Rotate all odd numbers right and all even numbers left in an Array of 1 to N
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:
- 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:
C++
#include<bits/stdc++.h>
using namespace std;
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;
}
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;
}
void rotate( int arr[])
{
left_rotate(arr);
right_rotate(arr);
for ( int i = 0; i < 6; i++)
{
cout << (arr[i]) << " " ;
}
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6 };
rotate(arr);
}
|
Java
import java.io.*;
import java.util.*;
import java.lang.*;
class GFG {
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;
}
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;
}
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] + " " );
}
}
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 3 , 4 , 5 , 6 };
rotate(arr);
}
}
|
Python3
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
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
def rotate(arr):
left_rotate(arr)
right_rotate(arr)
for i in range ( len (arr)):
print (arr[i], end = " " )
arr = [ 1 , 2 , 3 , 4 , 5 , 6 ]
rotate(arr);
|
C#
using System;
class GFG{
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;
}
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;
}
public static void rotate( int [] arr)
{
left_rotate(arr);
right_rotate(arr);
for ( int i = 0; i < arr.Length; i++)
{
Console.Write(arr[i] + " " );
}
}
public static void Main()
{
int [] arr = { 1, 2, 3, 4, 5, 6 };
rotate(arr);
}
}
|
Javascript
<script>
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 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 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>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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