Sort Version Numbers with a Dataclass

Your tool lists releases such as 1.2.0, 1.10.0 and 0.9.9, oldest first. Sorting the strings gets the order wrong. "1.10.0" < "1.2.0" is True, because strings compare one character at a time. Sort records of three integers instead.

solve([[1, 0, 10], [1, 0, 2], [0, 9, 9], [1, 0, 2]])
# ["0.9.9", "1.0.2", "1.0.2", "1.0.10"]

What to write

Fill in the Version dataclass:

• Give it three int fields in this order: major, minor, patch.
• Decorate it with @dataclass(slots=True, frozen=True, order=True). Leave the other options at their defaults, so all three fields take part in comparisons.

order=True compares major first, then minor, then patch, so 1.99.99 sorts before 2.0.0. frozen=True stops code from changing a version after you make it. slots=True stores the fields in fixed slots instead of a dictionary on each object, so you cannot add other attributes to a Version. The article Save Memory with slots=True covers it later in this track.

Then fill in solve(versions):

• Each inner list holds three non-negative integers, in the order major, minor, patch.
• Build a Version from each list and sort the versions from lowest to highest.
• Return each version as a string in the form "major.minor.patch", such as "1.0.10", with no spaces and no zero-padding.
• Keep duplicates, and leave the input list unchanged.

Your code must pass mypy --strict.

Edge cases

• An empty list returns an empty list.
• You do not need to check for malformed input.

Sort the Version records first, then turn them into strings. If you turn them into strings first and sort those, "1.0.10" sorts before "1.0.2".