Open In App

Rotate an Array – Clockwise or Right

Last Updated : 30 Oct, 2024
Comments
Improve
Suggest changes
Like Article
Like
Report
Try it on GfG Practice
redirect icon

Rotations in the array is defined as the process of rearranging the elements in an array by shifting each element to a new position. This is mostly done by rotating the elements of the array clockwise or counterclockwise.

Table of Content

Types of Rotations in Array

1. Right Rotation (or Clockwise)

Here, The array elements are shifted towards the right. 

left-array-rotation

2. Left Rotation (Or Counter Clockwise)

Here, The array elements are shifted towards the left.

right-arrow-Rotation

In this article, we will discuss about right rotation of the array. You can refer to Left rotate an array by d positions to know about the left rotation of the array.

How to implement rotations in an array?

There are several ways to implement array rotations. Some of the approaches are mentioned below. Here we are considering right rotation. The movements will be just the opposite for left rotation.

Input: arr[] = {1, 2, 3, 4, 5, 6}, d = 2
Output: {5, 6, 1, 2, 3, 4}
Explanation: After first right rotation, arr[] becomes {6, 1, 2, 3, 4, 5} and after the second rotation, arr[] becomes {5, 6, 1, 2, 3, 4}

Input: arr[] = {1, 2, 3}, d = 4
Output: {3, 1, 2}
Explanation: The array is rotated as follows:

  • After first left rotation, arr[] = {3, 1, 2}
  • After second left rotation, arr[] = {2, 3, 1}
  • After third left rotation, arr[] = {1, 2, 3}
  • After fourth left rotation, arr[] = {3, 1, 2}

1. Rotate one by one

At each iteration, shift the elements by one position to the right in a circular fashion (the last element becomes the first). Perform this operation d times to rotate the elements to the right by d positions.

Illustration:

Let us take arr[] = {1, 2, 3, 4, 5, 6}, d = 2.

First Step:
        => Rotate to right by one position.
        => arr[] = {6, 1, 2, 3, 4, 5}
Second Step:
        => Rotate again to right by one position
        => arr[] = {5, 6, 1, 2, 3, 4}
Rotation is done 2 times.
So the array becomes arr[] = {5, 6, 1, 2, 3, 4}

C++ 
// C++ Program to right rotate the array by d positions
// by rotating one element at a time

#include <bits/stdc++.h>
using namespace std;

// Function to right rotate array by d positions
void rotateArr(vector<int>& arr, int d) {
    int n = arr.size();
  
    // Repeat the rotation d times
    for (int i = 0; i < d; i++) {
      
        // Right rotate the array by one position
        int last = arr[n - 1];
        for (int j = n - 1; j > 0; j--) {
            arr[j] = arr[j - 1];
        }
        arr[0] = last;
    }
}

int main() {
    vector<int> arr = { 1, 2, 3, 4, 5, 6 };
    int d = 2;

    rotateArr(arr, d);

    for (int i = 0; i < arr.size(); i++)
        cout << arr[i] << " ";

    return 0;
}
C 
// C Program to right rotate the array by d positions
// by rotating one element at a time

#include <stdio.h>

// Function to right rotate array by d positions
void rotateArr(int arr[], int n, int d) {
    
    // Repeat the rotation d times
    for (int i = 0; i < d; i++) {
      
        // Right rotate the array by one position
        int last = arr[n - 1];
        for (int j = n - 1; j > 0; j--) {
            arr[j] = arr[j - 1];
        }
        arr[0] = last;      
    }
}

int main() {
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int d = 2;

    rotateArr(arr, n, d);

    for (int i = 0; i < n; i++)
        printf("%d ", arr[i]);
    return 0;
}
Java 
// Java Program to right rotate the array by d positions
// by rotating one element at a time

import java.util.Arrays;

class GfG {

    // Function to right rotate array by d positions
    static void rotateArr(int[] arr, int d) {
        int n = arr.length;

        // Repeat the rotation d times
        for (int i = 0; i < d; i++) {

            // Right rotate the array by one position
            int last = arr[n - 1];
            for (int j = n - 1; j > 0; j--) {
                arr[j] = arr[j - 1];
            }
            arr[0] = last;
        }
    }

