What is the time complexity of Compare Version Numbers?

The optimal solution has a time complexity of O(max(n, m)).

What is the space complexity of Compare Version Numbers?

The space complexity is O(n + m).

What data structures are used to solve Compare Version Numbers?

This problem uses Two Pointers, String. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Compare Version Numbers?

Insight 1: The problem requires splitting the version strings into individual revisions. Insight 2: We must handle cases where one version string has fewer revisions than the other by treating missing revisions as 0.

Which companies ask LeetCode 165. Compare Version Numbers in interviews?

This problem is frequently asked at ByteDance, Cisco, Nextdoor, TikTok, Zoho.

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Compare Version Numbers - LeetCode 165 Solution

Compare Version Numbers - Complete Solution Guide

Compare Version Numbers is LeetCode problem 165, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given two version strings , version1 and version2 , compare them. A version string consists of revisions separated by dots '.' . The value of the revision is its integer conversion ignoring leading zeros. To compare version strings, compare their revision values in left-to-right order . If one of the version strings has fewer revisions, treat the missing revision values as 0 . Return the following: If version1 < version2 , return -1. If version1 > version2 , return 1. Otherwise, return 0. Exampl

Detailed Explanation

The problem asks us to compare two version strings, `version1` and `version2`. These strings consist of revisions separated by dots (`.`). We need to compare the revisions numerically, treating missing revisions as 0. Leading zeros in revisions should be ignored (e.g., '1.01' is the same as '1.1'). The function should return -1 if `version1` < `version2`, 1 if `version1` > `version2`, and 0 if they are equal. The version strings contain only digits and dots, and revisions are valid 32-bit integers.

Solution Approach

The solution involves splitting both version strings into arrays of revisions. Then, we iterate through the revisions, comparing corresponding revisions after converting them to integers. If one version string has fewer revisions, we treat the missing revisions as 0. We stop when we find a difference between corresponding revisions or when we have processed all revisions.

Step-by-Step Algorithm

Step 1: Split both version strings `version1` and `version2` into arrays of revision strings using the '.' delimiter.
Step 2: Determine the maximum length of the two revision arrays.
Step 3: Iterate from 0 to `max_len` - 1. In each iteration, get the integer value of the i-th revision from each array. If the index `i` is out of bounds for a particular revision array, treat its value as 0.
Step 4: Compare the integer values of the revisions from `version1` and `version2`. If the revision from `version1` is less than the revision from `version2`, return -1. If it's greater, return 1.
Step 5: If the loop completes without finding any differences, it means the versions are equal. Return 0.

Key Insights

Insight 1: The problem requires splitting the version strings into individual revisions.
Insight 2: We must handle cases where one version string has fewer revisions than the other by treating missing revisions as 0.
Insight 3: Leading zeros need to be ignored when comparing individual revisions.

Complexity Analysis

Time Complexity: O(max(n, m))

Space Complexity: O(n + m)

Topics

This problem involves: Two Pointers, String.

Companies

Asked at: ByteDance, Cisco, Nextdoor, TikTok, Zoho.

Compare Version Numbers | LeetCode 165

Medium

Two Pointers String

Time: O(max(n, m))Space: O(n + m)

Solution Code

Available in:

class Solution {
    public int compareVersion(String version1, String version2) {
        String[] revs1 = version1.split("\\.");
        String[] revs2 = version2.split("\\.");
        
        int n1 = revs1.length;
        int n2 = revs2.length;
        
        int max_len = Math.max(n1, n2);
        
        for (int i = 0; i < max_len; i++) {
            int val1 = i < n1 ? Integer.parseInt(revs1[i]) : 0;
            int val2 = i < n2 ? Integer.parseInt(revs2[i]) : 0;
            
            if (val1 < val2) {
                return -1;
            } else if (val1 > val2) {
                return 1;
            }
        }
        
        return 0;
    }
}

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Problem Info

Description

Problem 165: Compare Version Numbers - Detailed explanation and solution approaches.

Topics

Two Pointers String

Companies

ByteDanceCiscoNextdoorTikTokZoho

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:165

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Split both version strings `version1` and `version2` into arrays of revision strings using the '.' delimiter.
2
Step 2: Determine the maximum length of the two revision arrays.
3
Step 3: Iterate from 0 to `max_len` - 1. In each iteration, get the integer value of the i-th revision from each array. If the index `i` is out of bounds for a particular revision array, treat its value as 0.
4
Step 4: Compare the integer values of the revisions from `version1` and `version2`. If the revision from `version1` is less than the revision from `version2`, return -1. If it's greater, return 1.
5
Step 5: If the loop completes without finding any differences, it means the versions are equal. Return 0.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(max(n, m))

💾

Space Complexity

O(n + m)

📝

Why?

The splitting of the version strings into arrays takes O(n) and O(m) space respectively. The comparison loop iterates a maximum of max(n, m) times, with constant-time operations inside. Thus, the time complexity is dominated by the loop and is O(max(n, m)). The space complexity is determined by the storage of revision strings after splitting and therefore O(n + m).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Two Pointers without splitting (Optimized Space)

Use two pointers, one for each version string, and iterate through the strings character by character. Extract the numbers by iterating until you encounter a '.'. This eliminates the need to store revisions in separate arrays.

Time: O(max(n, m)), where n and m are the lengths of version strings. This approach iterates through the strings at most once.Space: O(1), This is because it uses constant extra space, avoiding the creation of new arrays for revisions.

Trade-offs: This approach saves space, especially for very long version strings with many revisions. However, the code may be slightly more complex to read and implement compared to the splitting approach.

✨

Key Insights

1Insight 1: The problem requires splitting the version strings into individual revisions.
2Insight 2: We must handle cases where one version string has fewer revisions than the other by treating missing revisions as 0.
3Insight 3: Leading zeros need to be ignored when comparing individual revisions.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: `version1` = '1.0', `version2` = '1.0.0'. The solution correctly handles this by treating the missing revision in `version1` as 0, resulting in a return value of 0.
Edge case 2: `version1` = '1.01', `version2` = '1.001'. The solution correctly ignores leading zeros and compares 1 with 1, resulting in a return value of 0.
Edge case 3: Empty revisions are not possible according to the problem constraints, but code should not break if they occurred (though splitting on '.' may produce empty strings and atoi/parseInt will parse them as 0).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle the case where one version string has fewer revisions than the other. Remember to treat missing revisions as 0.

✗Common mistake 2: Not properly handling leading zeros when comparing individual revisions. Use `int()` or `Integer.parseInt()` to remove leading zeros implicitly during the integer conversion.

✗Common mistake 3: Using `strtok` incorrectly in C. `strtok` modifies the original string, which could be problematic if the input string is needed later. Also, `strtok` is not thread-safe.

✗Common mistake 4: Forgetting to use `\.` to escape the dot character in Java's `split()` method, as `.` is a special character in regular expressions.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Software versioning: Comparing versions of software libraries or applications to determine compatibility or update requirements.

🏦Real-world scenario 2: Database schema migrations: Tracking and comparing database schema versions to manage upgrades and downgrades.

🎮Real-world scenario 3: Configuration management: Managing configurations of systems and comparing versions of configuration files to detect changes and ensure consistency.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 13, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Compare Version Numbers"?

For this medium-level Two Pointers problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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