Given an array, right rotate it by k elements.
After K=3 rotation
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); }}// This code is contributed by Gitanjali. |
Output:
8 9 10 1 2 3 4 5 6 7
Time Complexity: O(N)
Auxiliary Space: O(1)
Please refer complete article on Reversal algorithm for right rotation of an array for more details!
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