MediumString

Validate IP Address

1 approach, code in every language

Given a string, decide whether it is a valid IPv4 address, a valid IPv6 address, or neither. An IPv4 address is four decimal groups separated by dots, where each group is a number from 0 to 255 with no leading zeros (except the single digit 0). An IPv6 address is eight groups separated by colons, where each group has one to four hexadecimal digits and both letter cases are allowed.

Return "IPv4" for a valid IPv4 string, "IPv6" for a valid IPv6 string, and "Neither" for anything that does not fit either format. Empty groups and trailing separators are invalid.

Example 1

Input: queryIP = "10.0.1.38"

Output: "IPv4"

Four groups, each between 0 and 255 with no leading zeros, so it is a valid IPv4 address.

Example 2

Input: queryIP = "2001:0db8:85a3:0:0:8a2e:0370:7334"

Output: "IPv6"

Eight colon-separated groups, each with one to four hexadecimal digits, so it is a valid IPv6 address.

Constraints

1 ≤ queryIP.length ≤ 40
queryIP consists of English letters, digits, and the characters '.' and ':'
The answer is exactly one of "IPv4", "IPv6", or "Neither"

Split on separators and validate each group

Try to parse the string as IPv4 by splitting on dots, otherwise try IPv6 by splitting on colons, checking every group against its own rules.

Loading animation…

Intuition

The two formats are distinguished by their separator, so we can test each independently and return the first one that fully passes. The subtle part is the group rules: IPv4 groups must be one to three digits, at most 255, and cannot carry a leading zero, while IPv6 groups must be one to four hexadecimal digits. Rejecting empty groups also rules out trailing or doubled separators, which is exactly where naive splits tend to go wrong.

Algorithm walkthrough

1Split the string on '.' and require exactly four parts to consider IPv4.
2For each IPv4 part, verify it is one to three digits, has no leading zero unless it is a single '0', and is numerically at most 255.
3If IPv4 fails, split on ':' and require exactly eight parts to consider IPv6.
4For each IPv6 part, verify it is one to four characters and every character is a hexadecimal digit in either case.
5Return "IPv4" or "IPv6" for the format that passes, and "Neither" if both checks fail.

Implementation

class Solution:
    def validIPAddress(self, queryIP: str) -> str:
        def is_ipv4(ip: str) -> bool:
            parts = ip.split('.')
            if len(parts) != 4:
                return False
            for part in parts:
                if not part.isdigit():
                    return False
                if len(part) > 1 and part[0] == '0':
                    return False
                if int(part) > 255:
                    return False
            return True

        def is_ipv6(ip: str) -> bool:
            parts = ip.split(':')
            if len(parts) != 8:
                return False
            hex_digits = "0123456789abcdefABCDEF"
            for part in parts:
                if len(part) < 1 or len(part) > 4:
                    return False
                for ch in part:
                    if ch not in hex_digits:
                        return False
            return True

        if is_ipv4(queryIP):
            return "IPv4"
        if is_ipv6(queryIP):
            return "IPv6"
        return "Neither"

Time

O(n)

Space

O(n)

n is the length of the input; splitting and scanning each group touches every character a constant number of times.

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