# Block swap algorithm for array rotation

Hello Everyone,

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

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:

• C++

#include <bits/stdc++.h>

using namespace std;

/*Prototype for utility functions */

void printArray( int arr[], int size);

void swap( int arr[], int fi, int si, int d);

void leftRotate( int arr[], 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, 0, n - d, d);

return ;

}

/* If A is shorter*/

if (d < n - d)

{

swap(arr, 0, n - d, d);

leftRotate(arr, d, n - d);

}

else /* If B is shorter*/

{

swap(arr, 0, d, n - d);

leftRotate(arr + n - d, 2 * d - n, d); /*This is tricky*/

}

}

/*UTILITY FUNCTIONS*/

/* function to print an array */

void printArray( int arr[], int size)

{

int i;

for (i = 0; i < size; i++)

cout << arr[i] << " " ;

cout << endl;

}

/*This function swaps d elements starting at index fi

with d elements starting at index si */

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

int main()

{

int arr[] = {1, 2, 3, 4, 5, 6, 7};

leftRotate(arr, 2, 7);

printArray(arr, 7);

return 0;

}

// This code is contributed by rathbhupendra

Output:

3 5 4 6 7 1 2

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

• C

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)