Exercise 7: Remove All Occurrences of a Value — Possible Solution
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numbers = [1, 2, 3, 2, 4, 2, 5]

index = 0
while index < len(numbers):
    if numbers[index] == 2:
        del numbers[index]
        # Do NOT increment index here — the next element has just
        # slid down into this same position.
    else:
        index += 1

print(numbers)

Output:
[1, 3, 4, 5]

WHY THIS WORKS AS AN ANSWER
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This is the single trickiest gotcha in the whole exercise set. When
del numbers[index] removes an element, every element after it shifts
one position to the LEFT to fill the gap — so the element that was at
index + 1 is now sitting at index. If the loop naively did index += 1
after every deletion (the same as it does for a non-match), it would
skip over that shifted element entirely, checking it never at all.

The fix is to only advance the index when nothing was deleted. When a
match is found and removed, index deliberately stays put, so the very
next iteration re-checks that same position — which now holds a
different value. This is exactly why the increment in a while loop
can't be written on "autopilot" at the bottom of the loop the way it
often can in a simple counting loop: sometimes the correct behaviour
is to NOT advance, and getting that wrong here doesn't cause an
infinite loop, it causes a silent, wrong answer (a leftover 2 in the
result) — arguably a worse bug to have, since nothing crashes or hangs
to reveal it.