# Compare Version Numbers | InterviewBit | Solution Explained

I explain the solution to the problem Compare Version Numbers on InterviewBit in detail. Problem Link.

## 0. Compare Version Numbers: Problem Discussion

### 0.0. IO Format

- Given: two strings which are version numbers,
**ver1**and**ver2** - Goal:
- If
**ver1**>**ver2**return**1**, - If
**ver1**<**ver2**return**-1**, - otherwise return
**0**.

- If

### 0.1. Examples

## 1. Compare Version Numbers: Observations & Logic

### 1.0. Intuitions

- the versions are strings, and we need a way to compare the versions
- converting strings to integers will help us compare them
- we know that a version has different numbers split by ‘.’, so we can
- split the string by ‘.’ character
- map all the string values to an integer
- get the list

- we can do all that in python using one single line:
`list(map(int, ver.split('.')))`

All right! We have extracted the information out of the input. Now what?

We need a way to compare two items in the two lists, simultaneously. So, we can use the zip operator in python3.

class Solution: def compareVersion(self, ver1, ver2): ver1 = list(map(int, ver1.split('.'))) ver2 = list(map(int, ver2.split('.'))) for v1, v2 in zip(ver1, ver2): if v1 > v2: return 1 elif v2 > v1: return -1 return 0

### 1.1. Implementation Issue

Here’s the thing though, zip only works well if the length of the iterables are the same. That is, only when len(v1) == len(v2) list.

In the case where there is a smaller list, zip will terminate with it.

For example, if we `zip([1,2,3], [1,2,3,7])`

we will terminate at 3! We will never get a chance to compare 7. So what do we do?

Here lies a key point: if one version is shorter in length to the other, we can append dummy zeroes to it!

For example, if we have a tricky case like `[1,3,3,0,0,7]`

and `[1,3,3]`

, we need to be able to **pad** zeroes like so:

One way to not do this manually is to use `zip_longest`

from itertools, as the implementation below will show.

## 2. Compare Version Numbers: Implementation

### 2.0. Code

from itertools import zip_longest class Solution: def compareVersion(self, ver1, ver2): ver1 = list(map(int, ver1.split('.'))) ver2 = list(map(int, ver2.split('.'))) for v1, v2 in zip_longest(ver1, ver2, fillvalue=0): if v1 > v2: return 1 elif v2 > v1: return -1 return 0

### 2.1. Code (without zip_longest)

class Solution: def compareVersion(self, ver1, ver2): ver1 = list(map(int, ver1.split('.'))) ver2 = list(map(int, ver2.split('.'))) i, j = 0, 0 m, n = len(ver1), len(ver2) if m > n: ver2.extend([0]*(m-n)) elif n > m: ver1.extend([0]*(n-m)) for v1, v2 in zip(ver1, ver2): if v1 > v2: return 1 elif v2 > v1: return -1 return 0

### 2.2. Complexity Analysis

- Time:
`O(N)`

, where N is the number of elements in the longest list, which can be the same order as the number of characters in the string:`1.1.1 --> [1,1,1]`

- Space:
`O(N)`

, since we have to store the lists of integers of both the versions separately

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Please write the proof.

This solution need an amendment for 2 digit number.