Efficient algorithm to find first available name

Question

I have an array containing names of items. I want to give the user the option to create items without specifying their name, so my program will have to supply a unique default name, like "Item 1".

The challenge is that the name has to be unique so i have to check all the array for that default name, and if there is an item with the same name i have to change the name to be "Item 2" and so on until i find an available name.

The obvious solution will be something like that:

String name = "Item ";
for (int i = 0; !isAvailable(name + i) ; i++);

My problem with that algorithm is that it runs at O(N^2).

I wonder if there is a known (or new) more efficient algorithm to solve this case.

In other words my question is this: Is there any algorithm that finds the first greater-than-zero number that dose NOT exist in a given array, and runs at less that N^2?

Answer 1

You can certainly do it in O(N) time, N being the number of items in your array:

• One of "Item 1", "Item 2", ... "Item N+1" must be free, so create an array of N+1 flags.
• Traverse the items, and for each name if it is of form "Item k" with 0 < k <= N+1, set that flag.
• Scan the flag array for the first clear flag.

Additional memory requirement is N+1 bits, which certainly beats any data structure that actually stores all N names.

Answer 2

Yes, there is.

First sort the array. Then run through it and return the first element whose value is not equal to its index (plus 1). The sort is O(n log n), the final step is O(n), so the entire thing is O(n log n).

If you put all items into a hash table, you can do it in O(n) at the cost of some space and an additional O(1) step at creation of new items. Since each element needs to be visited, O(n) is clearly optimal.

I'd be interested to see if there's an O(n) way to do this, without using any "persistent" data structure like the hash table. (And assuming unbounded integers, otherwise a bucket sort could be used as an O(n) sorting algorithm).