为什么二和问题的hash解是O(n)？

Question

Sorry to be a pain. I know it's a classic and I have checked, but I still can't understand why the time cost is O(n).

Here's my code.

def two_sum(L, sum):
    targets = {}
    for i in range(len(L)):
        if L[i] in targets:
            return (targets[L[i]], i)
        else: targets[sum - L[i]] = i

While I get that L is only iterated over once and that the value look up for a given key is O(1), isn't the search for an actual key in a dict O(n)? Doesn't that mean that for each value in L there is an O(n) look up cost to find if that value is key in the dict, making it a total of O(n^2)?

Apologies if I'm being stupid but when I searched for an answer everyone just seems to be saying the lookup cost in a hash table is O(1), so it's only O(n) * O(1) = O(n).

Edit: Just to clarify, the dict size in this problem grows with n.

Answer 1

Ryan Wilson's link along with interjay's wisdom helped me realise how this works.

It doesn't iterate over a list of keys when you call key in dict, it just hashes the key and sees if it gets a valid answer!

Cheers everyone.