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 , Print rightmost Kth element (a[N + i -K]). Otherwise,
- Print array after ‘K’ elements (a[i – K]).
- If i , 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!