == 0){
                sign = -sign;
            }
        }

        return permutation;

    }
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