Given a permutation arrays A[] consisting of N numbers in range [1, N], the task is to left rotate all the even numbers and right rotate all the odd numbers of the permutation and print the updated permutation.
Note: N is always even.
Examples:
Input: A = {1, 2, 3, 4, 5, 6, 7, 8}
Output: {7, 4, 1, 6, 3, 8, 5, 2}
Explanation:
Even element = {2, 4, 6, 8}
Odd element = {1, 3, 5, 7}
Left rotate of even number = {4, 6, 8, 2}
Right rotate of odd number = {7, 1, 3, 5}
Combining Both odd and even number alternatively.
Input: A = {1, 2, 3, 4, 5, 6}
Output: {5, 4, 1, 6, 3, 2}
Approach:
- It is clear that the odd elements are always on even index and even elements are always laying on odd index.
- To do left rotation of even number we choose only odd indices.
- To do right rotation of odd number we choose only even indices.
- Print the updated array.
Below is the implementation of the above approach:
Java
// Java program to implement // the above approach import java.io.*; import java.util.*; import java.lang.*; class GFG { // function to left rotate static void left_rotate( int [] arr) { int last = arr[ 1 ]; for ( int i = 3 ; i < arr.length; i = i + 2 ) { arr[i - 2 ] = arr[i]; } arr[arr.length - 1 ] = last; } // function to right rotate static void right_rotate( int [] arr) { int start = arr[arr.length - 2 ]; for ( int i = arr.length - 4 ; i >= 0 ; i = i - 2 ) { arr[i + 2 ] = arr[i]; } arr[ 0 ] = start; } // Function to rotate the array public static void rotate( int arr[]) { left_rotate(arr); right_rotate(arr); for ( int i = 0 ; i < arr.length; i++) { System.out.print(arr[i] + " " ); } } // Driver code public static void main(String[] args) { int arr[] = { 1 , 2 , 3 , 4 , 5 , 6 }; rotate(arr); } } |
5 4 1 6 3 2
Time Complexity: O(N)
Auxiliary Space: O(1)
Please refer complete article on Rotate all odd numbers right and all even numbers left in an Array of 1 to N for more details!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!