Challenge 1: Remove Duplicates, Keep Order — Possible Solution
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numbers = [3, 1, 3, 2, 1, 4, 2]

seen = set()
result = []

for n in numbers:
    if n not in seen:
        seen.add(n)
        result.append(n)

print(result)

Output:
[3, 1, 2, 4]

WHY THIS WORKS AS AN ANSWER
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A plain set(numbers) would remove duplicates correctly but lose the
original order entirely, since the chapter's own set section describes
sets as unordered — that's the exact trade-off the hint is pointing at.

This solution uses two collections together, each for what it's best
at: seen is a set used purely for its fast "n in seen" membership test
(the chapter's own set section highlights this speed), while result
is a list used purely to preserve insertion order, since lists (unlike
sets) remember the order items were added.

For each number, n not in seen checks whether it's been added before.
If not, it's added to both seen (so future duplicates of it are
caught) and result (so it appears exactly once, in its original
first-seen position). Duplicate values later in the list — like the
second 3, second 1, second 2 — are simply skipped because they're
already in seen.