# Program for array rotation

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 (Use temp array)

```Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7
1) Store 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.

Implementation:

## C/C++

```/*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 i, temp;
temp = arr[0];
for (i = 0; i < n-1; i++)
arr[i] = arr[i+1];
arr[i] = temp;
}

/* utility function to print an array */
void printArray(int arr[], int size)
{
int i;
for(i = 0; i < size; 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);
getchar();
return 0;
}
```

## Java

```class RotateArray
{
/*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 i, temp;
temp = arr[0];
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[i] = temp;
}

/* utility function to print an array */
void printArray(int arr[], int size)
{
int i;
for (i = 0; i < size; 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
```

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}
```

Implementation:

## C/C++

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

/*Fuction 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;
for (i = 0; i < gcd(d, n); 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 size)
{
int i;
for(i = 0; i < size; i++)
printf("%d ", arr[i]);
}

/*Fuction 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

```class RotateArray
{
/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
int i, j, k, temp;
for (i = 0; i < gcd(d, n); i++)
{
/* move i-th values of blocks */
temp = arr[i];
j = i;
while (1 != 0)
{
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] + " ");
}

/*Fuction 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
```

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

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