# Program for array rotation

• Difficulty Level : Easy
• Last Updated : 09 Nov, 2021

Write a function rotate(ar[], d, n) that rotates arr[] of size n by d elements.

Rotation of the above array by 2 will make array

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

METHOD 1 (Using temp array)

```Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7
1) Store the first d elements in a temp array
temp[] = [1, 2]
2) Shift rest of the arr[]
arr[] = [3, 4, 5, 6, 7, 6, 7]
3) Store back the d elements
arr[] = [3, 4, 5, 6, 7, 1, 2]```

Time complexity : O(n)
Auxiliary Space : O(d)

METHOD 2 (Rotate one by one)

```leftRotate(arr[], d, n)
start
For i = 0 to i < d
Left rotate all elements of arr[] by one
end```

To rotate by one, store arr[0] in a temporary variable temp, move arr[1] to arr[0], arr[2] to arr[1] …and finally temp to arr[n-1]
Let us take the same example arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2
Rotate arr[] by one 2 times
We get [2, 3, 4, 5, 6, 7, 1] after first rotation and [ 3, 4, 5, 6, 7, 1, 2] after second rotation.
Below is the implementation of the above approach :

## C++

```// C++ program to rotate an array by
// d elements
#include <bits/stdc++.h>
using namespace std;

/*Function to left Rotate arr[] of
size n by 1*/
void leftRotatebyOne(int arr[], int n)
{
int temp = arr[0], i;
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];

arr[n-1] = temp;
}

/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}

/* utility function to print an array */
void printArray(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}

/* Driver program to test above functions */
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);

// Function calling
leftRotate(arr, 2, n);
printArray(arr, n);

return 0;
}
```

## C

```// C program to rotate an array by
// d elements
#include <stdio.h>

/* Function to left Rotate arr[] of size n by 1*/
void leftRotatebyOne(int arr[], int n);

/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
int i;
for (i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}

void leftRotatebyOne(int arr[], int n)
{
int temp = arr[0], i;
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n-1] = temp;
}

/* utility function to print an array */
void printArray(int arr[], int n)
{
int i;
for (i = 0; i < n; i++)
printf("%d ", arr[i]);
}

/* Driver program to test above functions */
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
leftRotate(arr, 2, 7);
printArray(arr, 7);
return 0;
}
```

## Java

```// Java program to rotate an array by
// d elements

class RotateArray {
/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}

void leftRotatebyOne(int arr[], int n)
{
int i, temp;
temp = arr[0];
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n-1] = temp;
}

/* utility function to print an array */
void printArray(int arr[], int n)
{
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}

// Driver program to test above functions
public static void main(String[] args)
{
RotateArray rotate = new RotateArray();
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
rotate.leftRotate(arr, 2, 7);
rotate.printArray(arr, 7);
}
}

// This code has been contributed by Mayank Jaiswal
```

## Python3

```# Python3 program to rotate an array by
# d elements
# Function to left rotate arr[] of size n by d*/
def leftRotate(arr, d, n):
for i in range(d):
leftRotatebyOne(arr, n)

# Function to left Rotate arr[] of size n by 1*/
def leftRotatebyOne(arr, n):
temp = arr[0]
for i in range(n-1):
arr[i] = arr[i + 1]
arr[n-1] = temp

# utility function to print an array */
def printArray(arr, size):
for i in range(size):
print ("% d"% arr[i], end =" ")

# Driver program to test above functions */
arr = [1, 2, 3, 4, 5, 6, 7]
leftRotate(arr, 2, 7)
printArray(arr, 7)

# This code is contributed by Shreyanshi Arun
```

## C#

```// C# program for array rotation
using System;

class GFG {
/* Function to left rotate arr[]
of size n by d*/
static void leftRotate(int[] arr, int d,
int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}

static void leftRotatebyOne(int[] arr, int n)
{
int i, temp = arr[0];
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];

arr[n-1] = temp;
}

/* utility function to print an array */
static void printArray(int[] arr, int size)
{
for (int i = 0; i < size; i++)
Console.Write(arr[i] + " ");
}

// Driver code
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 5, 6, 7 };
leftRotate(arr, 2, 7);
printArray(arr, 7);
}
}

// This code is contributed by Sam007
```

