CyclicRotation

Rotate an array to the right by a given number of steps.

Question:

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:

class Solution { public int[] solution(int[] A, int 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.


Solution:

Solution to Codility's Cyclic Rotation problem which is from the Codility Lesson 2: Arrays and, is solved in Java 8 with 100% performance and correctness scores. The goal here is to rotate an array to the right by a given number of steps. You can find the question of this CyclicRotation problem in the Codility website. 


package Codility.Lesson2;

import java.util.Arrays;

public class CyclicRotation {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[] A = { 5, 6, 7, 8, 4, 2, 7, 1 };
		int K = 2;
		int[] result = Solution(A, K);
		System.out.println(Arrays.toString(result));
	}

	public static int[] Solution(int[] A, int K) {
		int temp = 0;
		int len = A.length;

		if (len < 1) {
			return A;
		}

		for (int i = 0; i < K; i++) {
			temp = A[len - 1];
			for (int j = A.length - 1; j > 0; j--) {
				A[j] = A[j - 1];
			}
			A[0] = temp;
		}
		return A;
	}
}