Arrays: Program to rotate an array to right

To rotate and array to right means to shift the elements to right.
The approach for the right rotation is an easy-to-understand method where the 'k' is the number of rotations that should be performed.
The array is rotated to right by shifting the elements of the arrays to the next element in the array.
In right rotation, the last element of the array will be shifted to the first position in the array.

Approach

1. To understand the concept better, let's take an example of an array of five elements array = {10, 20, 30, 40, 50}
2. If we rotate this array to the right by one index, every element would be moved to the right by one place.
3. The last element after moving to the right will come at the first place of the array. So after the right rotation
   the array will look like array = {50, 10, 20, 30, 40}

Complexity

Time Complexity : O(n*k), where n is the number of elements in the array and k is the number of times we need to perform right rotation on the array
Since k can be at most n-1(k = k%n), the time complexity in the worst case is O(n^2)
Space Complexity : O(1) as we are not using extra space

Below is a sample program to illustrate Array Rotation.
We have created a class, precisely rotate right, which takes an array and number of rotations(k) as parameters.

Java Code

    import java.util.Arrays;

    class RotateRight{
       
        public static void rightRotation(int[] arr, int k) {
            k = k % arr.length;
            for(int i = 0; i < k; i++) {	
                  int j, last;		
                  last = arr[arr.length - 1];	//stores last element of the array
                  for(j = arr.length - 1; j > 0; j--) {		
                    arr[j] = arr[j - 1];	    //shift array by one position
                }		
                  arr[0] = last;        //last element is added to first position 
            }
        }
        public static void main(String[] args) {
            int[] array = {10, 20, 30, 40, 50};
     
            int k = 1;   // no of times to rotate the array
     
            System.out.println("Original Array");
            System.out.println(Arrays.toString(array));
            rightRotation(array,k);
            System.out.println("After right rotation Array");
            System.out.println(Arrays.toString(array));
      }
    }

Output

Original Array :
[10, 20, 30, 40, 50]
Array after right rotation :
[50, 10, 20, 30, 40]

Alternate Approach

import java.util.Arrays;

class RotateRight{

    public static void rotate(int[] nums, int k) {       
        // if k is more than length of array, k = k % length of array is same as running the 
        // rotation k time. so 5 % 3 = 2.
        // So rotating array 2 times is same as rotating 5 times when k is more than length
            if(k > nums.length){
                k = k % nums.length;
            }
        
        // this is the reversal algorithm
        // rotate part of array from start until leaving the last k elements
        // then rotate the last k elements
        // then rotate the complete array
            reverse(0, nums.length-1-k, nums);
            reverse(nums.length-k, nums.length-1, nums);
            reverse(0, nums.length-1, nums);
       
    }
    
    public static void reverse(int start, int end, int[] arr){
        
        while(start < end){
            int temp = arr[start];
            arr[start] = arr[end];
            arr[end] = temp;
            
            start++;
            end--;
        }
    }

    public static void main(String args[]){
        int [] nums = {1,2,3,4,5,6,7}; 
        int k = 3;
        System.out.println("Original Array");
        System.out.println(Arrays.toString(nums));
        rotate(nums,k);
        System.out.println("After right rotation " + k + " times");
        System.out.println(Arrays.toString(nums));

    }
    
}

Output

Original Array
[1, 2, 3, 4, 5, 6, 7]
After right rotation 3 times
[5, 6, 7, 1, 2, 3, 4]