Write a function rotate(ar[], d, n) that rotates arr[] of size n by d elements using block swap algorithm
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.
#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;
}
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++
- C
- Java
- Python3
- C#
- Javascript
// 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);
}
Time Complexity: O(n)
Important Notice
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