Defanging an IP Address - Practice Coding Problems

Defanging an IP Address - Problem

String Easy

Given a valid (IPv4) IP address, return a defanged version of that IP address.

A defanged IP address replaces every period "." with "[.]".

Input & Output

Example 1 — Basic Case

$ Input: address = "1.1.1.1"

› Output: "1[.]1[.]1[.]1"

💡 Note: Replace each dot with [.]: 1.1.1.1 becomes 1[.]1[.]1[.]1

Example 2 — Standard IP

$ Input: address = "255.100.50.0"

› Output: "255[.]100[.]50[.]0"

💡 Note: Each dot is replaced with [.] while numbers remain unchanged

Example 3 — Different Numbers

$ Input: address = "192.168.1.1"

› Output: "192[.]168[.]1[.]1"

💡 Note: Three dots become three [.] replacements

Constraints

The given address is a valid IPv4 address.

Visualization

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Understanding the Visualization

Input

Valid IPv4 address with dots

Process

Replace each dot with [.]

Output

Defanged IP address

Key Takeaway

🎯 Key Insight: Simple string replacement - replace all dots with [.] brackets

Asked in

a Amazon 15 G Google 8

The key insight is that this is a simple string replacement problem. We need to replace every dot with [.]. The best approach is using built-in replace functions. Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Character by Character Replacement	O(n)	O(n)	Iterate through each character and build result manually
Built-in String Replace	O(n)	O(n)	Use language built-in string replacement functions

Character by Character Replacement — Algorithm Steps

Iterate through each character in the IP address
If character is '.', append '[.]' to result
If character is not '.', append the character to result
Return the built result string

Visualization

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Step-by-Step Walkthrough

Input String

Start with original IP address

Character Processing

Check each character, replace dots with [.]

Result

Build defanged IP address

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

char* solution(char* address) {
    static char result[100];
    int j = 0;
    for (int i = 0; address[i] != '\0'; i++) {
        if (address[i] == '.') {
            result[j++] = '[';
            result[j++] = '.';
            result[j++] = ']';
        } else {
            result[j++] = address[i];
        }
    }
    result[j] = '\0';
    return result;
}

int main() {
    char address[20];
    fgets(address, sizeof(address), stdin);
    address[strcspn(address, "\n")] = 0;
    char* result = solution(address);
    printf("%s\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string of length n

✓ Linear Growth

Space Complexity

O(n)

Result string can be up to 3 times the original length

⚡ Linearithmic Space

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Low Frequency

~5 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Character by Character Replacement — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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