1139EasyString

Wrapping the Dots of an Address

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

The string address is a valid IPv4 address: four groups of decimal digits joined by single dots, as in 192.168.0.1.

Return the address with every dot swapped for the three characters [.].

Examples

Example 1

Input: address = "192.168.0.1"
Output: "192[.]168[.]0[.]1"

Each of the three dots becomes the bracketed form and the digit groups pass through untouched.

Example 2

Input: address = "0.0.0.0"
Output: "0[.]0[.]0[.]0"

All four groups are a single zero, and the three dots are wrapped just the same.

Example 3

Input: address = "12.34.56.78"
Output: "12[.]34[.]56[.]78"

Only the dots change, so all eight digits come through exactly as they were.

Constraints

address is a valid IPv4 address

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def defang_i_paddr(address: str) -> str:

Java

public String defangIPaddr(String address)

The intended approach

Time O(n) over the address.Space O(n) for the pieces.

Nothing about the address needs checking or splitting apart: the four groups come through exactly as they are and only the dots change. So walk the characters once, pushing the three-character replacement whenever a dot turns up and the character itself otherwise. Collecting the pieces in a list and joining them at the end matters for longer strings, since adding to a string one piece at a time copies everything already built each time round.

Pythonentry point: defang_i_paddr

def defang_i_paddr(address):
    pieces = []
    for ch in address:
        if ch == ".":
            pieces.append("[.]")
        else:
            pieces.append(ch)
    return "".join(pieces)

The obvious version

Cut the address apart at its dots into the digit groups, then stitch the groups back together putting the three-character replacement between each neighbouring pair. Costs a fresh string for every join.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_defang_i_paddr(address):
    parts = address.split(".")
    out = parts[0]
    for i in range(1, len(parts)):
        out = out + "[.]" + parts[i]
    return out

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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