Recursive way to count distinct number of elements in array

Dorki :

Let's assume I have this array: ARR = {5, 7, 3, 3, 7, 5} and I also have the size ( = 6 in this example ) so the recursive function should return 3.

this is the declaration of the function/method:

int f(arr, size);

I tried this thing:

count = 0;
if(size == 1)
   return 1;
if(x[i] != f(arr, size-1)
    count++;

return count;

but it doesn't work, as f(arr, size-1) doesn't walk through all the elements of the array and compare.

hopefully you guys could help!

Sweeper :

Here's one way to do it:

private static int f(int[] arr, int size) {
    if (size <= 1) return 0; // there can't be duplicates if there are not even 2 elements!
    return f(arr, size - 1) + (inArray(arr, size - 1, arr[size - 1]) ? 1 : 0);
}

private static boolean inArray(int[] arr, int size, int elem) {
    if (size == 0) return false;
    return arr[size - 1] == elem || inArray(arr, size - 1, elem);
}

Basically the logic is this:

• The size indicates the first N elements in arr that we actually care about.
• If size is less than 2, we know there can't be any duplicates, so return 0.
• Now for the recursion case, we return either 1 or 0 depending on whether the last element is in the rest of the array, plus whatever number of duplicates in the rest of the array ("the last element" means array[size - 1] and "the rest" means calling the function with size - 1.)
• To determine whether an element is in an array, I used a recursive method as well, with a similar idea (check if the last element is elem, then check if the rest contains elem).