Codility 2.2 Arrays CyclicRotation

2 min readJan 16, 2018

Problem

A zero-indexed array A consisting of N integers is given. Rotation of the array means that each element is shifted right by one index, and the last element of the array is moved to the first place. For example, the rotation of array A = [3, 8, 9, 7, 6] is [6, 3, 8, 9, 7] (elements are shifted right by one index and 6 is moved to the first place).

The goal is to rotate array A K times; that is, each element of A will be shifted to the right K times.

Write a function:

def solution(A, K)

that, given a zero-indexed array A consisting of N integers and an integer K, returns the array A rotated K times.

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]

For another example, given

A = [0, 0, 0] K = 1

the function should return [0, 0, 0]

Given

A = [1, 2, 3, 4] K = 4

the function should return [1, 2, 3, 4]

Assume that:

N and K are integers within the range [0..100];
each element of array A is an integer within the range [−1,000..1,000].

In your solution, focus on correctness. The performance of your solution will not be the focus of the assessment.

Solving Process

Assumption, Test Coverage: The array length will be divided by the length of A in case it is bigger than the array length and display the array from K index of array till the end and from the beginning to the end.

It is like a Caesar’s enigma. Moving the index numbers for the alphabets with K numbers further. For example a = 1, b =2 , and c = 3 and such. If we move by 1 (K). z = 1, a =2, b = 3 ,c =4 and such.

Code Solution(mine)

def solution(A, K):     
   if K == 0 or len(A)/K ==0:
      return A
   if K > len(A):
      K = K/len(A)
   A = A[len(A)-K:len(A)] + (A[0:len(A)-K])
   return A

Big-O Calculation

def solution(A, K):     
   if K == 0 or len(A)/K ==0:
      return A
   if K > len(A):
      K = K/len(A)
   A = A[len(A)-K:len(A)] + (A[0:len(A)-K])
   return A
#O(1)

Other People’s code solution

def solution(A, K):if len(A) == 0:return AK = K % len(A)return A[-K:] + A[:-K]O(1)

Learning Points

Negative Index can be used to manipulate the array. Wow. This is what I need.