Compare Version Numbers

compare_version_numbers.py

"""
Compare two version numbers version1 and version2.
If version1 > version2 return 1; if version1 < version2 return -1;otherwise return 0.

You may assume that the version strings are non-empty and contain only digits and the . character.

The . character does not represent a decimal point and is used to separate number sequences.

For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of
the second first-level revision.

You may assume the default revision number for each level of a version number to be 0. For example, version number
3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level
revision number are both 0.

Example 1:
Input: version1 = "0.1", version2 = "1.1"
Output: -1

Example 2:
Input: version1 = "1.0.1", version2 = "1"
Output: 1

Example 3:
Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

Example 4:
Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”

Example 5:
Input: version1 = "1.0", version2 = "1.0.0"
Output: 0

Explanation: The first version number does not have a third level revision number, which means its third level revision
number is default to "0"
Time complexity: O(N+M+max(N,M)), where N and M are lengths of input strings. First O(N+M) will be for the splitting.
Space complexity: O(N+M) to store arrays nums1 and nums2
"""


class Solution:
    def compareVersion(self, version1: str, version2: str) -> int:
        nums1 = version1.split(".")
        nums2 = version2.split(".")
        n1, n2 = len(nums1), len(nums2)
        for i in range(max(n1, n2)):
            i1 = int(nums1[i]) if i < n1 else 0
            i2 = int(nums2[i]) if i < n2 else 0
            if i1 != i2:
                return 1 if i1 > i2 else -1
        return 0