Discussion on: Java Daily Coding Problem #001

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I’d use a set and iterate over the list and for each element I’d check if the set contains that element and if it does, it means we found a positive answer and if not, add k - arr[i] to the set.

for (int number: given){
   if (set.contains(number))
      return true;
    set.add(k - number);
}
 return false;

Thank you for sharing!

Trung Nguyen • Apr 21 '19 • Edited

I thought the same and this is way more efficient runtime.

Andrew (he/him) • Apr 21 '19

Very clever! I like this answer.

Andrew (he/him) • Apr 21 '19

What is the Big-O of this solution, though?

In the worst case, if we have...

2 elements: we will do a contains() on a set with 0 elements, then an add(), then a contains() on a set with 1 element, then an add().

3 elements: all of the above, plus a contains() on a set with 2 elements, then an add()

Assuming we skip the last add() when we know we've reached the end of the list, worst-case would be calling contains() on N sets of length 0 through N-1, plus calling add() N-1 times.

This is definitely less space-efficient than my solution, because of the set, but is it more time-efficient? What do you think?

Andrei Gatej • Apr 21 '19

I think it depends on how set works under the hood.
At first, because there is only one for loop, one might think it’s only O(n).

But how does the `contains()’ method actually work? I have just read here and it says that set will internally iterate its elements and compare each element to the object passed as parameter.

Another way of making sure that there isn’t another for loop going on behind the scenes is by using a map. Because accessing an item requires O(1) time.(as far as I know)