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

  • Difficulty Level : Easy
  • Last Updated : 26 Mar, 2021

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:

C++




// C++ program to implement
// the above approach
#include<bits/stdc++.h>
using namespace std;
 
// function to left rotate
void left_rotate(int arr[])
{
    int last = arr[1];
    for (int i = 3; i < 6; i = i + 2)
    {
        arr[i - 2] = arr[i];
    }
    arr[6 - 1] = last;
}
 
// function to right rotate
void right_rotate(int arr[])
{
    int start = arr[6 - 2];
    for (int i = 6- 4; i >= 0; i = i - 2)
    {
        arr[i + 2] = arr[i];
    }
    arr[0] = start;
}
 
// Function to rotate the array
void rotate(int arr[])
{
    left_rotate(arr);
    right_rotate(arr);
    for (int i = 0; i < 6; i++)
    {
        cout << (arr[i]) << " ";
    }
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
 
    rotate(arr);
}
 
// This code is contributed by rock_cool

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);
    }
}

Python3




# Python3 program for the above approach
 
# Function to left rotate
def left_rotate(arr):
     
    last = arr[1];
    for i in range(3, len(arr), 2):
        arr[i - 2] = arr[i]
         
    arr[len(arr) - 1] = last
 
# Function to right rotate
def right_rotate(arr):
     
    start = arr[len(arr) - 2]
    for i in range(len(arr) - 4, -1, -2):
        arr[i + 2] = arr[i]
         
    arr[0] = start
 
# Function to rotate the array
def rotate(arr):
     
    left_rotate(arr)
    right_rotate(arr)
    for i in range(len(arr)):
        print(arr[i], end = " ")
 
# Driver code
arr = [ 1, 2, 3, 4, 5, 6 ]
 
rotate(arr);
 
# This code is contributed by sanjoy_62

C#




// C# program to implement
// the above approach
using System;
 
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++)
    {
        Console.Write(arr[i] + " ");
    }
}
 
// Driver code
public static void Main()
{
    int[] arr = { 1, 2, 3, 4, 5, 6 };
 
    rotate(arr);
}
}
 
// This code is contributed by chitranayal

Javascript




<script>
 
    // Javascript program to implement
    // the above approach
     
    // function to left rotate
    function left_rotate(arr)
    {
        let last = arr[1];
        for (let i = 3; i < 6; i = i + 2)
        {
            arr[i - 2] = arr[i];
        }
        arr[6 - 1] = last;
    }
 
    // function to right rotate
    function right_rotate(arr)
    {
        let start = arr[6 - 2];
        for (let i = 6- 4; i >= 0; i = i - 2)
        {
            arr[i + 2] = arr[i];
        }
        arr[0] = start;
    }
 
    // Function to rotate the array
    function rotate(arr)
    {
        left_rotate(arr);
        right_rotate(arr);
        for (let i = 0; i < 6; i++)
        {
            document.write(arr[i] + " ");
        }
    }
 
      let arr = [ 1, 2, 3, 4, 5, 6 ];
   
    rotate(arr);
     
</script>
Output:
5 4 1 6 3 2

Time Complexity: O(N) 
Auxiliary Space: O(1)
 




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