Java 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

METHOD 1 (Using 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.

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}

Java

class RotateArray
{
    /*Function to left rotate arr[] of size 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 (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

Output :

3 4 5 6 7 1 2 

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

Please refer complete article on Program for array rotation for more details!