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

        rotateArr(arr, d);

        for (int i = 0; i < arr.length; i++) {
            System.out.print(arr[i] + " ");
        }
    }
}
Python 
# Python Program to right rotate the array by d positions
# by rotating one element at a time

# Function to right rotate array by d positions
def rotateArr(arr, d):
    n = len(arr)
  
    # Repeat the rotation d times
    for _ in range(d):
      
        # Right rotate the array by one position
        last = arr[n - 1]
        for i in range(n - 1, 0, -1):
            arr[i] = arr[i - 1]
        arr[0] = last

if __name__ == "__main__":
    arr = [1, 2, 3, 4, 5, 6]
    d = 2

    rotateArr(arr, d)

    # Print the rotated array
    for i in range(len(arr)):
        print(arr[i], end=" ")
C# 
// C# Program to right rotate the array by d positions
// by rotating one element at a time

using System;

class GfG {
    
    // Function to right rotate array by d positions
    static void rotateArr(int[] arr, int d) {
        int n = arr.Length;
      
        // Repeat the rotation d times
        for (int i = 0; i < d; i++) {
            
            // Right rotate the array by one position
            int last = arr[n - 1];
            for (int j = n - 1; j > 0; j--)
                arr[j] = arr[j - 1];
            arr[0] = last;      
        }
    }

    static void Main() {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int d = 2;

        rotateArr(arr, d);

        // Print the rotated array
        for (int i = 0; i < arr.Length; i++) {
            Console.Write(arr[i] + " ");
        }
        Console.WriteLine();
    }
}
JavaScript 
// JavaScript Program to right rotate the array by d positions
// by rotating one element at a time

// Function to right rotate array by d positions
function rotateArr(arr, d) {
    let n = arr.length;
  
    // Repeat the rotation d times
    for (let i = 0; i < d; i++) {
      
        // Right rotate the array by one position
        let last = arr[n - 1];
        for (let j = n - 1; j > 0; j--)
            arr[j] = arr[j - 1];
        arr[0] = last;      
    }
}

let arr = [1, 2, 3, 4, 5, 6];
let d = 2;

rotateArr(arr, d);

console.log(arr.join(' '));

Output
5 6 1 2 3 4

Time Complexity: O(n * d)
Auxiliary Space: O(1)

2. Using Temporary Array

The idea is to use a temporary array of size n, where n is the length of the original array. If we right rotate the array by d positions, the last d elements will be in the beginning and the first (n – d) elements will be at the end.

  • Copy the last d elements of the original array into the first d positions of the temporary array
  • Then copy the first n – d elements of the original array to the end of temporary array.
  • Finally, copy all the elements of temporary array back into the original array.

Illustration for Right Rotation by 2 positions:


Below is the implementation of the above approach

C++ 
// C++ Program to right rotate the array by d positions
// using temporary array

#include <bits/stdc++.h>
using namespace std;

// Function to rotate vector
void rotateArr(vector<int>& arr, int d) {
    int n = arr.size();

    // Handle case when d > n
    d %= n;
  
    // Storing rotated version of array
    vector<int> temp(n);

    // Copy last d elements to the front of temp
    for (int i = 0; i < d; i++)
        temp[i] = arr[n - d + i];

    // Copy the first n - d elements to the back of temp
    for (int i = 0; i < n - d; i++)
        temp[i + d] = arr[i];

    // Copying the elements of temp in arr
    // to get the final rotated vector
    for (int i = 0; i < n; i++)
        arr[i] = temp[i];
}

int main() {
    vector<int> arr = { 1, 2, 3, 4, 5, 6 };
    int d = 2;

    rotateArr(arr, d);

    // Print the rotated vector
    for (int i = 0; i < arr.size(); i++)
        cout << arr[i] << " ";

    return 0;
}
C 
// C Program to right rotate the array by d positions
// using temporary array

#include <stdio.h>
#include <stdlib.h>

// Function to rotate array by d positions using a temporary array
void rotateArr(int arr[], int n, int d) {
    
    // Handle case when d > n
    d %= n;
  
    // Storing rotated version of array
    int temp[n];

    // Copy last d elements to the front of temp
    for (int i = 0; i < d; i++)
        temp[i] = arr[n - d + i];

