[Solved] 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.

Question

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.

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.

Constraints:

  • queryIP consists only of English letters, digits and the characters '.' and ':'.

Python Solution

class Solution:
    def validIPAddress(self, a: str) -> str:
        def isIPv4(s):
            try : return str(int(s))==s and 0<=int(s)<=255
            except: return False
        def isIPv6(s):
            if len(s)>4:return False
            try: return int(s,16)>=0 and s[0]!='-'
            except: return False

        if a.count('.')==3 and all(isIPv4(x) for x in a.split('.')):
            return('IPv4')
        if a.count(':')==7 and all(isIPv6(x) for x in a.split(':')):
            return('IPv6')
        return('Neither')
Abhishek Sharma
Abhishek Sharma

Started my Data Science journey in my 2nd year of college and since then continuously into it because of the magical powers of ML and continuously doing projects in almost every domain of AI like ML, DL, CV, NLP.

Articles: 520

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