Problem

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 "x1.x2.x3.x4" where 0 <= xi <= 255 and xi 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 "x1:x2:x3:x4:x5:x6:x7:x8" where:

  • 1 <= xi.length <= 4
  • xi is a hexadecimal string which may contain digits, lowercase English letter ('a' to 'f') and upper-case English letters ('A' to 'F').
  • Leading zeros are allowed in xi.

For example, “2001:0db8:85a3:0000:0000:8a2e:0370:7334" and “2001:db8:85a3:0:0:8A2E:0370:7334" are valid IPv6 addresses, while “2001:0db8:85a3::8A2E:037j:7334" and “02001:0db8:85a3:0000:0000:8a2e:0370:7334" are invalid IPv6 addresses.

Examples

Example 1:

Input: queryIP = "172.16.254.1"
Output: "IPv4"
Explanation: This is a valid IPv4 address, return "IPv4".

Example 2:

Input: queryIP = "2001:0db8:85a3:0:0:8A2E:0370:7334"
Output: "IPv6"
Explanation: This is a valid IPv6 address, return "IPv6".

Example 3:

Input: queryIP = "256.256.256.256"
Output: "Neither"
Explanation: This is neither a IPv4 address nor a IPv6 address.

Solution

Method 1 - String

To determine whether a given string queryIP is a valid IPv4 address, a valid IPv6 address, or neither, we need to implement the following checks:

  1. Check for IPv4:

    • An IPv4 address consists of four decimal numbers separated by dots.
    • Each section must be between 0 and 255.
    • No leading zeros are allowed, except for the number 0 itself.

  2. Check for IPv6:

    • An IPv6 address consists of eight groups of four hexadecimal digits separated by colons.
    • Each group must be between 1 and 4 characters long.
    • Leading zeros are allowed.
    • Valid characters include digits (0-9) and letters (a-f, A-F).

Approach

  1. Identify and Split:
    • Use the presence of dots (.) and colons (:) to differentiate between IPv4 and IPv6.
    • Split the string based on dots for IPv4 and colons for IPv6 to check the sections.
  2. Validation Functions:
    • Create helper functions to validate each section of IPv4 and IPv6 addresses.
  3. Return Result:
    • Return “IPv4” if valid IPv4 address, “IPv6” if valid IPv6 address, and “Neither” if none.

Code

Java
public class Solution {
    public String validIPAddress(String queryIP) {
        if (queryIP.chars().filter(ch -> ch == '.').count() == 3) {
            return validateIPv4(queryIP);
        } else if (queryIP.chars().filter(ch -> ch == ':').count() == 7) {
            return validateIPv6(queryIP);
        } else {
            return "Neither";
        }
    }

    private String validateIPv4(String IP) {
        String[] parts = IP.split("\\.", -1);
        if (parts.length != 4) return "Neither";
        for (String part : parts) {
            if (part.length() == 0 || part.length() > 3) return "Neither";
            if (part.charAt(0) == '0' && part.length() != 1) return "Neither";
            for (char ch : part.toCharArray()) {
                if (!Character.isDigit(ch)) return "Neither";
            }
            if (Integer.parseInt(part) > 255) return "Neither";
        }
        return "IPv4";
    }

    private String validateIPv6(String IP) {
        String[] parts = IP.split(":", -1);
        if (parts.length != 8) return "Neither";
        String hexDigits = "0123456789abcdefABCDEF";
        for (String part : parts) {
            if (part.length() == 0 || part.length() > 4) return "Neither";
            for (char ch : part.toCharArray()) {
                if (hexDigits.indexOf(ch) == -1) return "Neither";
            }
        }
        return "IPv6";
    }
}
Python
class Solution:
    def validIPAddress(self, queryIP: str) -> str:
        if queryIP.count('.') == 3:
            return self.validateIPv4(queryIP)
        elif queryIP.count(':') == 7:
            return self.validateIPv6(queryIP)
        else:
            return "Neither"
    
    def validateIPv4(self, IP: str) -> str:
        parts = IP.split(".")
        if len(parts) != 4:
            return "Neither"
        for part in parts:
            if len(part) == 0 or len(part) > 3:
                return "Neither"
            if part[0] == '0' and len(part) != 1:
                return "Neither"
            if not part.isdigit() or int(part) > 255:
                return "Neither"
        return "IPv4"
    
    def validateIPv6(self, IP: str) -> str:
        parts = IP.split(":")
        if len(parts) != 8:
            return "Neither"
        hex_digits = "0123456789abcdefABCDEF"
        for part in parts:
            if len(part) == 0 or len(part) > 4:
                return "Neither"
            for ch in part:
                if ch not in hex_digits:
                    return "Neither"
        return "IPv6"

Complexity

  • ⏰ Time complexity: O(n), where n is the length of the input string. Each character is processed once.
  • 🧺 Space complexity: O(1), excluding the space required for storing result strings.