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.

Source: leetcode | Difficulty: EASY
Tags: String algorithms

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:
		new_string = ''
		for s in address:
			if s == '.':
				new_string += '[.]'
			else:
				new_string += s
		return new_string