Data Modelling & AI Data Structure & Algorithm

Print array after it is right rotated K times | Set 2

23 June 2025

0

Given an array arr[] of size N and a value K, the task is to print the array rotated by K times to the right.

Examples:

Input: arr = {1, 3, 5, 7, 9}, K = 2
Output: 7 9 1 3 5

Input: arr = {1, 2, 3, 4, 5}, K = 4
Output: 2 3 4 5 1

Algorithm: The given problem can be solved by reversing subarrays. Below steps can be followed to solve the problem:

Reverse all the array elements from 1 to N -1
Reverse the array elements from 1 to K – 1
Reverse the array elements from K to N -1

C++

// C++ implementation for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to rotate the array
// to the right, K times
void RightRotate(int Array[], int N, int K)
{
 
    // Reverse all the array elements
    reverse(Array, Array + N);
 
    // Reverse the first k elements
    reverse(Array, Array + K);
 
    // Reverse the elements from K
    // till the end of the array
    reverse(Array + K, Array + N);
 
    // Print the array after rotation
    for (int i = 0; i < N; i++) {
 
        cout << Array[i] << " ";
    }
 
    cout << endl;
}
 
// Driver code
int main()
{
 
    // Initialize the array
    int Array[] = { 1, 2, 3, 4, 5 };
 
    // Find the size of the array
    int N = sizeof(Array) / sizeof(Array[0]);
 
    // Initialize K
    int K = 4;
 
    // Call the function and
    // print the answer
    RightRotate(Array, N, K);
 
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;
class GFG
{
   
    // Function to rotate the array
    // to the right, K times
    static void RightRotate(int[] Array, int N, int K)
    {
 
        // Reverse all the array elements
        for (int i = 0; i < N / 2; i++) {
            int temp = Array[i];
            Array[i] = Array[N - i - 1];
            Array[N - i - 1] = temp;
        }
 
        // Reverse the first k elements
        for (int i = 0; i < K / 2; i++) {
            int temp = Array[i];
            Array[i] = Array[K - i - 1];
            Array[K - i - 1] = temp;
        }
 
        // Reverse the elements from K
        // till the end of the array
        for (int i = 0; i < (N-K) / 2; i++) {
            int temp = Array[(i + K)];
            Array[(i + K)] = Array[(N - i - 1)];
            Array[(N - i - 1)] = temp;
        }
 
        // Print the array after rotation
        for (int i = 0; i < N; i++) {
 
            System.out.print(Array[i] + " ");
        }
 
        System.out.println();
    }
 
    // Driver code
    public static void main(String[] args)
    {
       
        // Initialize the array
        int Array[] = { 1, 2, 3, 4, 5 };
 
        // Find the size of the array
        int N = Array.length;
 
        // Initialize K
        int K = 4;
 
        // Call the function and
        // print the answer
        RightRotate(Array, N, K);
    }
}
 
// This code is contributed by maddler.

Python3

# Python program for the above approach
import math
 
# Function to rotate the array
# to the right, K times
def RightRotate(Array, N, K):
 
    # Reverse all the array elements
    for i in range(math.ceil(N / 2)):
        temp = Array[i]
        Array[i] = Array[N - i - 1]
        Array[N - i - 1] = temp
 
    # Reverse the first k elements
    for i in range(math.ceil(K / 2)):
        temp = Array[i]
        Array[i] = Array[K - i - 1]
        Array[K - i - 1] = temp
 
    # Reverse the elements from K
    # till the end of the array
    for i in range(math.ceil((N-K) / 2)):
        temp = Array[(i + K)]
        Array[(i + K)] = Array[(N - i - 1)]
        Array[(N - i - 1)] = temp
 
    # Print the array after rotation
    for i in range(N):
        print(Array[i], end=" ")
 
 
# Driver Code
arr = [1, 2, 3, 4, 5]
N = len(arr)
K = 4
 
# Call the function and
# print the answer
RightRotate(arr, N, K)
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
class GFG
{
    // Function to rotate the array
    // to the right, K times
    static void RightRotate(int []Array, int N, int K)
    {
 
        // Reverse all the array elements
        for (int i = 0; i < N / 2; i++) {
            int temp = Array[i];
            Array[i] = Array[N - i - 1];
            Array[N - i - 1] = temp;
        }
 
        // Reverse the first k elements
        for (int i = 0; i < K / 2; i++) {
            int temp = Array[i];
            Array[i] = Array[K - i - 1];
            Array[K - i - 1] = temp;
        }
 
        // Reverse the elements from K
        // till the end of the array
        for (int i = 0; i < (N-K) / 2; i++) {
            int temp = Array[(i + K)];
            Array[(i + K)] = Array[(N - i - 1)];
            Array[(N - i - 1)] = temp;
        }
 
        // Print the array after rotation
        for (int i = 0; i < N; i++) {
 
            Console.Write(Array[i] + " ");
        }
    }
 
    // Driver code
    public static void Main()
    {
       
        // Initialize the array
        int []Array = { 1, 2, 3, 4, 5 };
 
        // Find the size of the array
        int N = Array.Length;
 
        // Initialize K
        int K = 4;
 
        // Call the function and
        // print the answer
        RightRotate(Array, N, K);
    }
}
// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
// Javascript program for the above approach
 
// Function to rotate the array
// to the right, K times
function RightRotate(Array, N, K)
{
 
    // Reverse all the array elements
    for (let i = 0; i < N / 2; i++) {
        let temp = Array[i];
        Array[i] = Array[N - i - 1];
        Array[N - i - 1] = temp;
    }   
 
