Rotate An Integer Array with an O(n) algorithm

Question

Write a function that rotates an integer array by a given number k. k elements from the end should move to the beginning of array, and all other elements should move to right to make the space.

The rotation should be done in-place.

Algorithm should not run in more than O(n), where n is the size of array.

Also a constant memory must be used to perform the operation.

For example,

if array is initialized with elements arr = {1, 2, 3, 4, 5, 6, 7, 8, 9}

rotate(arr, 3) will result the elements to be {7, 8, 9, 1, 2, 3, 4, 5, 6}

rotate(arr, 6) will result the {4, 5, 6, 7, 8, 9, 1, 2, 3}

Answer 1

C (104)

void reverse(int* a, int* b)
{
    while (--b > a) {
        *b ^= *a;
        *a ^= *b;
        *b ^= *a;
        ++a;
    }
}

void rotate(int *arr, int s_arr, int by)
{
    reverse(arr, arr+s_arr);
    reverse(arr, arr+by);
    reverse(arr+by, arr+s_arr);
}

Minified:

v(int*a,int*b){while(--b>a){*b^=*a;*a^=*b;*b^=*a++;}}r(int*a,int s,int y){v(a,a+s);v(a,a+y);v(a+y,a+s);}

Answer 2

APL (4)

¯A⌽B

• A is the number of places to rotate
• B is the name of the array to be rotated

I'm not sure if APL actually required it, but in the implementation I've seen (the internals of) this would take time proportional to A, and constant memory.

Answer 3

Here is a long winded C version of Colin's idea.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int gcd(int a, int b) {
  int t;
  if (a < b) {
    t = b; b = a; a = t;
  }
  while (b != 0) {
    t = a%b;
    a = b;
    b = t;
  }
  return a;
}

double arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
int s_arr = sizeof(arr)/sizeof(double);

/* We assume 1 <= by < s_arr */
void rotate(double *arr, int s_arr, int by) {
  int i, j, f;
  int g = gcd(s_arr,by);
  int n = s_arr/g;
  double t_in, t_out;

  for (i=0; i<g; i++) {
    f = i;
    t_in = arr[f + s_arr - by];
    for (j=0; j<n; j++) {
      t_out = arr[f];
      arr[f] = t_in;
      f = (f + by) % s_arr;
      t_in = t_out;
    }
  }
}

void print_arr(double *arr, int s_arr) {
  int i;
  for (i=0; i<s_arr; i++) printf("%g ",arr[i]);
  puts("");
}

int main() {
  double *temp_arr = malloc(sizeof(arr));
  int i;

  for (i=1; i<s_arr; i++) {
    memcpy(temp_arr, arr, sizeof(arr));
    rotate(temp_arr, s_arr, i);
    print_arr(temp_arr, s_arr);
  }
}

Answer 4

C

Not sure what the criteria is, but since I had fun with the algorithm, here is my entry:

void rotate(int* b, int size, int shift)
{
    int *done;
    int *p;
    int i;
    int saved;
    int c;

    p = b;
    done = p;
    saved = *p;
    for (i = 0; i < size; ++i) {
        c = saved;
        p += shift;
        if (p >= b+size) p -= size;
        saved = *p;
        *p = c;
        if (p == done) {
            p += 1;
            done = p;
            saved = *p;
        }
    }
}

I'll golf it for a good measure too; 126 bytes, can be made shorter:

void r(int*b,int s,int n){int*d,*p,i,t,c;d=p=b;t=*p;for(i=0;i<s;++i){c=t;p+=n;if(p>=b+s)p-=s;t=*p;*p=c;if(p==d){d=++p;t=*p;}}}

Answer 5

I don't see very many C++ solutions here, so I figured I'd try this one since it doesn't count characters.

