Given an array, the task is to cyclically rotate the array clockwise by one time.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}
Output: arr[] = {5, 1, 2, 3, 4}
Input: arr[] = {2, 3, 4, 5, 1}
Output: {1, 2, 3, 4, 5}
Approach 1: To solve the problem follow the below idea:
Assign every element with its previous element and first element with the last element .
Illustrations:
Consider an array: arr[] = {1, 2, 3, 4, 5}
- Initialize last element in variable ‘last_el’ that is 5
- Then, iterate from n-1 to 1 and assign arr[i] = arr[i-1]
- arr[4] = arr[3]
- arr[3] = arr[2]
- arr[2] = arr[1]
- arr[1] = arr[0]
- Assign arr[0] = last_el
- Thus the required array will be {5, 1, 2, 3, 4}
Follow the steps to solve the problem:
- Store the last element in any temp variable
- Shift every element one position ahead
- Assign first value = last value (stored in temp variable).
Below is the implementation for the above approach:
C++14
#include <iostream>
using namespace std;
void rotate( int arr[], int n)
{
int last_el = arr[n - 1];
for ( int i = n - 1; i > 0; i--) {
arr[i] = arr[i - 1];
}
arr[0] = last_el;
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << "Given array is \n" ;
for ( int i = 0; i < n; i++)
cout << arr[i] << ' ' ;
rotate(arr, n);
cout << "\nRotated array is\n" ;
for ( int i = 0; i < n; i++)
cout << arr[i] << ' ' ;
return 0;
}
|
C
#include <stdio.h>
void rotate( int arr[], int n)
{
int last_el = arr[n - 1];
for ( int i = n - 1; i > 0; i--)
arr[i] = arr[i - 1];
arr[0] = last_el;
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5 }, i;
int n = sizeof (arr) / sizeof (arr[0]);
printf ( "Given array is\n" );
for (i = 0; i < n; i++)
printf ( "%d " , arr[i]);
rotate(arr, n);
printf ( "\nRotated array is\n" );
for (i = 0; i < n; i++)
printf ( "%d " , arr[i]);
return 0;
}
|
Java
import java.util.Arrays;
public class Test {
static int arr[] = new int [] { 1 , 2 , 3 , 4 , 5 };
static void rotate()
{
int last_el = arr[arr.length - 1 ], i;
for (i = arr.length - 1 ; i > 0 ; i--)
arr[i] = arr[i - 1 ];
arr[ 0 ] = last_el;
}
public static void main(String[] args)
{
System.out.println( "Given Array is" );
System.out.println(Arrays.toString(arr));
rotate();
System.out.println( "Rotated Array is" );
System.out.println(Arrays.toString(arr));
}
}
|
Python3
def rotate(arr, n):
last_el = arr[n - 1 ]
for i in range (n - 1 , 0 , - 1 ):
arr[i] = arr[i - 1 ]
arr[ 0 ] = last_el
arr = [ 1 , 2 , 3 , 4 , 5 ]
n = len (arr)
print ( "Given array is" )
for i in range ( 0 , n):
print (arr[i], end = ' ' )
rotate(arr, n)
print ( "\nRotated array is" )
for i in range ( 0 , n):
print (arr[i], end = ' ' )
|
C#
using System;
public class Test {
static int [] arr = new int [] { 1, 2, 3, 4, 5 };
static void rotate()
{
int last_el = arr[arr.Length - 1], i;
for (i = arr.Length - 1; i > 0; i--)
arr[i] = arr[i - 1];
arr[0] = last_el;
}
public static void Main()
{
Console.WriteLine( "Given Array is" );
string Original_array = string .Join( " " , arr);
Console.WriteLine(Original_array);
rotate();
Console.WriteLine( "Rotated Array is" );
string Rotated_array = string .Join( " " , arr);
Console.WriteLine(Rotated_array);
}
}
|
Javascript
<script>
function rotate(arr, n)
{
var last_el = arr[n-1], i;
for (i = n-1; i > 0; i--)
arr[i] = arr[i-1];
arr[0] = last_el;
}
var arr = [1, 2, 3, 4, 5];
var n = arr.length;
document.write( "Given array is <br>" );
for ( var i = 0; i< n; i++)
document.write(arr[i] + " " );
rotate(arr, n);
document.write( "<br>Rotated array is <br>" );
for ( var i = 0; i < n; i++)
document.write(arr[i] + " " );
</script>
|
PHP
<?php
function rotate(& $arr , $n )
{
$last_el = $arr [ $n - 1];
for ( $i = $n - 1;
$i > 0; $i --)
$arr [ $i ] = $arr [ $i - 1];
$arr [0] = $last_el ;
}
$arr = array (1, 2, 3, 4, 5);
$n = sizeof( $arr );
echo "Given array is \n" ;
for ( $i = 0; $i < $n ; $i ++)
echo $arr [ $i ] . " " ;
rotate( $arr , $n );
echo "\nRotated array is\n" ;
for ( $i = 0; $i < $n ; $i ++)
echo $arr [ $i ] . " " ;
?>
|
Output
Given array is
1 2 3 4 5
Rotated array is
5 1 2 3 4
Time Complexity: O(n), as we need to iterate through all the elements. Where n is the number of elements in the array.
