understanding the cyclic rotation codility challenge?

Question

I want to start by saying thank you for the help first.

I am tackling the cyclic rotation problem where you have to shift the contents of an list/array to the right and effectively wrapping the elements around so for example:

For example, given

 A = [3, 8, 9, 7, 6]
 K = 3

the function should return [9, 7, 6, 3, 8]. Three rotations were made:

 [3, 8, 9, 7, 6] -> [6, 3, 8, 9, 7]
 [6, 3, 8, 9, 7] -> [7, 6, 3, 8, 9]
 [7, 6, 3, 8, 9] -> [9, 7, 6, 3, 8]

my code is as follows:

def solution(A, K):
    new_list = []
    first_index = 0
    for i in range(0, K):
        for num in range(0, len(A) - 1):
            new_list.insert(0, A.pop())
            print('new list {}'.format(new_list))
            print('old list {}'.format(A))
            if len(new_list) == 3:
                new_list.insert(len(new_list), A.pop(first_index))
                print(new_list)

after 3 rotations I get the list as A = [8, 9, 7, 6, 3] so to me it appears to place the last element from A to the front of the new_list.

So any help or points in the right direction would be helpful thank you again.

Asked By: user2152012

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Source

Answer 1

You can simply do it with this code.

def solution(A, K):
    K = K % len(A)
    return A[-K:] + A[:-K]

You can check more ways to do this here.
Efficient way to rotate a list in python

Answered By: Lead Developer

Answer 2

So I realised I only had to loop around K times and check for an empty list. And what I got in the end is:

def solution(A, K):
    for i in range(0, K): # will perform K iterations of the below code
        if A == []: # check if list is empty
            return A # return A if A is the empty list
        A.insert(0, A.pop()) # inserts at the first index of A the last element of A
    return A # will return the list A

Answered By: user2152012

Answer 3

Another Solution using deque from the collections module. It has an in-built rotate function.

from collections import deque
def solution(A, K):
    m=deque(A)
    m.rotate(K)
    return list(m)

Answered By: God Is One

Answer 4

def slice(A):
    B = []
    for i  in range(0,len(A)) :
        B.append(A[-1+i])
    return B 

def solution(A, K):
    for i in range (1,K+1):
        A = slice(A)
    return A

Answered By: seyyah

Answer 5

def solution(A,K):        
    for k in np.arange(K):
        B=[]
        for i in range(len(A)):
            B.append(A[i-1])
        A=B
    return B

Answered By: DanielWarda

Answer 6

The solution I found seems more of a clean code I guess

def solution(A, K):

        n = len(A)`enter code here`
        K = K % n # Ensure K is within the range of array size 
        rotated_arr = A[K:] + A[:K] # Rotating from end to the starting
        return rotated_arr
    A = [1, 2, 3, 4]
    K = int(input())
    rotated_arr = solution(A, K)
    print(rotated_arr)

Answered By: Farood Codes

understanding the cyclic rotation codility challenge?

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