This is true "in-place" rotation, so uses 0 extra space (except technically swap and 3 ints) and since the loop is exactly N, also fulfills the O(N) complexity.

template <class T, size_t N>
void rot(std::array<T,N>& x, int shift)
{
        size_t base=0;
        size_t cur=0; 
        for (int i = 0; i < N; ++i)
        {
                cur=(cur+shift)%N; // figure out where we are going
                if (cur==base)     // exact multiple so we have to hit the mods when we wrap
                {
                        cur++;
                        base++;
                }
                std::swap(x.at(base), x.at(cur)); // use x[base] as holding area
        }
}

Answer 6

If you perform each of the possible cycles of rotations by n in turn (there are GCD(n, len(arr)) of these), then you only need a single temporary copy of an array element and a few state variables. Like this, in Python:

from fractions import gcd

def rotate(arr, n):
    total = len(arr)
    cycles = gcd(n, total)
    for start in range(0, cycles):
        cycle = [i % total for i in range(start, abs(n * total) / cycles, n)]
        stash = arr[cycle[-1]]
        for j in reversed(range(1, len(cycle))):
            arr[cycle[j]] = arr[cycle[j - 1]]
        arr[cycle[0]] = stash

Answer 7

C (137 characters)

#include <stdio.h>

void rotate(int * array, int n, int k) {
    int todo = (1<<n+1)-1;
    int i = 0, j;
    int tmp = array[0];

    while (todo) {
        if (todo & 1<<i) {
            j = (i-k+n)%n;
            array[i] = todo & 1<<j ? array[j] : tmp;
            todo -= 1<<i;
            i = j;
        } else tmp = array[++i];
    }
}

int main() {
    int a[] = {1,2,3,4,5,6,7,8,9};
    rotate(a, 9, 4);
    for (int i=0; i<9;i++) printf("%d ", a[i]);
    printf("\n");
}

Function rotate minified to 137 characters:

void r(int*a,int n,int k){int m=(1<<n+1)-1,i=0,j,t=a[0];while(m)if(m&1<<i){j=(i-k+n)%n;a[i]=(m&1<<j)?a[j]:t;m-=1<<i;i=j;}else t=a[++i];}

Answer 8

Factor has a built-in type for rotatable arrays, <circular>, so this is actually a O(1) operation:

: rotate ( circ n -- )
    neg swap change-circular-start ;

IN: 1 9 [a,b] <circular> dup 6 rotate >array .
{ 4 5 6 7 8 9 1 2 3 }
IN: 1 9 [a,b] <circular> dup 3 rotate >array .
{ 7 8 9 1 2 3 4 5 6 }

A less cheatish Factor equivalent of Ben Voigt's impressive C solution:

: rotate ( n s -- ) 
    reverse! swap cut-slice [ reverse! ] bi@ 2drop ;

IN: 7 V{ 0 1 2 3 4 5 6 7 8 9 } [ rotate ] keep .
V{ 3 4 5 6 7 8 9 0 1 2 }

Answer 9

JavaScript 45

Went for golf anyway because I like golf. It is at maximum O(N) as long as t is <= size of the array.

function r(o,t){for(;t--;)o.unshift(o.pop())}

To handle t of any proportion in O(N) the following can be used (weighing in at 58 characters):

function r(o,t){for(i=t%o.length;i--;)o.unshift(o.pop())}

Doesn't return, edits the array in place.

Answer 10

REBEL - 22

/_(( \d+)+)( \d+)/$3$1

Input: k expressed as a unary integer using _ as a digit, followed by a space, then a space-delimited array of integers.

Output: A space, then the array rotated.

Example:

___ 1 2 3 4 5/_(( \d+)+)( \d+)/$3$1

Final state:

 3 4 5 1 2

Explanation:

At each iteration, it replaces one _ and an array [array] + tail with tail + [array].