    // Copy the first n - d elements to the back of temp
    for (int i = 0; i < n - d; i++)
        temp[i + d] = arr[i];

    // Copying the elements of temp in arr to get the 
    // final rotated array
    for (int i = 0; i < n; i++)
        arr[i] = temp[i];
}

int main() {
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int d = 2;

    rotateArr(arr, n, d);

    // Print the rotated array
    for (int i = 0; i < n; i++)
        printf("%d ", arr[i]);

    return 0;
}
Java 
// Java Program to right rotate the array by d positions
// using temporary array

import java.util.Arrays;

class GfG {

    // Function to rotate array by d positions using a temporary array
    static void rotateArr(int[] arr, int d) {
        int n = arr.length;

        // Handle case when d > n
        d %= n;

        // Storing rotated version of array
        int[] temp = new int[n];

        // Copy last d elements to the front of temp
        for (int i = 0; i < d; i++)
            temp[i] = arr[n - d + i];

        // Copy the first n - d elements to the back of temp
        for (int i = 0; i < n - d; i++)
            temp[i + d] = arr[i];

        // Copying the elements of temp in arr to get the 
        // final rotated array
        for (int i = 0; i < n; i++)
            arr[i] = temp[i];
    }

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

        rotateArr(arr, d);

        // Print the rotated array
        for (int i = 0; i < arr.length; i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.println();
    }
}
Python 
# Python Program to right rotate the array by d positions
# using temporary array

def rotateArr(arr, d):
    n = len(arr)

    # Handle case when d > n
    d %= n

    # Storing rotated version of array
    temp = [0] * n

    # Copy last d elements to the front of temp
    for i in range(d):
        temp[i] = arr[n - d + i]

    # Copy the first n - d elements to the back of temp
    for i in range(n - d):
        temp[i + d] = arr[i]

    # Copying the elements of temp in arr
    # to get the final rotated array
    for i in range(n):
        arr[i] = temp[i]

if __name__ == "__main__":
    arr = [1, 2, 3, 4, 5, 6]
    d = 2

    rotateArr(arr, d)

    # Print the rotated array
    print(' '.join(map(str, arr)))
C# 
// C# Program to right rotate the array by d positions
// using temporary array

using System;

class GfG {
    
    // Function to rotate array
    static void rotateArr(int[] arr, int d) {
        int n = arr.Length;

        // Handle case when d > n
        d %= n;

        // Storing rotated version of array
        int[] temp = new int[n];

        // Copy last d elements to the front of temp
        for (int i = 0; i < d; i++)
            temp[i] = arr[n - d + i];

        // Copy the first n - d elements to the back of temp
        for (int i = 0; i < n - d; i++)
            temp[i + d] = arr[i];

        // Copying the elements of temp in arr
        // to get the final rotated array
        for (int i = 0; i < n; i++)
            arr[i] = temp[i];
    }

    static void Main() {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int d = 2;

        rotateArr(arr, d);

        // Print the rotated array
        for (int i = 0; i < arr.Length; i++)
            Console.Write(arr[i] + " ");
        Console.WriteLine();
    }
}
JavaScript 
// JavaScript Program to right rotate the array by d positions
// using temporary array

// Function to rotate array
function rotateArr(arr, d) {
    let n = arr.length;

    // Handle case when d > n
    d %= n;
  
    // Storing rotated version of array
    let temp = new Array(n);

    // Copy last d elements to the front of temp
    for (let i = 0; i < d; i++) {
        temp[i] = arr[n - d + i];
    }

    // Copy the first n - d elements to the back of temp
    for (let i = 0; i < n - d; i++) {
        temp[i + d] = arr[i];
    }

    // Copying the elements of temp in arr
    // to get the final rotated array
    for (let i = 0; i < n; i++) {
        arr[i] = temp[i];
    }
}

let arr = [1, 2, 3, 4, 5, 6];
let d = 2;

rotateArr(arr, d);

// Print the rotated array
console.log(arr.join(' '));

Output
5 6 1 2 3 4

Time complexity: O(n), where n is the size of input array arr[]. 
Auxiliary Space: O(n)

3.  Juggling Algorithm

The idea behind Juggling Algorithm is that instead of moving one by one, we can use the concept of cycles. Each cycle is independent and represents a group of elements that will shift among themselves during the rotation. If the starting index of a cycle is i, then the next elements will be present at indices (i + d) % n, (i + 2d) % n, (i + 3d) % n … and so on till we reach back to index i.