Auxiliary Space: O(1), as we are using constant space.
Approach 2: To solve the problem follow the below idea:
We can use two pointers, As we know in cyclic rotation we will bring last element to first and shift rest in forward direction, we can do this by swapping every element with last element till we get to the last point.
Follow the steps to solve the problem:
- Take two pointers i and j which point to first and last element of array respectively.
- Start swapping arr[i] and arr[j] and keep j fixed and i moving towards j.
- Repeat above step till i is not equal to j.
Below is the implementation for the above idea:
C++14
#include <iostream>
using namespace std;
void rotate( int arr[], int n)
{
int i = 0, j = n - 1;
while (i != j) {
swap(arr[i], arr[j]);
i++;
}
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5 }, i;
int n = sizeof (arr) / sizeof (arr[0]);
cout << "Given array is \n" ;
for (i = 0; i < n; i++)
cout << arr[i] << " " ;
rotate(arr, n);
cout << "\nRotated array is\n" ;
for (i = 0; i < n; i++)
cout << arr[i] << " " ;
return 0;
}
|
C
#include <stdio.h>
void swap( int * x, int * y)
{
int temp = *x;
*x = *y;
*y = temp;
}
void rotate( int arr[], int n)
{
int i = 0, j = n - 1;
while (i != j) {
swap(&arr[i], &arr[j]);
i++;
}
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5 }, i;
int n = sizeof (arr) / sizeof (arr[0]);
printf ( "Given array is\n" );
for (i = 0; i < n; i++)
printf ( "%d " , arr[i]);
rotate(arr, n);
printf ( "\nRotated array is\n" );
for (i = 0; i < n; i++)
printf ( "%d " , arr[i]);
return 0;
}
|
Java
import java.util.Arrays;
public class Test {
static int arr[] = new int [] { 1 , 2 , 3 , 4 , 5 };
static void rotate()
{
int i = 0 , j = arr.length - 1 ;
while (i != j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i++;
}
}
public static void main(String[] args)
{
System.out.println( "Given Array is" );
System.out.println(Arrays.toString(arr));
rotate();
System.out.println( "Rotated Array is" );
System.out.println(Arrays.toString(arr));
}
}
|
Python3
def rotate(arr, n):
i = 0
j = n - 1
while i ! = j:
arr[i], arr[j] = arr[j], arr[i]
i = i + 1
pass
arr = [ 1 , 2 , 3 , 4 , 5 ]
n = len (arr)
print ( "Given array is" )
for i in range ( 0 , n):
print (arr[i], end = ' ' )
rotate(arr, n)
print ( "\nRotated array is" )
for i in range ( 0 , n):
print (arr[i], end = ' ' )
|
C#
using System;
class GFG {
static int [] arr = new int [] { 1, 2, 3, 4, 5 };
static void rotate()
{
int n = arr[arr.Length - 1];
int i = 0, j = n - 1;
while (i != j) {
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
i++;
}
}
public static void Main()
{
Console.WriteLine( "Given Array is" );
string Original_array = string .Join( " " , arr);
Console.WriteLine(Original_array);
rotate();
Console.WriteLine( "Rotated Array is" );
string Rotated_array = string .Join( " " , arr);
Console.WriteLine(Rotated_array);
}
}
|
Javascript
<script>
function rotate(arr, n){
var i = 0
var j = n-1
while (i != j){
let temp;
temp = arr[i];
arr[i] = arr[j];
arr[j]= temp;
i =i+1
}
}
var arr = [1, 2, 3, 4, 5];
var n = arr.length;
document.write( "Given array is <br>" );
for ( var i = 0; i< n; i++)
document.write(arr[i] + " " );
rotate(arr, n);
document.write( "<br>Rotated array is <br>" );
for ( var i = 0; i < n; i++)
document.write(arr[i] + " " );
</script>
|
Output
Given array is
1 2 3 4 5
Rotated array is
5 1 2 3 4
Time Complexity: O(n), as we need to iterate through all the elements. Where n is the number of elements in the array.
