Strictly speaking, the time complexity of 'stop at 5' would be:

O(min(N, f(k))) where f(k) is the number of elements needed to inspect for the given inputs (implying 5 <= f(k) <= N). This means the 'stop at 5' solution is better in all cases except the worst case.

The overhead of checking the count is practically nothing and I struggle to think of a scenario (extremely small dictionaries?) where it isn't worth it to keep the count and stop.

Imo;

O(N) vs O(N) - in worst case where autocomplete words are at the end of the list, runtime is same for both.

Ω(N) vs Ω(1) - in the best case where the autocomplete words are at the start of the list, it may be worth it to stop looping when we have found 5.

Is it? O(n) versus O(1)? Not?

Yes, this iterates through every element of the list. I don't think the perfomance gained from returning after 5 results is worth it in this case, the algorithm is still of the same time complexity, and I think it looks cleaner this way.

I don't think the perfomance gained from returning after 5 results is worth it in this case, the algorithm is still of the same time complexity, and I think it looks cleaner this way.

Solution is elegant but what about performance? Does this iterate through every element in the list even after 5 are found?

The assignment says the double and triple have to be the same digit

I dont understand how this works. Can someone explain?