Java Program for Block swap algorithm 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

Algorithm : 

Initialize A = arr[0..d-1] and B = arr[d..n-1]
1) Do following until size of A is equal to size of B

  a)  If A is shorter, divide B into Bl and Br such that Br is of same 
       length as A. Swap A and Br to change ABlBr into BrBlA. Now A
       is at its final place, so recur on pieces of B.  

   b)  If A is longer, divide A into Al and Ar such that Al is of same 
       length as B Swap Al and B to change AlArB into BArAl. Now B
       is at its final place, so recur on pieces of A.

2)  Finally when A and B are of equal size, block swap them.

Recursive Implementation:

Java

import java.util.*;

class GFG
{
    // Wrapper over the recursive function leftRotateRec()
    // It left rotates arr[] by d.
    public static void leftRotate(int arr[], int d,
                                                int n)
    {
        leftRotateRec(arr, 0, d, n);
    }

    public static void leftRotateRec(int arr[], int i,
                                  int d, int n)
    {
        /* Return If number of elements to be rotated
        is zero or equal to array size */
        if(d == 0 || d == n)
            return;

        /*If number of elements to be rotated
        is exactly half of array size */
        if(n - d == d)
        {
            swap(arr, i, n - d + i, d);
            return;
        }

        /* If A is shorter*/
        if(d < n - d)
        {
            swap(arr, i, n - d + i, d);
            leftRotateRec(arr, i, d, n - d);
        }
        else /* If B is shorter*/
        {
            swap(arr, i, d, n - d);
            leftRotateRec(arr, n - d + i, 2 * d - n, d); /*This is tricky*/
        }
    }

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

/*This function swaps d elements
starting at index fi with d elements
starting at index si */
public static void swap(int arr[], int fi,
                        int si, int d)
{
    int i, temp;
    for(i = 0; i < d; i++)
    {
        temp = arr[fi + i];
        arr[fi + i] = arr[si + i];
        arr[si + i] = temp;
    }
}

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

// This code is contributed by codeseeker

Output:

3 5 4 6 7 1 2

Time Complexity: O(N), where N represents the size of the given array.
Auxiliary Space: O(N) due to recursive stack space.

Iterative Implementation:
Here is iterative implementation of the same algorithm. Same utility function swap() is used here.

Java

//Java code for above implementation
static void leftRotate(int arr[], int d, int n)
{
int i, j;
if(d == 0 || d == n)
    return;
i = d;
j = n - d;
while (i != j)
{
    if(i < j) /*A is shorter*/
    {
    swap(arr, d-i, d+j-i, i);
    j -= i;
    }
    else /*B is shorter*/
    {
    swap(arr, d-i, d, j);
    i -= j;
    }
    // printArray(arr, 7);
}
/*Finally, block swap A and B*/
swap(arr, d-i, d, i);
}

Time Complexity: O(N), where N represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Please see following posts for other methods of array rotation: 
https://www.geeksforgeeks.org/array-rotation/amp/
https://www.geeksforgeeks.org/program-for-array-rotation-continued-reversal-algorithm/amp/

Please write comments if you find any bug in the above programs/algorithms or want to share any additional information about the block swap algorithm.

Please refer complete article on Block swap algorithm for array rotation for more details!