165. Compare Version Numbers - Detailed Explanation

Problem Statement

You are given two version numbers, version1 and version2, represented as strings. A version number is composed of one or more revision numbers separated by dots '.'. For example, "1.0.1" and "0.1" are valid version numbers.

Your task is to compare these version numbers:

• Return 1 if version1 > version2
• Return -1 if version1 < version2
• Return 0 if the two versions are equal

Note:

• The revision numbers are compared as integers, so "1.01" and "1.001" are considered equal because both revisions represent the integer 1.
• If a version number does not specify a revision at a particular level, treat it as 0. For example, "1.0" is equal to "1.0.0".

Example 1

• Input: version1 = "1.01", version2 = "1.001"
• Output: 0
• Explanation: After ignoring leading zeros, both versions become "1.1" and are equal.

Example 2

• Input: version1 = "1.0", version2 = "1.0.0"
• Output: 0
• Explanation: The missing revision in version1 is treated as 0. Both versions represent "1.0.0".

Example 3

• Input: version1 = "0.1", version2 = "1.1"
• Output: -1
• Explanation: Comparing the first revision, 0 < 1, so version1 is smaller.

Constraints

• The version strings are non-empty.
• The version strings contain only digits and dots.
• The dots separate revision numbers; there will be no leading, trailing, or consecutive dots.

Hints

1. Splitting the Version String:
   Think about how you can break down the version string into its revision numbers. Splitting the string by the dot character (".") is a natural choice.

2. Integer Comparison:
   Once split, convert the string parts into integers to naturally handle the removal of any leading zeros. This way, "001" and "1" both become the integer 1.

3. Handling Unequal Lengths:
   If the two version strings have a different number of revisions, treat the missing revisions in the shorter version as 0. For example, compare "1.0" with "1.0.0" by treating the missing third revision in "1.0" as 0.

Approach Overview

There are a couple of approaches to solve this problem:

1. Direct Comparison Approach (Optimal)

Idea:

• Step 1: Split both version strings by the dot (".") to obtain arrays of revision numbers.
• Step 2: Convert these revisions to integers.
• Step 3: Iterate over the maximum length of both arrays. At each index, compare the corresponding revisions (using 0 if an index is out of bounds).
• Step 4: As soon as a difference is found, return 1 or -1 accordingly. If no differences are found after checking all revisions, return 0.

Time Complexity: O(n) where n is the number of revision parts (the length of the longer version string split array).
Space Complexity: O(n) for the arrays of revision numbers.

2. Two-Pointer Approach (Variation)

Idea:

• Instead of splitting the entire string at once, you can use two pointers to parse each revision number on the fly.
• Compare the parsed integers as you move through the strings.
• This approach avoids extra space for arrays but is more complex to implement.

Time Complexity: O(n)
Space Complexity: O(1) (ignoring the input space)

For simplicity and clarity, the direct comparison approach is easier to implement and understand, making it ideal for coding interviews.

Detailed Step-by-Step Walkthrough (Direct Comparison Approach)

1. Split the Version Strings:

   • For version1, use split('.') to create an array of strings.
   • For version2, do the same.
   • For example, if version1 is "1.0.1" and version2 is "1", then you get:
     • version1 parts: ["1", "0", "1"]
     • version2 parts: ["1"]

2. Convert and Compare:

   • Convert each revision from a string to an integer.
   • Determine the maximum length between the two arrays. In the above example, the maximum length is 3.
   • Loop from index 0 to max_length - 1:
     • At each index i, if a version does not have a revision (i.e., index out of bounds), treat it as 0.
     • Compare the integers:
       • If the revision from version1 is greater than that of version2, return 1.
       • If it is smaller, return -1.
   • If all revisions are equal, return 0.

3. Example Walkthrough:
   Consider comparing "1.0.1" vs. "1":

   • Index 0: Compare 1 vs. 1 → equal.
   • Index 1: Compare 0 vs. 0 (since version2 has no index 1, treat as 0) → equal.
   • Index 2: Compare 1 vs. 0 (again, version2 has no index 2) → 1 > 0, so return 1.

Code Implementation

Python Code