Auxiliary Space: O(1), as we are using constant space.
Approach 3: To solve the problem follow the below idea:
We can use Reversal Algorithm also , reverse first n-1 elements and then whole array which will result into one right rotation.
Follow the steps to solve the problem:
- Reverse the array two times.
- First time we will reverse the first n-1(n=size of array) elements.
- Finally, we will get our rotated array by reversing the entire array.
Below is the implementation for the above idea:
C++14
#include <iostream>
using namespace std;
void rotate( int arr[], int n, int k)
{
int i, j;
for (i = 0, j = n - k - 1; i < j; i++, j--) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
for (i = 0, j = n - 1; i < j; i++, j--) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
int k = 1;
int i, j;
cout << "Given array is \n" ;
for (i = 0; i < n; i++)
cout << arr[i] << " " ;
rotate(arr, n, k);
cout << "\nRotated array is\n" ;
for (i = 0; i < n; i++)
cout << arr[i] << " " ;
return 0;
}
|
Java
import java.util.*;
class Main {
public static void main(String[] args)
{
int [] arr = { 1 , 2 , 3 , 4 , 5 };
int n = arr.length;
int k = 1 ;
int i, j;
System.out.println( "Given array is " );
for (i = 0 ; i < n; i++)
System.out.print(arr[i] + " " );
for (i = 0 , j = n - k - 1 ; i < j; i++, j--) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
for (i = 0 , j = n - 1 ; i < j; i++, j--) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
System.out.println( "\nRotated array is" );
for (i = 0 ; i < n; i++)
System.out.print(arr[i] + " " );
}
}
|
Python3
arr = [ 1 , 2 , 3 , 4 , 5 ]
n = len (arr)
k = 1
i, j = 0 , 0
print ( "Given array is" )
for i in range (n):
print (arr[i], end = " " )
for i, j in zip ( range ( 0 , (n - k) / / 2 ), range (n - k - 1 , (n - k) / / 2 - 1 , - 1 )):
temp = arr[i]
arr[i] = arr[j]
arr[j] = temp
for i, j in zip ( range ( 0 , n / / 2 ), range (n - 1 , n / / 2 - 1 , - 1 )):
temp = arr[i]
arr[i] = arr[j]
arr[j] = temp
print ( "\nRotated array is" )
for i in range (n):
print (arr[i], end = " " )
|
C#
using System;
class Program {
static void Main( string [] args)
{
int [] arr = { 1, 2, 3, 4, 5 };
int n = arr.Length;
int k = 1;
int i, j;
Console.WriteLine( "Given array is:" );
for (i = 0; i < n; i++)
Console.Write(arr[i] + " " );
for (i = 0, j = n - k - 1; i < j; i++, j--) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
for (i = 0, j = n - 1; i < j; i++, j--) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
Console.WriteLine( "\nRotated array is:" );
for (i = 0; i < n; i++)
Console.Write(arr[i] + " " );
Console.ReadLine();
}
}
|
Javascript
let arr = [1, 2, 3, 4, 5];
let n = arr.length;
let k = 1;
let i, j;
console.log( "Given array is" );
for (i = 0; i < n; i++) {
process.stdout.write(arr[i] + " " );
}
for (i = 0, j = n - k - 1; i < j; i++, j--) {
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
for (i = 0, j = n - 1; i < j; i++, j--) {
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
console.log( "\nRotated array is" );
console.log();
for (i = 0; i < n; i++) {
process.stdout.write(arr[i] + " " );
}
|
Output
Given array is
1 2 3 4 5
Rotated array is
5 1 2 3 4
Time Complexity: O(n), as we are reversing the array. Where n is the number of elements in the array.
Auxiliary Space: O(1), as we are using constant space.
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Last Updated :
11 Sep, 2023
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