## PHP

```<?php
// PHP program to rotate an array
// by d elements

/*Function to left Rotate arr[]
of size n by 1*/
function leftRotatebyOne(&\$arr, \$n)
{
\$temp = \$arr[0];
for (\$i = 0; \$i < \$n - 1; \$i++)
\$arr[\$i] = \$arr[\$i + 1];

\$arr[\$n-1] = \$temp;
}

/*Function to left rotate arr[]
of size n by d*/
function leftRotate(&\$arr, \$d, \$n)
{
for (\$i = 0; \$i < \$d; \$i++)
leftRotatebyOne(\$arr, \$n);
}

/* utility function to print
an array */
function printArray(&\$arr, \$n)
{
for (\$i = 0; \$i < \$n; \$i++)
echo \$arr[\$i] . " ";
}

// Driver Code
\$arr = array( 1, 2, 3, 4, 5, 6, 7 );
\$n = sizeof(\$arr);

// Function calling
leftRotate(\$arr, 2, \$n);
printArray(\$arr, \$n);

// This code is contributed
// by ChitraNayal
?>

```

## Javascript

```<script>

// JavaScript program to rotate an array by
// d elements

/* Function to left rotate arr of size n by d */
function leftRotate(arr , d , n) {
for (i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}

function leftRotatebyOne(arr , n) {
var i, temp;
temp = arr[0];
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n - 1] = temp;
}

/* utility function to print an array */
function printArray(arr , n) {
for (i = 0; i < n; i++)
document.write(arr[i] + " ");
}

// Driver program to test above functions

var arr = [ 1, 2, 3, 4, 5, 6, 7 ];
leftRotate(arr, 2, 7);
printArray(arr, 7);

// This code is contributed by todaysgaurav

</script>
```

Output :

`3 4 5 6 7 1 2 `

Time complexity : O(n * d)
Auxiliary Space : O(1)
METHOD 3 (A Juggling Algorithm)
This is an extension of method 2. Instead of moving one by one, divide the array in different sets
where number of sets is equal to GCD of n and d and move the elements within sets.
If GCD is 1 as is for the above example array (n = 7 and d =2), then elements will be moved within one set only, we just start with temp = arr[0] and keep moving arr[I+d] to arr[I] and finally store temp at the right place.
Here is an example for n =12 and d = 3. GCD is 3 and

```Let arr[] be {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

a) Elements are first moved in first set – (See below
diagram for this movement)```

```          arr[] after this step --> {4 2 3 7 5 6 10 8 9 1 11 12}

b)    Then in second set.
arr[] after this step --> {4 5 3 7 8 6 10 11 9 1 2 12}

c)    Finally in third set.
arr[] after this step --> {4 5 6 7 8 9 10 11 12 1 2 3}```

Below is the implementation of the above approach :

## C++

```// C++ program to rotate an array by
// d elements
#include <bits/stdc++.h>
using namespace std;

/*Function to get gcd of a and b*/
int gcd(int a, int b)
{
if (b == 0)
return a;

else
return gcd(b, a % b);
}

/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
/* To handle if d >= n */
d = d % n;
int g_c_d = gcd(d, n);
for (int i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
int temp = arr[i];
int j = i;

while (1) {
int k = j + d;
if (k >= n)
k = k - n;

if (k == i)
break;

arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}

// Function to print an array
void printArray(int arr[], int size)
{
for (int i = 0; i < size; i++)
cout << arr[i] << " ";
}

/* Driver program to test above functions */
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);

// Function calling
leftRotate(arr, 2, n);
printArray(arr, n);

return 0;
}
```

