/*
Description:
Note: This kata is a translation of this (Java) one: http://www.codewars.com/kata/rotate-array. I have not translated this first one as usual because I did not solved it, and I fear not being able to solve it (Java is not my cup of... tea). @cjmcgraw, if you want to use my translation on your kata feel free to use it.

Create a function named "rotate" that takes an array and returns a new one with the elements inside rotated n spaces.

If n is greater than 0 it should rotate the array to the right. If n is less than 0 it should rotate the array to the left. If n is 0, then it should return the array unchanged.

Example:

var data = [1, 2, 3, 4, 5];

rotate(data, 1) // => [5, 1, 2, 3, 4]
rotate(data, 2) // => [4, 5, 1, 2, 3]
rotate(data, 3) // => [3, 4, 5, 1, 2]
rotate(data, 4) // => [2, 3, 4, 5, 1]
rotate(data, 5) // => [1, 2, 3, 4, 5]

rotate(data, 0) // => [1, 2, 3, 4, 5]

rotate(data, -1) // => [2, 3, 4, 5, 1]
rotate(data, -2) // => [3, 4, 5, 1, 2]
rotate(data, -3) // => [4, 5, 1, 2, 3]
rotate(data, -4) // => [5, 1, 2, 3, 4]
rotate(data, -5) // => [1, 2, 3, 4, 5]
Furthermore the method should take ANY array of objects and perform this operation on them:

rotate(['a', 'b', 'c'], 1)     // => ['c', 'a', 'b']
rotate([1.0, 2.0, 3.0], 1)     // => [3.0, 1.0, 2.0]
rotate([true, true, false], 1) // => [false, true, true]
Finally the rotation shouldn't be limited by the indices available in the array. Meaning that if we exceed the indices of the array it keeps rotating.

Example:

var data = [1, 2, 3, 4, 5]

rotate(data, 7)     // => [4, 5, 1, 2, 3]
rotate(data, 11)    // => [5, 1, 2, 3, 4]
rotate(data, 12478) // => [3, 4, 5, 1, 2]
*/
function rotate(array, n) {
  n = n % array.length;
  return array.slice(-n).concat(array.slice(0, -n));
}
