# LeetCode 165. Compare Version Numbers

## Description

https://leetcode.com/problems/compare-version-numbers/

Given two version numbers, `version1` and `version2`, compare them.

Version numbers consist of one or more revisions joined by a dot `'.'`. Each revision consists of digits and may contain leading zeros. Every revision contains at least one character. Revisions are 0-indexed from left to right, with the leftmost revision being revision 0, the next revision being revision 1, and so on. For example `2.5.33` and `0.1` are valid version numbers.

To compare version numbers, compare their revisions in left-to-right order. Revisions are compared using their integer value ignoring any leading zeros. This means that revisions `1` and `001` are considered equal. If a version number does not specify a revision at an index, then treat the revision as `0`. For example, version `1.0` is less than version `1.1` because their revision 0s are the same, but their revision 1s are `0` and `1` respectively, and `0 < 1`.

Return the following:

• If `version1 < version2`, return `-1`.
• If `version1 > version2`, return `1`.
• Otherwise, return `0`.

Example 1:

```Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both "01" and "001" represent the same integer "1".
```

Example 2:

```Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: version1 does not specify revision 2, which means it is treated as "0".
```

Example 3:

```Input: version1 = "0.1", version2 = "1.1"
Output: -1
Explanation: version1's revision 0 is "0", while version2's revision 0 is "1". 0 < 1, so version1 < version2.
```

Example 4:

```Input: version1 = "1.0.1", version2 = "1"
Output: 1
```

Example 5:

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

Constraints:

• `1 <= version1.length, version2.length <= 500`
• `version1` and `version2` only contain digits and `'.'`.
• `version1` and `version2` are valid version numbers.
• All the given revisions in `version1` and `version2` can be stored in a 32-bit integer.

## Explanation

split version string by ‘.’ and then compare each split

## Python Solution

``````class Solution:
def compareVersion(self, version1: str, version2: str) -> int:

version1_list = version1.split('.')
version2_list = version2.split('.')

l1 = len(version1_list)
l2 = len(version2_list)

for i in range(0, max(l1, l2)):
v1 = int(version1_list[i]) if i < l1 else 0
v2 = int(version2_list[i]) if i < l2 else 0

if v1 < v2:
return -1
elif v1 > v2:
return 1

return 0``````
• Time Complexity: ~N
• Space Complexity: ~1