## Rotation of Array

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

```Input: [1,2,3,4,5,6,7] and k = 3
Output: [5,6,7,1,2,3,4]

Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]```

1. Using Temporary Array:

We use a temporary array in which we place every element of the input array at its correct position.The `ith` element of the input array should be placed in the `(i + k)%(input_array.length) th` position of the temporary array.

``````/**
* Rotation by temp array
* @param nums
* @param k
*/
public static void rotate_2(int[] nums, int k) {
int[] a = new int[nums.length];
for (int i = 0; i < nums.length; i++) {
a[(i + k) % nums.length] = nums[i];
}
// Copy the temp array to the original array
for (int i = 0; i < nums.length; i++) {
nums[i] = a[i];
}
}``````

The Time Complexity is `O(n)` and Space Complexity is also `O(n)`.

2. By Reversing the Array:

In this approach, we firstly reverse all the elements of the array. Then, reversing the first k elements followed by reversing the rest `n−k` elements gives us the required result.

```Original List  : 1 2 3 4 5 6 7
After reversing all numbers : 7 6 5 4 3 2 1
After reversing first k numbers : 5 6 7 4 3 2 1
After revering last n-k numbers : 5 6 7 1 2 3 4 ```
``````/**
* In place rotation
* @param nums
* @param k
*/
public static void rotate_1(int[] nums, int k) {
int len = nums.length;
// If given k in greater than array length
k = k % len;
reverse(nums, 0, nums.length - 1);
reverse(nums, 0, k - 1);
reverse(nums, k, nums.length - 1);
}

private static void reverse(int[] arr, int start, int end) {
while (start < end) {
int temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}``````

Complexity Analysis

• Time complexity: `O(n)`. `n` elements are reversed a total of three times.
• Space complexity: `O(1)`. No extra space is used.
Arrays Algorithm