93. Restore IP Addresses

[ Problem ]

A valid IP address consists of exactly four integers separated by single dots. Each integer is between 0 and 255 (inclusive) and cannot have leading zeros.

• For example, "0.1.2.201" and "192.168.1.1" are valid IP addresses, but "0.011.255.245", "192.168.1.312" and "192.168@1.1" are invalid IP addresses.

Given a string s containing only digits, return all possible valid IP addresses that can be formed by inserting dots into s. You are not allowed to reorder or remove any digits in s. You may return the valid IP addresses in any order.

 

Example 1:

Input: s = "25525511135"

Output: ["255.255.11.135","255.255.111.35"]

Example 2:

Input: s = "0000"

Output: ["0.0.0.0"]

Example 3:

Input: s = "101023"

Output: ["1.0.10.23","1.0.102.3","10.1.0.23","10.10.2.3","101.0.2.3"]

 

Constraints:

• 1 <= s.length <= 20
• s consists of digits only.

[ Solution ]

class Solution:
    def restoreIpAddresses(self, s: str) -> List[str]:
        ret = []

        def isValid(ss: str) -> bool:
            if len(ss) > 1 and ss[0] == '0':
                return False
            elif len(ss) == 3 and int(ss) > 255:
                return False
            else:
                return True

        def findIpAddr(octet: int, remain: str, res: str):
            if octet == 5:
                if len(remain) == 0:
                    ret.append(res[1:])
                return
            for idx in range(min(3, len(remain))):
                octetIp = remain[:idx+1]
                sub = remain[idx+1::]
                if isValid(octetIp):
                    findIpAddr(octet+1, sub, res + "." + octetIp)

        findIpAddr(1, s, "")

        return ret

 

 

The ip in an octet can't be above 255, so there needs to check whether it is above 255 or not. Moreover ip can't start with '0' if it is above 10.