Codility Lesson2 - CyclicRotation

Codility Lesson2 - CyclicRotation

https://app.codility.com/demo/results/trainingX872UE-GT9/

Test results - Codility

 

문제

An 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 an 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.

 

 

정답1

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def rotation(A):
    length = len(A)
    newA = [0] * length

    for i in range(1,length):
        newA[i]= A[i-1]
    
    newA[0]=A[length-1]
    return newA

def solution(A, K):
    # Implement your solution here

    for j in range(K):
        newA= rotation(A)     
        A=newA

    return newA

 

 

결과1

->  원소가 없을 때, K가 0일 때를 고려하지 않음. A의 갯수만큼 []*len(A)하기 때문에 K가 0일 때를 언급 하지 않으면 [0]만 반환

---

 

정답-Retry

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

def rotation(A):
    length = len(A)
    newA = [0] * length

    for i in range(1,length):
        newA[i]= A[i-1]
    
    newA[0]=A[length-1]
    return newA

def solution(A, K):
    # Implement your solution here
    length = len(A)
    newA = [0] * length

    if K==0:
        newA=A
    elif length >=1:
        for j in range(K):
            newA= rotation(A)     
            A=newA
    
    return newA

 

결과-Retry