Problems › Python › validation
Compare two version strings
A deploy tool decides whether the version on the server is behind the one being released.
- Versions are dot-separated numbers: 1.2.10 has parts 1, 2 and 10.
- Compare part by part numerically, so 1.10 is above 1.2 — not below it, as a text sort would have it.
- A missing part counts as zero, so 1.2 and 1.2.0 are the same version.
- Leading zeros mean nothing: 1.01 is 1.1.
- Return -1 when the first is older, 1 when it is newer, 0 when they match.
compare_versions(first: string, second: string) → int
Where you start
def compare_versions(first: str, second: str) -> int:
Worked examples
| Call | Result |
|---|---|
compare_versions("1.2.10", "1.10.2") | -1 |
compare_versions("1.2", "1.2.0") | 0 |
compare_versions("2.0", "1.9.9") | 1 |
compare_versions("1.0.0", "1.0.0") | 0 |
Hint
Walk both to the length of the longer one, reading a missing part as zero.
Reference solution in Python
def compare_versions(first: str, second: str) -> int:
a = first.split('.')
b = second.split('.')
for i in range(max(len(a), len(b))):
x = int(a[i]) if i < len(a) else 0
y = int(b[i]) if i < len(b) else 0
if x != y:
return -1 if x < y else 1
return 0