Problem: Rotate an array of n elements to the right by k steps. For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].

rotate

// Problem: Rotate an array of n elements to the right by k steps.
// For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].
// How many different ways do you know to solve this problem?

var array = [1,2,3,4,5,6,7];
var result = [5,6,7,1,2,3,4];

// Solution 1 - Intermediate Array
function rotate(nums, k) {
	var result = [],
		i,
		j = 0;

	if (k > nums.length) {
		k = k % nums.length;
	}

	for (i = 0; i < k; i++) {
		result[i] = nums[nums.length - k + i];
	}

	for (i = k; i < nums.length; i++) {
		result[i] = nums[j];
		j++;
	}

	return result;
}
rotate(array, 3);

// Solution 2 - Bubble Rotate
function IllegalArgumentException(value) {
   this.value = value;
   this.message = "does not conform to the expected format for a zip code";
   this.toString = function() {
	  return this.value + this.message;
   };
}
function rotate(arr, order) {
	if (arr == null || order < 0) {
		throw  new IllegalArgumentException("Illegal argument!");
	}

	var i,
		j,
		temp;
	for (i = 0; i < order; i++) {
		for (j = arr.length - 1; j > 0; j--) {
			temp = arr[j];
			arr[j] = arr[j - 1];
			arr[j - 1] = temp;
		}
	}

	return arr;
}
rotate(array, 3);

// Solution 3 - Reversal
function rotate(arr, order) {
	order = order % arr.length;

	if (arr == null || order < 0) {
		throw new IllegalArgumentException("Illegal argument!");
	}

	// length of first part
	var length = arr.length - order; 
 
	reverse(arr, 0, length - 1);
	reverse(arr, length, arr.length - 1);
	reverse(arr, 0, arr.length - 1);

    return arr;
}
function reverse(arr, left, right) {
	if (arr == null || arr.length == 1) {
		return;
	}

	while(left < right) {
		var temp = arr[left];
		arr[left] = arr[right];
		arr[right] = temp;
		left++;
		right--;
	}

    return arr;
}

rotate(array, 3);