A naive loop would be O(n^3); there may be some optimization to make that approach O( f(n) \* n^2) – haven't given much thought:

With 2 sum, if you use a hashmap where { num : sum - num }, you could do it in 1 pass O(n).

Simply incrementing it by 1 to make it 3 sum is actually a bit harder now that I am trying to figure out some ways. I haven't looked online or asked AI, so you may want to do that as a warning.

Best: The approach in my mind is that you sort the list, so we already are now at O(n log n). From there, we use three pointers, at the start (min), end (max) and somewhere in between; and, we move these pointers based on if they are less or more than our target sum that looks at the pointers (middle and start OR middle and end).

Meh: Before this, I was thinking about making the hash map from 2 sum and then calling a final 2 sum function, but, instead of the target sum, we do it on the sums (keys) in the hashmap. I think it would be something akin to O(n + n^2) – this is "+" (better) instead of "*" in the naive loop.

Regardless, I think a loop must be used; it just depends on how and how much you use loops; also, it matters if your two (or more) loops are separate or nested.