C++ 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:

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 4 5 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.

C++




// C++ code for above implementation
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);
}
  
// This code is contributed by Shivani


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 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!