There is an old trick for rotating elements in an array (I first saw it in Programming Pearls)

Say you want to rotate an array to the left by three elements.

First reverse the first three elements, next reverse the remaining elements, and then reverse the entire array.

Starting Array:
1 2 3 4 5 6 7

After reversing the first three elements
3 2 1 4 5 6 7

After reversing the remaining elements
3 2 1 7 6 5 4

Finally reverse the entire array to get the final rotated array
4 5 6 7 1 2 3

Reversing portions of the array can be done in place so you don t need any extra memory.

You can leave the data in place, and have a "base index" member to indicate where the array should start. You then need to use this to adjust the index when accessing the array. The array itself should be private, and only accessed through accessor functions that do the adjustment. Something like this:

class quack
{
public:
    explicit quack(int size) : items(new Item[size]), size(size), base(0) {}
    ~quack() {delete [] items;}

    void rotate(int n)      {base = (base + n) % size;}
    Item &operator[](int i) {return items[(base + i) % size];}

private:
    quack(quack const&) = delete;          // or implement these if you want
    void operator=(quack const&) = delete; // the container to be copyable

    Item *items;
    int   size;
    int   base;
};

although I d call it something like RotatableArray, rather than quack.

As usual, if you really have to physically rotate the elements, the correct answer for C++ would be to use std::rotate, which does exactly what you want to do.

If you have to implement it manually (as a practice assignment), take a look at these slides for algorithms from John Bentley s "Programming Pearls".

doing the rotations one by one is really not the way to go. If you are doing anything more than 2 or 3 rotations it gets really slow really quick.

edit: as a final thought... putting the elements in a (double) linked looped list (so the final element points to the first), would require for a rotate to only move the head pointer a few elements. (The head pointer being a pointer to indicate which element in the looped list is the beginning).

this is by far the quickest (and easiest) way to do a rotate on a list of elements

Really the way to do it is to use indexes instead of pointers.

int to = 0;
int from = (to + nRotations) % count;
if (to == from)
    return;

for (int i=0; i < count; i++) {
   swap(from, to);
   from = advance(from);
   to = advance(to);
}

// ...
static inline int advance(int n, int count) { return (n + 1) % count; }

I may have an alternate solution to rotating the array inline. Rather than the old trick of reversing sets of elements, as proposed earlier, this approach works as follows:

Initialization:

(Note q = amount to shift left, n = length of array)

1. Determine the first source element, which is located at x1=q%n
2. The destination element is at x2=0
3. char, ch1, is the ar[x1] element
4. char, ch2, is the ar[x2] element

Loop on i=0 to n-1, where n = length of array

1. Overwrite the destination element, ar[x2] with ch1
2. Set ch1 = ch2
3. Set x1 = x2
4. Set x2 = x2 - q
5. If x2 is negative due to the above subtraction, add n to it
6. ch2 = ar[x2]

The following may help explain how this works.

Example, rotate to the left by 2 characters:

a b c d e f g

c d e f g a b

x1    ch1    x2    ch2
2     c      0     a
0     a      5     f
5     f      3     d
3     d      1     b
1     b      6     g
6     g      4     e
4     e      2     c

As you can see, this requires no more than n iterations, so it is linear time algorithm that also rotates inline (requires no additional storage other than the few temporary variables).

Here is a function that implements the above algorithm so you can try it out:

void rotate(char *ar, int q)
{
    if (strlen(ar) < 2)
    {
        return;
    }

    if (q <= 0)
    {
        return;
    }

    char ch1;
    char ch2;
    int x1;
    int x2;
    int i;
    int n;

    n = strlen(ar);

    q %= n;

    if (q == 0)
    {
        return;
    }

    x1 = q;
    ch1 = ar[x1];
    x2 = 0;
    ch2 = ar[x2];

    for (i=0;i<n;i++)
    {
        ar[x2] = ch1;
        ch1 = ch2;

        x1 = x2;

        x2 -= q;

        if (x2 < 0)
        {
            x2 += n;
        }

        ch2 = ar[x2];
    }
}

Here s one that I got by modifying the code here:

template<class It>
It rotate(It begin, It const middle, It end)
{
    typename std::iterator_traits<It>::difference_type i = 0, j;
    if (begin != middle && middle != end)
    {
        while ((i = std::distance(begin, middle)) !=
               (j = std::distance(middle,   end)))
        {
            It k = middle;
            std::advance(
                k,
                std::max(typename std::iterator_traits<It>::difference_type(),
                j - i));
            std::swap_ranges(k, end, begin);
            if (i > j) { std::advance(begin, j); }
            else { std::advance(end, -i); }
        }
    }
    return std::swap_ranges(middle - i, middle, middle);
}

Here is code for sdtom solution

void rotateArray(int arr[], int low, int high)
{
    while  (low<high) {
        int temp = arr[low];
        arr[low]=arr[high];
        arr[high]=temp;
        low++;
        high--;
    }
}
void rotate (int arr[], int k,int uperLimit)
{
    rotateArray(arr,0,k);
    rotateArray(arr,k+1,uperLimit-1);
    rotateArray(arr,0,uperLimit-1);

}

https://stackoverflow.com/q/17119000/2383578 : This is somewhat similar to what is discussed here. Rotating the array by a particular amount.

There are many ways to do array rotation by d places.

1. Block swap algorithm.
2. Juggling Algorithm.
3. Reversal Algorithm.

The link below is from Programming Pearls pdf. Check this out. Explained very clearly with code.

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