Description
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:
<li>The given <code>address</code> is a valid IPv4 address.</li>
Solutions
Solution 1: Direct Replacement
We can directly replace the '.' in the string with '[.]'.
The time complexity is O(n), where n is the length of the string. Ignoring the space consumption of the answer, the space complexity is O(1).
PythonJavaC++GoTypeScript
class Solution: def defangIPaddr(self, address: str) -> str: return address.replace('.', '[.]')(code-box)
