Cyclic Rotation Exercise from Codility (JavaScript/NodeJS)

CyclicRotation.js

/**
 * Shift items in the array to the right by number of indexes 
 * @param {array} A
 * @param {int} K
 * @return {array} 
 */
function solution(A, K) {
    // if A is meant to be shifted by its own length (K) then just return the array
    if (A.length === K || K === 0) {
        return A
    }
    
    // Run K number of times saving last element in the array as a temporary variable, adding it to the front of the array and removing the last element
    for (let i = 0; i < K; i++) {
        let lastElement = A[A.length - 1]
        A.unshift(lastElement)
        A.pop()
    }
    return A
}

My solution

function solution(A, K) {
  // edge case destroyer
  if (A.length === 0 || A.length === K || K === 0) return A;
  let finalArr = A;
  // add characters to the start of finalArr
  for (let i = 0; i < K; i++) {
  finalArr.unshift(A[A.length - 1 - i]);
  }
  // assign sliced array to variable
  finalArr = finalArr.slice(0, -K);
  console.log(finalArr);
  // return sliced array
  return finalArr;
}

This is my solution for 100%

function solution(A, K) {
    const size = A.length // sets length
    if(K>size) K = (K - (size*(Math.floor(K/size)))) // simplifies rotation (also realized you could do k = k % size)
    if(size === 0) return []; // if the string is empty return it as is
    
    const insertLeft = A.slice(-1 * K)
    const rotateRight = A.slice(0, -1 * K)
    const result = insertLeft.concat(rotateRight)
    return result
}

My solution

function solution(A, K) {
  // edge case destroyer
  if (A.length === 0 || A.length === K || K === 0) return A;
  let finalArr = A;
  // add characters to the start of finalArr
  for (let i = 0; i < K; i++) {
  finalArr.unshift(A[A.length - 1 - i]);
  }
  // assign sliced array to variable
  finalArr = finalArr.slice(0, -K);
  console.log(finalArr);
  // return sliced array
  return finalArr;
}

Solution from elendil7 fail performance tests