Right rotate given Array K times using Pointers
Given an array arr[] of size N and an integer K, the task is to right rotate the array K times.
Examples:
Input: arr[] = {1, 3, 5, 7, 9}, K = 2
Output: 7 9 1 3 5
Explanation: After 1st rotation – {9, 1, 3, 5, 7}
After 2nd rotation – {7, 9, 1, 3, 5}Input: {1, 2, 3, 4, 5, 6}, K = 2
Output: 5 6 1 2 3 4
Approach: The naive approach and approach based on reversing parts of the array is discussed here.
Pointer based approach: The base of this concept is the reversal algorithm for array rotation. The array is divided into two parts where the first part is of size (N-K) and the end part is of size K. These two parts are individually reversed. Then the whole array is reversed.
Below is the implementation of the above approach:
C++
// C++ code to implement above approach#include <iostream>using namespace std;// Function to print the arrayvoid print(int arr[], int N){ for (int i = 0; i < N; i++) cout << *(arr + i) << " ";}// Function to reverse the array// from start to end indexvoid reverse(int arr[], int start, int end){ int temp; int size = end - start; // Reversal based on pointer approach for (int i = 0; i < (size / 2); i++) { temp = *(arr + i + start); *(arr + i + start) = *(arr + start + size - i - 1); *(arr + start + size - i - 1) = temp; }}// Function to right rotate the array K timesvoid right(int arr[], int K, int N){ reverse(arr, 0, N - K); reverse(arr, N - K, N); reverse(arr, 0, N); print(arr, N);}// Driver codeint main(){ int arr[] = { 1, 2, 3, 4, 5, 6 }; int N = sizeof(arr) / sizeof(arr[0]); int K = 2; right(arr, K, N); return 0;} |
Java
// Java code to implement above approachimport java.util.*;class GFG{// Function to print the arraystatic void print(int arr[], int N){ for (int i = 0; i < N; i++) System.out.print(arr[i]+ " ");}// Function to reverse the array// from start to end indexstatic int[] reverse(int arr[], int start, int end){ int temp; int size = end - start; // Reversal based on pointer approach for (int i = 0; i < (size / 2); i++) { temp = arr[ i + start]; arr[i + start] = arr[start + size - i - 1]; arr[start + size - i - 1] = temp; } return arr;}// Function to right rotate the array K timesstatic void right(int arr[], int K, int N){ arr = reverse(arr, 0, N - K); arr = reverse(arr, N - K, N); arr = reverse(arr, 0, N); print(arr, N);}// Driver codepublic static void main(String[] args){ int arr[] = { 1, 2, 3, 4, 5, 6 }; int N = arr.length; int K = 2; right(arr, K, N);}}// This code is contributed by 29AjayKumar |
Python3
# Python code to implement above approach# Function to print arraydef print1(arr, N): for i in range(N): print(arr[i], end = " ");# Function to reverse the array# from start to end indexdef reverse(arr, start, end): temp = 0; size = end - start; # Reversal based on pointer approach for i in range(size//2): temp = arr[i + start]; arr[i + start] = arr[start + size - i - 1]; arr[start + size - i - 1] = temp; return arr;# Function to right rotate the array K timesdef right(arr, K, N): arr = reverse(arr, 0, N - K); arr = reverse(arr, N - K, N); arr = reverse(arr, 0, N); print1(arr, N);# Driver codeif __name__ == '__main__': arr = [1, 2, 3, 4, 5, 6]; N = len(arr); K = 2; right(arr, K, N); # This code is contributed by 29AjayKumar |
C#
// C# code to implement above approachusing System;class GFG { // Function to print the array static void print(int[] arr, int N) { for (int i = 0; i < N; i++) Console.Write(arr[i] + " "); } // Function to reverse the array // from start to end index static void reverse(int[] arr, int start, int end) { int temp; int size = end - start; // Reversal based on pointer approach for (int i = 0; i < (size / 2); i++) { temp = arr[i + start]; arr[i + start] = arr[start + size - i - 1]; arr[start + size - i - 1] = temp; } } // Function to right rotate the array K times static void right(int[] arr, int K, int N) { reverse(arr, 0, N - K); reverse(arr, N - K, N); reverse(arr, 0, N); print(arr, N); } // Driver code public static void Main() { int[] arr = { 1, 2, 3, 4, 5, 6 }; int N = arr.Length; int K = 2; right(arr, K, N); }}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript code to implement above approach // Function to print the array const print = (arr, N) => { for (let i = 0; i < N; i++) document.write(`${arr[i]} `); } // Function to reverse the array // from start to end index const reverse = (arr, start, end) => { let temp; let size = end - start; // Reversal based on pointer approach for (let i = 0; i < parseInt(size / 2); i++) { temp = arr[i + start]; arr[i + start] = arr[start + size - i - 1]; arr[start + size - i - 1] = temp; } } // Function to right rotate the array K times const right = (arr, K, N) => { reverse(arr, 0, N - K); reverse(arr, N - K, N); reverse(arr, 0, N); print(arr, N); } // Driver code let arr = [1, 2, 3, 4, 5, 6]; let N = arr.length; let K = 2; right(arr, K, N);// This code is contributed by rakeshsahni</script> |
Output
5 6 1 2 3 4
Time Complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!
