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 5Input: 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 timesvoid 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 codeint 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 approachimport math# Function to rotate the array# to the right, K timesdef 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 Codearr = [1, 2, 3, 4, 5]N = len(arr)K = 4# Call the function and# print the answerRightRotate(arr, N, K)# This code is contributed by Saurabh Jaiswal |
C#
// C# program for the above approachusing 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 timesfunction 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 Codelet arr = [ 1, 2, 3, 4, 5 ];let N = arr.length;let K =4;// Call the function and// print the answerRightRotate(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 arrayvoid 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 codeint 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 arrayvoid 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 codeint 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 codeconst 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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