Given an Array of size N and a values K, around which we need to right rotate the array. How to quickly print the right rotated array?
Examples :
Input: Array[] = {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: Array[] = {1, 2, 3, 4, 5}, K = 4.
Output: 2 3 4 5 1
Approach:
- We will first take mod of K by N (K = K % N) because after every N rotations array will become the same as the initial array.
- Now, we will iterate the array from i = 0 to i = N-1 and check,
- If i < K, Print rightmost Kth element (a[N + i -K]). Otherwise,
- Print array after ‘K’ elements (a[i – K]).
- If i < K, Print rightmost Kth element (a[N + i -K]). Otherwise,
Below is the implementation of the above approach.
Java
// Java Implementation of Right Rotation // of an Array K number of times import java.util.*; import java.lang.*; import java.io.*; class Array_Rotation { // Function to rightRotate array static void RightRotate(int a[], int n, int k) { // If rotation is greater // than size of array k=k%n; for(int i = 0; i < n; i++) { if(i<k) { // Printing rightmost // kth elements System.out.print(a[n + i - k] + " "); } else { // Prints array after // 'k' elements System.out.print(a[i - k] + " "); } } System.out.println(); } // Driver program public static void main(String args[]) { int Array[] = {1, 2, 3, 4, 5}; int N = Array.length; int K = 2; RightRotate(Array, N, K); } } // This code is contributed by M Vamshi Krishna |
Output:
4 5 1 2 3
Time complexity : O(n)
Auxiliary Space : O(1)
Please refer complete article on Print array after it is right rotated K times for more details!
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