Exercise 2 solution

    def dedupe(items):
        seen = set()
        out = []
        for x in items:
            if x not in seen:
                seen.add(x)
                out.append(x)
        return out

    print(dedupe([3, 1, 3, 2, 1]))    # [3, 1, 2]

Two structures:
  - a set (seen) for the "have I met this already?" test, which is one hash
    lookup per item instead of a scan of everything kept so far;
  - a list (out) to keep the first-seen order, because a set has no order.

Checking "x not in out" instead of using the set would work but would be
O(n squared) overall, which becomes very slow on long inputs.

Shorter alternative (values must be hashable): list(dict.fromkeys(items)),
since a dict keeps insertion order and its keys are unique.