Java Program for Reversal algorithm for right rotation of an array
Last Updated :
23 Jul, 2025
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!
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