    // Reverse the first k elements
    for (let i = 0; i < K / 2; i++) {
        let temp = Array[i];
        Array[i] = Array[K - i - 1];
        Array[K - i - 1] = temp;
    }
 
    // Reverse the elements from K
    // till the end of the array
    for (let i = 0; i < (N-K) / 2; i++) {
        let temp = Array[(i + K)];
        Array[(i + K)] = Array[(N - i - 1)];
        Array[(N - i - 1) % N] = temp;
    }
 
    // Print the array after rotation
    for (let i = 0; i < N; i++) {
        document.write(Array[i] + " ");
    }
}
 
// Driver Code
 
let arr = [ 1, 2, 3, 4, 5 ];
let N = arr.length;
let K =4;
 
// Call the function and
// print the answer
RightRotate(arr, N, K);
 
// This code is contributed by Samim Hossain Mondal.
</script>

Output

2 3 4 5 1

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

Print array after it is right rotated K times using Recursion

Step-by-step approach:

If “k” = 0, return the array “arr” and terminate the recursion.
Else, move the last element to the first position and rotate the array “arr” to the right by one position by moving each element one position to the right.
Recursively call the “rotateArray()” function with the same array “arr”, of size “n”, and k-1.
Return the rotated array “arr” and terminate the recursion.

Below is the implementation of the above Recursive approach:

C++

// C++ implementation for Print array after it is right
// rotated K times using Recursion
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to rotate the array
void rotateArray(int arr[], int n, int k)
{
    if (k == 0) {
        return;
    }
 
    // Rotate the array to the right by one position
    int temp = arr[n - 1];
    for (int i = n - 1; i > 0; i--) {
        arr[i] = arr[i - 1];
    }
    arr[0] = temp;
 
    // Recursively rotate the remaining elements k-1 times
    rotateArray(arr, n, k - 1);
}
 
// Driver code
int main()
{
    int arr[] = { 1, 3, 5, 7, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;
 
    rotateArray(arr, n, k);
 
    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
    cout << endl;
 
    return 0;
}
 
// This code is contributed by  Vaibhav Saroj

C

#include <stdio.h>
 
// Function to rotate the array
void rotateArray(int arr[], int n, int k) {
    if (k == 0) {
        return;
    }
     
    // rotate the array to the right by one position
    int temp = arr[n-1];
    for (int i = n-1; i > 0; i--) {
        arr[i] = arr[i-1];
    }
    arr[0] = temp;
     
    // recursively rotate the remaining elements k-1 times
    rotateArray(arr, n, k-1);
}
 
// Driver code
int main() {
    int arr[] = {1, 3, 5, 7, 9};
    int n = sizeof(arr)/sizeof(arr[0]);
    int k = 2;
     
    rotateArray(arr, n, k);
     
    for (int i = 0; i < n; i++) {
        printf("%d ", arr[i]);
    }
    printf("\n");
     
    return 0;
}
 
// This code is contributed by  Vaibhav Saroj

Java

import java.util.*;
 
public class Main {
     
    // Function to rotate the array
    public static void rotateArray(int[] arr, int n, int k) {
        if (k == 0) {
            return;
        }
         
        // rotate the array to the right by one position
        int temp = arr[n-1];
        for (int i = n-1; i > 0; i--) {
            arr[i] = arr[i-1];
        }
        arr[0] = temp;
         
        // recursively rotate the remaining elements k-1 times
        rotateArray(arr, n, k-1);
    }
     
    // Driver code
    public static void main(String[] args) {
        int[] arr = {1, 3, 5, 7, 9};
        int n = arr.length;
        int k = 2;
         
        rotateArray(arr, n, k);
         
        for (int i = 0; i < n; i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.println();
    }
}
 
// This code is contributed by  Vaibhav Saroj

C#

// C# code 
using System;
 
public class GFG {
     
    // Function to rotate the array
    public static void rotateArray(int[] arr, int n, int k) {
        if (k == 0) {
            return;
        }
         
        // rotate the array to the right by one position
        int temp = arr[n-1];
        for (int i = n-1; i > 0; i--) {
            arr[i] = arr[i-1];
        }
        arr[0] = temp;
         
        // recursively rotate the remaining elements k-1 times
        rotateArray(arr, n, k-1);
    }
     
    // Driver code
    public static void Main() {
        int[] arr = {1, 3, 5, 7, 9};
        int n = arr.Length;
        int k = 2;
         
        rotateArray(arr, n, k);
         
        for (int i = 0; i < n; i++) {
            Console.Write(arr[i] + " ");
        }
        Console.WriteLine();
    }
}
 
// This code is contributed by Utkarsh Kumar

Javascript

function rotateArray(arr, n, k) {
  if (k === 0) {
    return;
  }
   
  // rotate the array to the right by one position
  const temp = arr[n-1];
  for (let i = n-1; i > 0; i--) {
    arr[i] = arr[i-1];
  }
  arr[0] = temp;
   
  // recursively rotate the remaining elements k-1 times
  rotateArray(arr, n, k-1);
}
 
// Driver code
const arr = [1, 3, 5, 7, 9];
const n = arr.length;
const k = 2;
 
rotateArray(arr, n, k);
 
console.log(arr); // Output: [ 7, 9, 1, 3, 5 ]
 
 
// This code is contributed by  Vaibhav Saroj

Output

7 9 1 3 5

Time Complexity: O(k*n)
Auxiliary Space: O(k)

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Print array after it is right rotated K times | Set 2

C++

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Javascript

Print array after it is right rotated K times using Recursion

C++

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Java

C#

Javascript

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