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



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.

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.



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Improved By : AbhijeetPal

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