Algorithm to rotate an array in linear time

Question

How to rotate an integer array by i times using swap function only in linear time.

Answer 1

Let us say we have a function called arr_reverse(arr,i,j) which reverses the elements of the array arr between index i and j using the swap function.

Example:

arr = {1,2,3,4,5}  i = 0 j = 2 

then the function will return:

{3,2,1,4,5}   ^^^^^ 

Implementing this function is straight forward and is O(N).

Now let's use this function in rotating the array.

arr     = {1,2,3,4,5} // input array k       = 2 // amount of right rotation result  = {4,5,1,2,3} // expected result  l       = 5 // length of array.  Step 1: Call arr_reverse(arr,l-k,l-1) which is arr_reverse(arr,3,4) we get {1,2,3,5,4}                ^^^  Step 2: Call arr_reverse(arr,0,l-k-1) which is arr_reverse(arr,0,2) we get {3,2,1,5,4}         ^^^^^       Step 3: Call arr_reverse(arr,0,l-1) which is arr_reverse(arr,0,4) we get {4,5,1,2,3}          ^^^^^^^^^ 

The entire process makes use of arr_reverse 3 times, making it O(N)

Answer 2

You can do this in linear time by using a reverse() helper.

// rotate array of size=size, by n positions void rotate(int array[], int size, int n) {   // reverse array[0...size-1]   reverse(array, 0, size-1);    // reverse A[0...n-1]   reverse(array, 0, n-1);    // reverse A[n...size-1]   reverse(array, n, size-1); }  // reverse elements in the array[pos_from ... pos_to] void reverse(int array[], int pos_from, int pos_to) {    ... } 

Implementing reverse(int array[], int pos_from, int pos_to) using swaps is left as an exercise for the reader. Hint: This can be done in linear time.

Answer 3

Here's a better solution, of a different nature than the others. It involves fewer array swaps than the others. Python:

import fractions
# rotates an array in-place i positions to the left, in linear time
def rotate(arr,i):
    n = len(arr)
    reps = fractions.gcd(n,i)
    swaps = n / reps
    for start in xrange(reps):
        ix = start
        tmp = arr[ix]
        for s in xrange(swaps-1):
            previx = ix
            ix = (ix + i) % n
            arr[previx] = arr[ix]
        arr[ix] = tmp
    return arr

Answer 4

Using linear time O(2N+m), and constant space O(4). m = GCD(n, p)

It's up to 50% faster than the swapping approach, because swapping requires writing O(N) times to a temporary.

http://www.eis.mdx.ac.uk/staffpages/r_bornat/oldteaching/I2A/slides%209%20circshift.pdf

for (m=0, count=0; count!=n; m++) {
    type t=A[m];
    for (i=m, j=m+p; j!=m; i=j, j = j+p<n ? j+p : j+p-n, count++)
        A[i]=A[j];
    A[i]=t; count++;
}