Defanging an IP Address

easy string replace simulation

Problem

You are given a valid IPv4 address as a string. Defang it by replacing every period character "." with the string "[.]". Return the resulting string.

Inputaddress = "1.1.1.1"

Output"1[.]1[.]1[.]1"

Each of the three dots becomes the three-character token [.] while every other character is copied unchanged.

address

def defang_ip_addr(address):
    out = []
    for ch in address:
        if ch == '.':
            out.append('[.]')
        else:
            out.append(ch)
    return ''.join(out)

function defangIPaddr(address) {
  let out = "";
  for (const ch of address) {
    if (ch === ".") out += "[.]";
    else out += ch;
  }
  return out;
}

class Solution {
    public String defangIPaddr(String address) {
        StringBuilder out = new StringBuilder();
        for (char ch : address.toCharArray()) {
            if (ch == '.') out.append("[.]");
            else out.append(ch);
        }
        return out.toString();
    }
}

string defangIPaddr(string address) {
    string out;
    for (char ch : address) {
        if (ch == '.') out += "[.]";
        else out += ch;
    }
    return out;
}

Explanation

"Defanging" just means making an IP address harmless to click by turning every dot into the token [.]. The solution is a straightforward single scan that copies the string while swapping the dots.

We build the answer in a list (or string builder) called out. Walking character by character, whenever we see a period '.' we append the three-character string "[.]"; for any other character we append it unchanged.

Using a list and joining at the end avoids repeatedly recreating an immutable string, which keeps it efficient and clear.

Because each input character produces a fixed output, the work is linear in the length of the address.

Example: "1.1.1.1" becomes "1[.]1[.]1[.]1" — each of the three dots expands to [.] while the digits are copied as-is.

Time: O(n) Space: O(n)