Example:

___ 1 2 3 4 5
__ 5 1 2 3 4
_ 4 5 1 2 3
 3 4 5 1 2

Answer 11

Python 3

from fractions import gcd
def rotatelist(arr, m):
    n = len(arr)
    m = (-m) % n # Delete this line to change rotation direction
    for i0 in range(gcd(m, n)):
        temp = arr[i0]
        i, j = i0, (i0 + m) % n
        while j != i0:
            arr[i] = arr[j]
            i, j = j, (j + m) % n
        arr[i] = temp

Constant memory
O(n) time complexity

Answer 12

Java

public static void rotate(int[] arr, int by) {
    int n = arr.length;
    int i = 0;
    int j = 0;
    while (i < n) {
        int k = j;
        int value = arr[k];
        do {
            k = (k + by) % n;
            int tmp = arr[k];
            arr[k] = value;
            value = tmp;
            i++;
        } while (k != j);
        j++;
    }
}

Demo here.

Minified Javascript, 114:

function rotate(e,r){n=e.length;i=0;j=0;while(i<n){k=j;v=e[k];do{k=(k+r)%n;t=e[k];e[k]=v;v=t;i++}while(k!=j);j++}}

Answer 13

Haskell

This is actually θ(n), as the split is θ(k) and the join is θ(n-k). Not sure about memory though.

rotate 0 xs = xs
rotate n xs | n >= length xs = rotate (n`mod`(length xs)) xs
            | otherwise = rotate' n xs

rotate' n xs = let (xh,xt) = splitAt n xs in xt++xh

Answer 14

Perl 5, 42 bytes

sub r{$a=pop;map{unshift@$a,pop@$a}1..pop}

Try it online!

Subroutine takes the distance to rotate as the first parameter and a reference to the array as the second. Run time is constant based on the distance of the rotation. Array size does not affect run time. Array is modified in place by removing an element from the right and putting it on the left.

Answer 15

Python

def rotate(a, n): a[:n], a[n:] = a[-n:], a[:-n] 

Answer 16

python

   import copy
    def rotate(a, r):
        c=copy.copy(a);b=[]
        for i in range(len(a)-r):   b.append(a[r+i]);c.pop();return b+c

Answer 17

vb.net O(n) (not Constant memory)

Function Rotate(Of T)(a() As T, r As Integer ) As T()     
  Dim p = a.Length-r
  Return a.Skip(p).Concat(a.Take(p)).ToArray
End Function

Answer 18

Ruby

def rotate(arr, n)
  arr.tap{ (n % arr.size).times { arr.unshift(arr.pop) } }  
end

Answer 19

C (118)

Probably was a bit too lenient with some of the specifications. Uses memory proportional to shift % length. Also able to rotate in the opposite direction if a negative shift value is passed.

r(int *a,int l,int s){s=s%l<0?s%l+l:s%l;int *t=malloc(4*s);memcpy(t,a+l-s,4*s);memcpy(a+s,a,4*(l-s));memcpy(a,t,4*s);}

Answer 20

Python 2, 57

def rotate(l,n):
 return l[len(l)-n:len(l)]+l[0:len(l)-n]

If only l[-n:len(l)-n] worked like I'd expect it to. It just returns [] for some reason.

Answer 21

C++, 136

template<int N>void rotate(int(&a)[N],int k){auto r=[](int*b,int*e){for(int t;--e>b;t=*b,*b++=*e,*e=t);};r(a,a+k);r(a+k,a+N);r(a,a+N);}

Answer 22

def r(a,n): return a[n:]+a[:n]

Could someone please check if this actually meets the requirements? I think it does, but it I haven't studied CS (yet).

Answer 23

Java

Swap the last k elements with the first k elements, and then rotate the remaining elements by k. When you have fewer than k elements left at the end, rotate them by k % number of remaining elements. I don't think anyone above took this approach. Performs exactly one swap operation for every element, does everything in place.

public void rotate(int[] nums, int k) {
    k = k % nums.length; // If k > n, reformulate
    rotate(nums, 0, k);
}

private void rotate(int[] nums, int start, int k) {
    if (k > 0) {
        if (nums.length - start > k) { 
            for (int i = 0; i < k; i++) {
                int end = nums.length - k + i;
                int temp = nums[start + i];
                nums[start + i] = nums[end];
                nums[end] = temp;
            }
            rotate(nums, start + k, k); 
        } else {
            rotate(nums, start, k % (nums.length - start)); 
        }
    }
}