Java Program for Reversal algorithm for right rotation of an array

Given an array, right rotate it by k elements. 

After K=3 rotation 

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 
                6, 7, 8, 9, 10}
          k = 3
Output: 8 9 10 1 2 3 4 5 6 7

Input: arr[] = {121, 232, 33, 43 ,5}
           k = 2
Output: 43 5 121 232 33

Note : In the below solution, k is assumed to be smaller than or equal to n. We can easily modify the solutions to handle larger k values by doing k = k % n
Algorithm:

rotate(arr[], d, n)
  reverseArray(arr[], 0, n-1) ;
  reverse(arr[], 0, d-1);
  reverse(arr[], d, n-1);

Below is the implementation of above approach: 

Java

// Java program for right rotation of
// an array (Reversal Algorithm)
import java.io.*;

class GFG
{
    // Function to reverse arr[]
    // from index start to end
    static void reverseArray(int arr[], int start,
                             int end)
    {
        while (start < end)
        {
           int temp = arr[start];
           arr[start] = arr[end];
           arr[end] = temp;
           start++;
           end--;
         }
    }

    // Function to right rotate
    // arr[] of size n by d
    static void rightRotate(int arr[], int d, int n)
    {
       reverseArray(arr, 0, n - 1);
       reverseArray(arr, 0, d - 1);
       reverseArray(arr, d, n - 1);
    }

    // Function to print an array
    static void printArray(int arr[], int size)
    {
       for (int i = 0; i < size; i++)
          System.out.print(arr[i] + " ");
    }

    public static void main (String[] args)
    {
        int arr[] = {1, 2, 3, 4, 5,
                     6, 7, 8, 9, 10};

    int n = arr.length;
    int k = 3;

    rightRotate(arr, k, n);
    printArray(arr, n);

    }
}
// This code is contributed by Gitanjali.

Output:

8 9 10 1 2 3 4 5 6 7 

Time Complexity: O(N)

Auxiliary Space: O(1)

Please refer complete article on Reversal algorithm for right rotation of an array for more details!