Given a valid (IPv4) IP address, return a defanged version of that IP address.
A defanged IP address replaces every period "." with "[.]".
 
Example 1:
Input: address = "1.1.1.1"
Output: "1[.]1[.]1[.]1"
Example 2:
Input: address = "255.100.50.0"
Output: "255[.]100[.]50[.]0"

 
Constraints:

The given address is a valid IPv4 address.

Difficulty: introductory

Starter code:

class Solution:
    def defangIPaddr(self, address: str) -> str:
        

Solutions:
class Solution:
    def defangIPaddr(self, address: str) -> str:
        return address.replace('.', '[.]')


class Solution:
    def defangIPaddr(self, address: str) -> str:
        stringArr  = address.split('.')
        
        return '[.]'.join(stringArr)
        
        
        
        


class Solution:
    def defangIPaddr(self, address: str) -> str:
        return '[.]'.join(address.split('.'))

class Solution:
    def defangIPaddr(self, address: str) -> str:
        new_string =''
        for s in address:
            if s =='.':
                new_string+='[.]'
            else:
                new_string+=s
        return new_string


class Solution:
    def defangIPaddr(self, address: str) -> str:
        return address.replace('.','[.]')