Solution Explanation: Validate IP Address

This problem asks us to validate if a given string represents a valid IPv4 or IPv6 address. The solution uses a simulation approach, creating separate functions to validate each IP type.

Approach:

The solution cleverly separates the validation logic into two distinct functions: isIPv4 and isIPv6. This improves readability and maintainability.

isIPv4(s):

1. Handles trailing dots: It first checks if the input string ends with a dot (.), which is invalid for IPv4.
2. Splits the string: It splits the string by dots (.) into four parts. If the resulting array doesn't have exactly four elements, it's not a valid IPv4.
3. Validates each part: It iterates through each part, verifying:
   • It's not a number with a leading zero (e.g., "010").
   • It's a valid integer between 0 and 255 (inclusive). It utilizes a helper function (convert in Java/C++/Go/TypeScript, direct conversion in Python/Rust) to safely convert the string to an integer and handle potential overflow.
4. Returns true/false: If all parts pass the validation, it returns true; otherwise, false.

isIPv6(s):

1. Handles trailing colons: Similar to IPv4, it checks for invalid trailing colons (`:').
2. Splits the string: The string is split by colons (`:'). The resulting array must have eight elements for a valid IPv6.
3. Validates each part: Each part is validated to:
   • Have a length between 1 and 4 characters (inclusive).
   • Contain only hexadecimal digits (0-9, a-f, A-F). The code efficiently uses isxdigit in C++ or checks against a valid character set in other languages.
4. Returns true/false: Returns true if all parts pass, and false otherwise.

Main function validIPAddress(queryIP):

The main function simply calls isIPv4 and isIPv6 sequentially. If either function returns true, the corresponding IP type is returned; otherwise, "Neither" is returned.

Time Complexity: O(n), where n is the length of the input string. The splitting and validation processes iterate through the string linearly.

Space Complexity: O(n) in the worst case. This is due to the temporary arrays created during the splitting process. In the best case, the space complexity is O(1).

Code Examples (Python):

class Solution:
    def validIPAddress(self, queryIP: str) -> str:
        def is_ipv4(s: str) -> bool:
            ss = s.split(".")
            if len(ss) != 4:
                return False
            for t in ss:
                if len(t) > 1 and t[0] == "0":
                    return False
                try:
                    if not 0 <= int(t) <= 255:
                        return False
                except ValueError:
                    return False
            return True
 
        def is_ipv6(s: str) -> bool:
            ss = s.split(":")
            if len(ss) != 8:
                return False
            for t in ss:
                if not 1 <= len(t) <= 4:
                    return False
                if not all(c in "0123456789abcdefABCDEF" for c in t):
                    return False
            return True
 
        if is_ipv4(queryIP):
            return "IPv4"
        if is_ipv6(queryIP):
            return "IPv6"
        return "Neither"

This Python code efficiently implements the algorithm described above. The try-except block handles potential ValueError during integer conversion. Other language implementations follow a similar structure, adapting to the nuances of their respective syntax and libraries.