468-ValidateIPAddress.py

#  Write a function to check whether an input string is a valid IPv4 address or
#  IPv6 address or neither.

# IPv4 addresses are canonically represented in dot-decimal notation, which
# consists of four decimal numbers, each ranging from 0 to 255, separated by
# dots ("."), e.g.,172.16.254.1;

# Besides, leading zeros in the IPv4 is invalid. For example, the address
# 172.16.254.01 is invalid.

# IPv6 addresses are represented as eight groups of four hexadecimal digits,
# each group representing 16 bits. The groups are separated by colons (":"). For
# example, the address 2001:0db8:85a3:0000:0000:8a2e:0370:7334 is a valid one.
# Also, we could omit some leading zeros among four hexadecimal digits and some
# low-case characters in the address to upper-case ones, so
# 2001:db8:85a3:0:0:8A2E:0370:7334 is also a valid IPv6 address(Omit leading
# zeros and using upper cases).

# However, we don't replace a consecutive group of zero value with a single
# empty group using two consecutive colons (::) to pursue simplicity. For
# example, 2001:0db8:85a3::8A2E:0370:7334 is an invalid IPv6 address.

# Besides, extra leading zeros in the IPv6 is also invalid. For example, the
# address 02001:0db8:85a3:0000:0000:8a2e:0370:7334 is invalid.

# Note: You may assume there is no extra space or special characters in the
# input string. 

from collections import Counter
import re

class Solution:
    def validIPAddress(self, IP: str) -> str:
        if Counter(IP)["."] == 3:
            for num in IP.split("."):
                try:
                    int(num)
                except:
                    return "Neither"
                if int(num) > 255 or re.match('[^a-zA-Z0-9]', num) or num[0] == '0' and len(num) > 1:
                    return "Neither"
            return "IPv4"
        elif Counter(IP)[":"] == 7:
            for num in IP.split(":"):
                if len(num) > 4 or len(num) == 0:
                    return "Neither"
                for n in num:
                    if ord(n.lower()) > 102 or re.match('[^a-zA-Z0-9]', n):
                        return "Neither"
            return "IPv6"
        return "Neither"