So for any index i, we know that element at index i will move to index (i + d) % n. Now, we can simply rotate all elements in the same cycle without interfering with any other cycle.

Working of the above algorithm:


Below is the implementation of the algorithm:

C++ 
// C++ Program to right rotate the array by d positions
// using Juggling Algorithm

#include <bits/stdc++.h>
using namespace std;

// Function to rotate vector
void rotateArr(vector<int> &arr, int d) {
    int n = arr.size();

    // Handle the case where d > size of array
    d %= n;

    // Calculate the number of cycles in the rotation
    int cycles = __gcd(n, d);

    // Process each cycle
    for (int i = 0; i < cycles; i++) {

        // Start index of current cycle
        int currIdx = i;
        int currEle = arr[currIdx];

        // Rotate elements till we reach the start of cycle
        do {
            int nextIdx = (currIdx + d) % n;
            int nextEle = arr[nextIdx];
          
            // Update the element at next index with the current element
            arr[nextIdx] = currEle;
          
            // Update the current element to next element
            currEle = nextEle;
          
            // Move to the next index
            currIdx = nextIdx;
        } while (currIdx != i);
    }
}

int main()
{
    vector<int> arr = {1, 2, 3, 4, 5, 6};
    int d = 2;

    rotateArr(arr, d);

    // Print the rotated vector
    for (int i = 0; i < arr.size(); i++)
        cout << arr[i] << " ";

    return 0;
}
C 
// C Program to right rotate the array by d positions
// using Juggling Algorithm

#include <stdio.h>
#include <stdlib.h>

// Function to rotate array
void rotateArr(int *arr, int n, int d) {
    
    // Handle the case where d > size of array
    d %= n;

    // Calculate the number of cycles in the rotation
    int cycles = gcd(n, d);

    // Process each cycle
    for (int i = 0; i < cycles; i++) {

        // Start index of current cycle
        int currIdx = i;
        int currEle = arr[currIdx];

        // Rotate elements till we reach the start of cycle
        do {
            int nextIdx = (currIdx + d) % n;
            int nextEle = arr[nextIdx];
          
            // Update the element at next index with the current element
            arr[nextIdx] = currEle;
          
            // Update the current element to next element
            currEle = nextEle;
          
            // Move to the next index
            currIdx = nextIdx;
        } while (currIdx != i);
    }
}

// function to compute GCD
int gcd(int a, int b) {
    while (b) {
        int temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

int main() {
    int arr[] = {1, 2, 3, 4, 5, 6};
    int n = sizeof(arr) / sizeof(arr[0]);
    int d = 2;

    rotateArr(arr, n, d);

    // Print the rotated array
    for (int i = 0; i < n; i++)
        printf("%d ", arr[i]);

    return 0;
}
Java 
// Java Program to right rotate the array by d positions
// using Juggling Algorithm

import java.util.Arrays;

class GfG {

    // Function to rotate array
    static void rotateArr(int[] arr, int d) {
        int n = arr.length;

        // Handle the case where d > size of array
        d %= n;

        // Calculate the number of cycles in the rotation
        int cycles = gcd(n, d);

        // Process each cycle
        for (int i = 0; i < cycles; i++) {

            // Start index of current cycle
            int currIdx = i;
            int currEle = arr[currIdx];

            // Rotate elements till we reach the start of cycle
            do {
                int nextIdx = (currIdx + d) % n;
                int nextEle = arr[nextIdx];

                // Update the element at next index with the current element
                arr[nextIdx] = currEle;

                // Update the current element to next element
                currEle = nextEle;

                // Move to the next index
                currIdx = nextIdx;
            } while (currIdx != i);
        }
    }

    // function to compute GCD
    public static int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

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

        rotateArr(arr, d);

