Exercise 9: Invert a Dictionary — Possible Solution
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student_ids = {"Alice": 1001, "Bob": 1002, "Cleo": 1003, "Dev": 1001}

inverted = {}
names = list(student_ids.keys())
index = 0

while index < len(names):
    name = names[index]
    student_id = student_ids[name]

    if student_id in inverted:
        # Collision: this ID already maps to an earlier name.
        # Deliberate choice — keep the FIRST name seen, ignore later ones.
        pass
    else:
        inverted[student_id] = name

    index += 1

print(inverted)

Output:
{1001: 'Alice', 1002: 'Bob', 1003: 'Cleo'}

WHY THIS WORKS AS AN ANSWER
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Alice and Dev both have ID 1001, so a naive inversion — always doing
inverted[student_id] = name with no check — would silently let whichever
one is processed LAST overwrite the other, and since Dev comes after
Alice in the dictionary, Dev would win by pure accident of insertion
order. That's an easy mistake to ship without ever noticing it, since
nothing crashes.

This solution makes the collision an explicit, deliberate decision
instead: "if student_id in inverted" checks whether this ID has already
been claimed, and if so, keeps the first name seen (Alice) rather than
letting Dev silently overwrite it. Depending on the real requirement,
"keep the last one" or "store both names in a list" would be equally
valid alternative decisions — the important part is that the choice is
made on purpose, not left to whatever order the dictionary happens to
iterate in.