## C

```// C program to rotate an array by
// d elements
#include <stdio.h>

/* function to print an array */
void printArray(int arr[], int size);

/*Function to get gcd of a and b*/
int gcd(int a, int b);

/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
int i, j, k, temp;
/* To handle if d >= n */
d = d % n;
int g_c_d = gcd(d, n);
for (i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
temp = arr[i];
j = i;
while (1) {
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}

/*UTILITY FUNCTIONS*/
/* function to print an array */
void printArray(int arr[], int n)
{
int i;
for (i = 0; i < n; i++)
printf("%d ", arr[i]);
}

/*Function to get gcd of a and b*/
int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}

/* Driver program to test above functions */
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
leftRotate(arr, 2, 7);
printArray(arr, 7);
getchar();
return 0;
}
```

## Java

```// Java program to rotate an array by
// d elements
class RotateArray {
/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
/* To handle if d >= n */
d = d % n;
int i, j, k, temp;
int g_c_d = gcd(d, n);
for (i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
temp = arr[i];
j = i;
while (true) {
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}

/*UTILITY FUNCTIONS*/

/* function to print an array */
void printArray(int arr[], int size)
{
int i;
for (i = 0; i < size; i++)
System.out.print(arr[i] + " ");
}

/*Function to get gcd of a and b*/
int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}

// Driver program to test above functions
public static void main(String[] args)
{
RotateArray rotate = new RotateArray();
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
rotate.leftRotate(arr, 2, 7);
rotate.printArray(arr, 7);
}
}

// This code has been contributed by Mayank Jaiswal
```

## Python3

```# Python3 program to rotate an array by
# d elements
# Function to left rotate arr[] of size n by d
def leftRotate(arr, d, n):
d = d % n
g_c_d = gcd(d, n)
for i in range(g_c_d):

# move i-th values of blocks
temp = arr[i]
j = i
while 1:
k = j + d
if k >= n:
k = k - n
if k == i:
break
arr[j] = arr[k]
j = k
arr[j] = temp

# UTILITY FUNCTIONS
# function to print an array
def printArray(arr, size):
for i in range(size):
print ("% d" % arr[i], end =" ")

# Function to get gcd of a and b
def gcd(a, b):
if b == 0:
return a;
else:
return gcd(b, a % b)

# Driver program to test above functions
arr = [1, 2, 3, 4, 5, 6, 7]
n = len(arr)
d = 2
leftRotate(arr, d, n)
printArray(arr, n)

# This code is contributed by Shreyanshi Arun

```

## C#

```// C# program for array rotation
using System;

class GFG {
/* Function to left rotate arr[]
of size n by d*/
static void leftRotate(int[] arr, int d,
int n)
{
int i, j, k, temp;
/* To handle if d >= n */
d = d % n;
int g_c_d = gcd(d, n);
for (i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
temp = arr[i];
j = i;
while (true) {
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}

/*UTILITY FUNCTIONS*/
/* Function to print an array */
static void printArray(int[] arr, int size)
{
for (int i = 0; i < size; i++)
Console.Write(arr[i] + " ");
}

/* Function to get gcd of a and b*/
static int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}

// Driver code
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 5, 6, 7 };
leftRotate(arr, 2, 7);
printArray(arr, 7);
}
}

// This code is contributed by Sam007
```

## Javascript

```<script>

// JavaScript program to rotate an array by
// d elements

/*Function to get gcd of a and b*/
function gcd( a, b){
if (b == 0)
return a;
else
return gcd(b, a % b);
}

/*Function to left rotate arr[] of siz n by d*/
function leftRotate(arr, d, n){
/* To handle if d >= n */
d = d % n;
let g_c_d = gcd(d, n);
for (let i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
let temp = arr[i];
let j = i;

while (1) {
let k = j + d;
if (k >= n)
k = k - n;

if (k == i)
break;

arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}

// Function to print an array
function printArray(arr, size){
for (let i = 0; i < size; i++)
document.write(arr[i] +" ");
}

/* Driver program to test above functions */
let arr = [ 1, 2, 3, 4, 5, 6, 7 ];
let n = arr.length;
// Function calling
leftRotate(arr, 2, n);
printArray(arr, n);

</script>
```

Output :

`3 4 5 6 7 1 2 `

Time complexity : O(n)
Auxiliary Space : O(1)

Please see following posts for other methods of array rotation:
Block swap algorithm for array rotation
Reversal algorithm for array rotation