What is the time complexity of Validate IP Address?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Validate IP Address?

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

What are the key insights for solving Validate IP Address?

Insight 1: The problem can be solved by first identifying whether the input is potentially IPv4 or IPv6 based on the presence of '.' or ':' characters. Insight 2: String splitting is essential to extract the individual components of the IP address.

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Validate IP Address - LeetCode 468 Solution

Q: How to solve Validate IP Address?

The solution approach involves three main steps: 1. Determine the IP version (IPv4 or IPv6) by counting the occurrences of '.' and ':' characters. 2. Split the input string based on the delimiter ('.' for IPv4, ':' for IPv6). 3. Validate each part based on the rules of the corresponding IP version. If any validation fails, return 'Neither'. If all validations pass, return the IP version ('IPv4' or 'IPv6').

Q: Which companies ask LeetCode 468. Validate IP Address in interviews?

This problem is frequently asked at Cisco, Deutsche Bank, Flexport, Nvidia, ServiceNow and 3 more companies.

Validate IP Address - Complete Solution Guide

Validate IP Address is LeetCode problem 468, 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 a string queryIP , return "IPv4" if IP is a valid IPv4 address, "IPv6" if IP is a valid IPv6 address or "Neither" if IP is not a correct IP of any type. A valid IPv4 address is an IP in the form "x 1 .x 2 .x 3 .x 4 " where 0 <= x i <= 255 and x i cannot contain leading zeros. For example, "192.168.1.1" and "192.168.1.0" are valid IPv4 addresses while "192.168.01.1" , "192.168.1.00" , and "[email protected]" are invalid IPv4 addresses. A valid IPv6 address is an IP in the form "x 1 :x 2 :x 3 :x 4

Detailed Explanation

The problem requires validating if a given string `queryIP` is a valid IPv4 address, a valid IPv6 address, or neither. A valid IPv4 address consists of four decimal numbers (0-255) separated by dots ('.'), with no leading zeros. A valid IPv6 address consists of eight hexadecimal numbers (0-FFFF) separated by colons (':'). Leading zeros are allowed in IPv6 parts. The function should return 'IPv4', 'IPv6', or 'Neither' based on the input string.

Solution Approach

The solution approach involves three main steps: 1. Determine the IP version (IPv4 or IPv6) by counting the occurrences of '.' and ':' characters. 2. Split the input string based on the delimiter ('.' for IPv4, ':' for IPv6). 3. Validate each part based on the rules of the corresponding IP version. If any validation fails, return 'Neither'. If all validations pass, return the IP version ('IPv4' or 'IPv6').

Step-by-Step Algorithm

Step 1: Count the number of '.' and ':' characters in the input `queryIP`.
Step 2: If the count of '.' is 3, assume it's potentially an IPv4 address and call `validate_ipv4()`.
Step 3: If the count of ':' is 7, assume it's potentially an IPv6 address and call `validate_ipv6()`.
Step 4: If neither condition is met, return 'Neither'.
Step 5: In `validate_ipv4()`, split the string by '.' delimiter.
Step 6: Iterate through each part of the IPv4 address:
Step 7: Check if the part contains only digits. If not, return 'Neither'.
Step 8: Check for leading zeros in parts with length greater than 1. If present, return 'Neither'.
Step 9: Convert the part to an integer and check if it's between 0 and 255 (inclusive). If not, return 'Neither'.
Step 10: If all parts are valid, return 'IPv4'.
Step 11: In `validate_ipv6()`, split the string by ':' delimiter.
Step 12: Iterate through each part of the IPv6 address:
Step 13: Check if the part length is between 1 and 4 (inclusive). If not, return 'Neither'.
Step 14: Attempt to convert the part to an integer with base 16 (hexadecimal). Use a try-except block to handle `ValueError` if the conversion fails. If conversion fails, return 'Neither'.
Step 15: If all parts are valid, return 'IPv6'.

Key Insights

Insight 1: The problem can be solved by first identifying whether the input is potentially IPv4 or IPv6 based on the presence of '.' or ':' characters.
Insight 2: String splitting is essential to extract the individual components of the IP address.
Insight 3: Validation of each component requires checking for numeric or hexadecimal format, valid range (0-255 for IPv4, 0-FFFF for IPv6), and the absence of leading zeros in IPv4.
Insight 4: Error handling, especially using try-except blocks, is crucial for handling invalid input formats during number conversions.
Insight 5: The number of parts after splitting the IP address should exactly match the IP version. Four for IPv4, eight for IPv6.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: String.

Companies

Asked at: Cisco, Deutsche Bank, Flexport, Nvidia, ServiceNow, Sprinklr, Turing, X.

Validate IP Address | LeetCode 468

Medium

String

Time: O(1)Space: O(1)

Solution Code

Available in:

class Solution {
    public String validIPAddress(String queryIP) {
        if (queryIP.chars().filter(ch -> ch == '.').count() == 3) {
            return validate_ipv4(queryIP);
        } else if (queryIP.chars().filter(ch -> ch == ':').count() == 7) {
            return validate_ipv6(queryIP);
        } else {
            return "Neither";
        }
    }

    private String validate_ipv4(String ip) {
        String[] parts = ip.split("\\.");
        for (String part : parts) {
            if (!part.matches("\\d+")) {
                return "Neither";
            }
            if (part.length() > 1 && part.charAt(0) == '0') {
                return "Neither";
            }
            try {
                int num = Integer.parseInt(part);
                if (!(0 <= num && num <= 255)) {
                    return "Neither";
                }
            } catch (NumberFormatException e) {
                return "Neither";
            }
        }
        return "IPv4";
    }

    private String validate_ipv6(String ip) {
        String[] parts = ip.split(":");
        for (String part : parts) {
            if (!(1 <= part.length() && part.length() <= 4)) {
                return "Neither";
            }
            try {
                Integer.parseInt(part, 16);
            } catch (NumberFormatException e) {
                return "Neither";
            }
        }
        return "IPv6";
    }
}

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Problem Info

Description

Problem 468: Validate IP Address - Detailed explanation and solution approaches.