        // Print the rotated array
        for (int i = 0; i < arr.length; i++)
            System.out.print(arr[i] + " ");
    }
}
Python 
# Java Program to right rotate the array by d positions
# using Juggling Algorithm

from math import gcd

# Function to rotate list
def rotateArr(arr, d):
    n = len(arr)

    # Handle the case where d > size of array
    d %= n

    # Calculate the number of cycles in the rotation
    cycles = gcd(n, d)

    # Process each cycle
    for i in range(cycles):

        # Start index of current cycle
        currIdx = i
        currEle = arr[currIdx]

        # Rotate elements till we reach the start of cycle
        while True:
            nextIdx = (currIdx + d) % n
            nextEle = arr[nextIdx]

            # Update the element at next index with the current element
            arr[nextIdx] = currEle

            # Update the current element to next element
            currEle = nextEle

            # Move to the next index
            currIdx = nextIdx

            if currIdx == i:
                break

if __name__ == "__main__":
    arr = [1, 2, 3, 4, 5, 6]
    d = 2

    rotateArr(arr, d)

    # Print the rotated list
    for i in range(len(arr)):
        print(arr[i], end=" ")
C# 
// C# Program to right rotate the array by d positions
// using Juggling Algorithm

using System;

class GfG {
    
    // Function to rotate array
    static void RotateArr(int[] arr, int d) {
        int n = arr.Length;

        // Handle the case where d > size of array
        d %= n;

        // Calculate the number of cycles in the rotation
        int cycles = Gcd(n, d);

        // Process each cycle
        for (int i = 0; i < cycles; i++) {
            
            // Start index of current cycle
            int currIdx = i;
            int currEle = arr[currIdx];

            // Rotate elements till we reach the start of cycle
            do {
                int nextIdx = (currIdx + d) % n;
                int nextEle = arr[nextIdx];

                // Update the element at next index with the current element
                arr[nextIdx] = currEle;

                // Update the current element to next element
                currEle = nextEle;

                // Move to the next index
                currIdx = nextIdx;
            } while (currIdx != i);
        }
    }

    // function to compute GCD
    static int Gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    static void Main(string[] args) {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int d = 2;

        RotateArr(arr, d);

        // Print the rotated array
        for (int i = 0; i < arr.Length; i++)
            Console.Write(arr[i] + " ");
    }
}
JavaScript 
// JavaScript Program to right rotate the array by d positions
// using Juggling Algorithm

// Function to rotate array
function rotateArr(arr, d) {
    const n = arr.length;

    // Handle the case where d > size of array
    d %= n;

    // Calculate the number of cycles in the rotation
    const cycles = gcd(n, d);

    // Process each cycle
    for (let i = 0; i < cycles; i++) {

        // Start index of current cycle
        let currIdx = i;
        let currEle = arr[currIdx];

        // Rotate elements till we reach the start of cycle
        do {
            let nextIdx = (currIdx + d) % n;
            let nextEle = arr[nextIdx];

            // Update the element at next index with the current element
            arr[nextIdx] = currEle;

            // Update the current element to next element
            currEle = nextEle;

            // Move to the next index
            currIdx = nextIdx;
        } while (currIdx !== i);
    }
}

// function to compute GCD
function gcd(a, b) {
    while (b !== 0) {
        const temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

const arr = [1, 2, 3, 4, 5, 6];
const d = 2;

rotateArr(arr, d);

// Print the rotated array
console.log(arr.join(" "));

Output
5 6 1 2 3 4

4. The Reversal Algorithm

The idea is based on the observation that if we right rotate the array by d positions, the last d elements will be in the front and first (n – d) elements will be at the end.

  • First reverse all the elements of the array.
  • Then reverse first d elements.
  • Finally, reverse last (n – d) elements to get the final rotated array.

Illustration:


Below is the implementation of the above approach:

C++ 
// C++ Code to right rotate an array using Reversal Algorithm

#include <bits/stdc++.h>

using namespace std;

// Function to rotate an array by d elements to the right
void rotateArr(vector<int>& arr, int d) {
    int n = arr.size();
  
    // Handle the case where d > size of array
    d %= n;

  	// Reverse the entire array
    reverse(arr.begin(), arr.end());
  
    // Reverse the first d elements
    reverse(arr.begin(), arr.begin() + d);

    // Reverse the remaining n-d elements
    reverse(arr.begin() + d, arr.end());
}

int main() {
    vector<int> arr = { 1, 2, 3, 4, 5, 6 };
    int d = 2;
    
  	rotateArr(arr, d);
  
    for (int i = 0; i < arr.size(); i++) 
        cout << arr[i] << " ";
    return 0;
}
C 
// C Code to right rotate an array using Reversal Algorithm

