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Java Program to Rotate all odd numbers right and all even numbers left in an Array of 1 to N

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  • Last Updated : 27 Jan, 2022

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:

  1. It is clear that the odd elements are always on even index and even elements are always laying on odd index.
  2. To do left rotation of even number we choose only odd indices.
  3. To do right rotation of odd number we choose only even indices.
  4. 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);
    }
}

Output:

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!


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