Topics

String

Companies

CiscoDeutsche BankFlexportNvidiaServiceNowSprinklrTuringX

Stats

Difficulty:Medium

Languages:4

Companies:8

Problem #:468

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

15 steps

1
Step 1: Count the number of '.' and ':' characters in the input `queryIP`.
2
Step 2: If the count of '.' is 3, assume it's potentially an IPv4 address and call `validate_ipv4()`.
3
Step 3: If the count of ':' is 7, assume it's potentially an IPv6 address and call `validate_ipv6()`.
4
Step 4: If neither condition is met, return 'Neither'.
5
Step 5: In `validate_ipv4()`, split the string by '.' delimiter.
6
Step 6: Iterate through each part of the IPv4 address:
7
Step 7: Check if the part contains only digits. If not, return 'Neither'.
8
Step 8: Check for leading zeros in parts with length greater than 1. If present, return 'Neither'.
9
Step 9: Convert the part to an integer and check if it's between 0 and 255 (inclusive). If not, return 'Neither'.
10
Step 10: If all parts are valid, return 'IPv4'.
11
Step 11: In `validate_ipv6()`, split the string by ':' delimiter.
12
Step 12: Iterate through each part of the IPv6 address:
13
Step 13: Check if the part length is between 1 and 4 (inclusive). If not, return 'Neither'.
14
Step 14: Attempt to convert the part to an integer with base 16 (hexadecimal). Use a try-except block to handle `ValueError` if the conversion fails. If conversion fails, return 'Neither'.
15
Step 15: If all parts are valid, return 'IPv6'.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(1) because the number of parts in IPv4 and IPv6 addresses is fixed (4 and 8 respectively). The operations performed on each part, such as validation checks and string splitting, take constant time. The space complexity is O(1) because we're using a fixed amount of extra space, regardless of the input string's length. The size of the `parts` array is constant (4 for IPv4 and 8 for IPv6).

📖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

Regular Expressions

Use regular expressions to validate the IP address formats. Regular expressions can provide a concise way to define the valid patterns for IPv4 and IPv6 addresses.

Time: O(1)Space: O(1)

Trade-offs: While concise, regular expressions can be less readable and potentially less performant than the current approach. Debugging complex regular expressions can also be challenging. The current approach is generally faster and more easily understood, particularly for those less familiar with regex. However, very complex validation might be simplified via Regex.

✨

Key Insights

1Insight 1: The problem can be solved by first identifying whether the input is potentially IPv4 or IPv6 based on the presence of '.' or ':' characters.
2Insight 2: String splitting is essential to extract the individual components of the IP address.
3Insight 3: Validation of each component requires checking for numeric or hexadecimal format, valid range (0-255 for IPv4, 0-FFFF for IPv6), and the absence of leading zeros in IPv4.
4Insight 4: Error handling, especially using try-except blocks, is crucial for handling invalid input formats during number conversions.
5Insight 5: The number of parts after splitting the IP address should exactly match the IP version. Four for IPv4, eight for IPv6.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: IPv4 address with leading zeros (e.g., "192.168.01.1"). The solution explicitly checks for leading zeros and returns 'Neither'.
Edge case 2: IPv4 or IPv6 address with an invalid number of parts (e.g., "192.168.1" or "2001:db8::1"). The solution checks the count of '.' or ':' characters to ensure it is 3 or 7 respectively, before proceeding with validation. In C, it explicitly checks the `part_count` after splitting to match IPv4 and IPv6.
Edge case 3: IPv4 part greater than 255 (e.g., "256.256.256.256"). The solution converts each part to an integer and checks if it's within the valid range (0-255).
Edge case 4: IPv6 part with length greater than 4 (e.g., "2001:0db8:85a3:0000:0000:8a2e:0370:73345"). The solution checks the length of each part and returns 'Neither' if it's not within the valid range (1-4).
Edge case 5: IPv6 part with non-hexadecimal characters (e.g., "2001:0db8:85a3:0000:0000:8a2e:0370:733g"). The solution attempts to convert each part to an integer with base 16, which will raise a ValueError if the part contains non-hexadecimal characters. C uses isxdigit to determine hex characters.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check for leading zeros in IPv4 addresses.

✗Common mistake 2: Not handling exceptions (e.g., ValueError) when converting IPv6 parts to integers with base 16.

✗Common mistake 3: Not correctly validating the range of IPv4 parts (0-255).

✗Common mistake 4: Not correctly validating the length of IPv6 parts (1-4 characters).

✗Common mistake 5: Failing to check the exact number of components (4 for IPv4 and 8 for IPv6).

✅

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: Network security - Validating IP addresses is crucial for firewalls, intrusion detection systems, and other security appliances to filter and analyze network traffic.

🏦Real-world scenario 2: Web server logging - Web servers record the IP addresses of clients accessing the server. Validating these IP addresses can help ensure data integrity and identify potential malicious activity.

🎮Real-world scenario 3: Data validation - Validating IP addresses in user input or data feeds to ensure data quality and prevent errors in applications.

📱Real-world scenario 4: IP address geolocation - Determining the geographical location of a user based on their IP address requires valid IP address input.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

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

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: September 17, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Validate IP Address"?

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

🤖 Need help with this problem?