Block swap algorithm for array rotation

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

Array

Rotation of the above array by 2 will make array

ArrayRotation1

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

#include<stdio.h> 
  
/*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++) 
    printf("%d ", arr[i]); 
  printf("\n "); 
}  
  
/*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 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; 
}     

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); 
} 

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)

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

References:
http://www.cs.bell-labs.com/cm/cs/pearls/s02b.pdf

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.

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C

Java

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