Right rotation of an array using JAVA

Software Development
#1

Given an array, right rotate it by k elements.

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);

}

}

Output:

8 9 10 1 2 3 4 5 6 7