Program 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

METHOD 1 (Use temp array)

Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7
1) Store d elements in a temp array
   temp[] = [1, 2]
2) Shift rest of the arr[]
   arr[] = [3, 4, 5, 6, 7, 6, 7]
3) Store back the d elements
   arr[] = [3, 4, 5, 6, 7, 1, 2]

Time complexity O(n)
Auxiliary Space: O(d)

METHOD 2 (Rotate one by one)

leftRotate(arr[], d, n)
start
  For i = 0 to i < d
    Left rotate all elements of arr[] by one
end

To rotate by one, store arr[0] in a temporary variable temp, move arr[1] to arr[0], arr[2] to arr[1] …and finally temp to arr[n-1]

Let us take the same example arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2
Rotate arr[] by one 2 times
We get [2, 3, 4, 5, 6, 7, 1] after first rotation and [ 3, 4, 5, 6, 7, 1, 2] after second rotation.

/*Function to left Rotate arr[] of size n by 1*/ void leftRotatebyOne(int arr[], int n);   /*Function to left rotate arr[] of size n by d*/ void leftRotate(int arr[], int d, int n) {   int i;   for (i = 0; i < d; i++)     leftRotatebyOne(arr, n); }   void leftRotatebyOne(int arr[], int n) {   int i, temp;   temp = arr[0];   for (i = 0; i < n-1; i++)      arr[i] = arr[i+1];   arr[i] = temp; }   /* utility function to print an array */ void printArray(int arr[], int size) {   int i;   for(i = 0; i < size; i++)     printf("%d ", arr[i]); }   /* 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; }

Time complexity: O(n*d)
Auxiliary Space: O(1)

METHOD 3 (A Juggling Algorithm)
This is an extension of method 2. Instead of moving one by one, divide the array in different sets
where number of sets is equal to GCD of n and d and move the elements within sets.
If GCD is 1 as is for the above example array (n = 7 and d =2), then elements will be moved within one set only, we just start with temp = arr[0] and keep moving arr[I+d] to arr[I] and finally store temp at the right place.

Here is an example for n =12 and d = 3. GCD is 3 and

Let arr[] be {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

a) Elements are first moved in first set – (See below diagram for this movement)



          arr[] after this step --> {4 2 3 7 5 6 10 8 9 1 11 12}

b) Then in second set.
          arr[] after this step --> {4 5 3 7 8 6 10 11 9 1 2 12}

c) Finally in third set.
          arr[] after this step --> {4 5 6 7 8 9 10 11 12 1 2 3}

/* function to print an array */ void printArray(int arr[], int size);   /*Fuction to get gcd of a and b*/ int gcd(int a,int b);   /*Function to left rotate arr[] of siz n by d*/ void leftRotate(int arr[], int d, int n) {   int i, j, k, temp;   for (i = 0; i < gcd(d, n); i++)   {     /* move i-th values of blocks */     temp = arr[i];     j = i;     while(1)     {       k = j + d;       if (k >= n)         k = k - n;       if (k == i)         break;       arr[j] = arr[k];       j = k;     }     arr[j] = temp;   } }   /*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]); }   /*Fuction to get gcd of a and b*/ int gcd(int a,int b) {    if(b==0)      return a;    else      return gcd(b, a%b); }   /* 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; }

Time complexity: O(n)
Auxiliary Space: O(1)

Please see following posts for other methods of array rotation:
Block swap algorithm for array rotation
Reversal algorithm for array rotation

References:
http://www.cs.bell-labs.com/cm/cs/pearls/s02b.pdf

Please write comments if you find any bug in above programs/algorithms.