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:
function 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 = 3the 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 = 1the function should return [0, 0, 0]
Given
A = [1, 2, 3, 4] K = 4the 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.
function solution(A, K) {
// performance is not a concern so the simplest solution is probably the best
// rotate the array k times
for (let i = 0; i < K; ++i){
// loop through the array until assign the previous value and shuffle along
let previous = A[A.length -1]
for (let j = 0; j < A.length; ++j){
let temp = A[j]
A[j] = previous
previous = temp
}
}
return A
}function solution(A, K) {
// performance is not a concern so the simplest solution is probably the best
// just repeat the rotation process k times
for (let i = 0; i < K; ++i){
// loop through the array and assign the previous value to the current position and store the current value as the next previous. previous starts as the last value as it loops round
let previous = A[A.length -1]
for (let j = 0; j < A.length; ++j){
let temp = A[j]
A[j] = previous
previous = temp
}
}
return A
}function solution(A, K) {
// performance is not a concern so the simplest solution is probably the best
// just repeat the rotation process k times
for (let i = 0; i < K; ++i){
// loop through the array and assign the previous value to the current position and store the current value as the next previous. previous starts as the last value as it loops round
let previous = A[A.length -1]
for (let j = 0; j < A.length; ++j){
let temp = A[j]
A[j] = previous
previous = temp
}
}
return A
}The solution obtained perfect score.