#include <stdio.h>

// Function to reverse a portion of the array from start to end
void reverse(int arr[], int start, int end) {
    while (start < end) {
        int temp = arr[start];
        arr[start] = arr[end];
        arr[end] = temp;
        start++;
        end--;
    }
}

// Function to rotate an array by d elements to the right
void rotateArr(int arr[], int n, int d) {
    
    // Handle the case where d > size of array
    d %= n;

    // Reverse the entire array
    reverse(arr, 0, n - 1);

    // Reverse the first d elements
    reverse(arr, 0, d - 1);

    // Reverse the remaining n-d elements
    reverse(arr, d, n - 1);
}

int main() {
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int d = 2;

    rotateArr(arr, n, d);

    for (int i = 0; i < n; i++) 
        printf("%d ", arr[i]);
    
    return 0;
}
Java 
// Java Code to right rotate an array using Reversal Algorithm

import java.util.*;

class GfG {
    
    // Function to rotate an array by d elements to the right
    static void rotateArr(int[] arr, int d) {
        int n = arr.length;
        
        // Handle the case where d > size of array
        d %= n;

        // Reverse the entire array
        reverse(arr, 0, n - 1);
        
        // Reverse the first d elements
        reverse(arr, 0, d - 1);

        // Reverse the remaining n-d elements
        reverse(arr, d, n - 1);
    }

    // function to reverse a portion of the array
    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--;
        }
    }

    public static void main(String[] args) {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int d = 2;
        
        rotateArr(arr, d);
        
        for (int i = 0; i < arr.length; i++) 
            System.out.print(arr[i] + " ");
    }
}
Python 
# Python Code to right rotate an array using Reversal Algorithm

# Function to rotate an array by d elements to the right
def rotateArr(arr, d):
    n = len(arr)
    
    # Handle the case where d > size of array
    d %= n

    # Reverse the entire array
    arr.reverse()
    
    # Reverse the first d elements
    arr[:d] = reversed(arr[:d])

    # Reverse the remaining n-d elements
    arr[d:] = reversed(arr[d:])

if __name__ == "__main__":
    arr = [1, 2, 3, 4, 5, 6]
    d = 2
    
    rotateArr(arr, d)
    
    for i in range(len(arr)):
        print(arr[i], end=" ")
C# 
// C# Code to right rotate an array using Reversal Algorithm

using System;

class GfG {
    
    // Function to rotate an array by d elements to the right
    static void rotateArr(int[] arr, int d) {
        int n = arr.Length;
        
        // Handle the case where d > size of array
        d %= n;

        // Reverse the entire array
        Array.Reverse(arr);
        
        // Reverse the first d elements
        Array.Reverse(arr, 0, d);

        // Reverse the remaining n-d elements
        Array.Reverse(arr, d, n - d);
    }

    static void Main() {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int d = 2;
        
        rotateArr(arr, d);
        
        for (int i = 0; i < arr.Length; i++)
            Console.Write(arr[i] + " ");
    }
}
JavaScript 
// JavaScript Code to right rotate an array using Reversal Algorithm

// Function to rotate an array by d elements to the right
function rotateArr(arr, d) {
    let n = arr.length;
  
    // Handle the case where d > size of array
    d %= n;

    // Reverse the entire array
    arr.reverse();
  
    // Reverse the first d elements
    reverse(arr, 0, d - 1);

    // Reverse the remaining n-d elements
    reverse(arr, d, n - 1);
}

// function to reverse a portion of the array
function reverse(arr, start, end) {
    while (start < end) {
        let temp = arr[start];
        arr[start] = arr[end];
        arr[end] = temp;
        start++;
        end--;
    }
}

let arr = [1, 2, 3, 4, 5, 6];
let d = 2;

rotateArr(arr, d);

console.log(arr.join(" "));

Output
5 6 1 2 3 4

Time Complexity: O(n)
Auxiliary Space: O(1)



Next Article
Maximum Score by travelling from (1, 1) to (N, N) on grid

S
sahil_anower_gfg
Improve
Article Tags :
Practice Tags :

Similar Reads

We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy
Lightbox