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Defanging an IP Address

Easy

Google

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Topics:

Strings

Given a valid IPv4 address, your task is to defang it. Defanging an IP address involves replacing 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 provided address is guaranteed to be a valid IPv4 address.

Can you implement a function that takes an IPv4 address as a string and returns its defanged version? Describe the time and space complexity of your solution.

Defanging an IP Address

Easy

Google

6 views

Topics:

Strings

Given a valid IPv4 address, your task is to defang it. Defanging an IP address involves replacing 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 provided address is guaranteed to be a valid IPv4 address.

Can you implement a function that takes an IPv4 address as a string and returns its defanged version? Describe the time and space complexity of your solution.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What constitutes a valid IP address format? Can you provide a concrete example of an IP address I should expect as input?
Should I validate the input IP address to ensure it follows the standard IPv4 or IPv6 format before defanging?
What should the function return if the input is null, empty, or not a valid IP address?
Are leading or trailing spaces allowed in the input string, and should they be trimmed before processing?
Is it acceptable to modify the input string in place, or should I return a new string with the defanged IP address?

Brute Force Solution

Approach

The brute force method for defanging an IP address means we carefully examine each character in the address one by one. When we find a period, we replace it with the special "[.]" sequence. We repeat this process until the entire address has been checked.

Here's how the algorithm would work step-by-step:

Take the original IP address.
Look at the very first character.
If that character is a period, replace it with "[.]".
If that character is anything else, leave it as it is.
Move on to the next character in the IP address.
Again, check if it's a period; if so, replace it with "[.]", otherwise leave it.
Continue doing this for every single character in the IP address, one after the other.
Once you've gone through the entire IP address and made all the necessary replacements, you're done!

Code Implementation

def defang_i_p_address(ip_address):
    defanged_ip = ''
    for character in ip_address:

        # Need to replace periods with '[.]'
        if character == '.':

            defanged_ip += '[.]'

        # All other characters are kept
        else:

            defanged_ip += character
    return defanged_ip

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each character of the IP address once. The input size, denoted as n, is the length of the IP address string. For each of the n characters, a constant-time operation (comparison and potential replacement) is performed. Therefore, the time complexity is directly proportional to the length of the IP address, resulting in O(n).

Space Complexity

O(N) – The provided algorithm iterates through each character of the input IP address and potentially replaces periods with '[.]'. This implies creating a new string to store the modified IP address. In the worst case, if the IP address consists entirely of periods, each character will be replaced by '[.]', resulting in a new string of length 3N. Thus, the auxiliary space required is proportional to the length of the original IP address, denoted as N, where N is the length of the IP address.

Optimal Solution

Approach

The goal is to replace each period in the IP address with '[.]'. We can achieve this directly by going through the IP address and making the substitutions. This is efficient because we only look at each part of the IP address once.

Here's how the algorithm would work step-by-step:

Take the IP address as a string.
Go through the string, and whenever you find a period, replace it with '[.]'.
After going through the entire string and making all the replacements, you will have the defanged IP address.
Return the modified string.

Code Implementation

def defang_ip_address(ip_address):
    defanged_ip = ""

    # Iterate through the IP address string.
    for character in ip_address:
        # Replace periods with '[.]'
        if character == '.':

            #Defanged IP requires replacement
            defanged_ip += '[.]'

        else:
            defanged_ip += character

    # Return the defanged IP address.
    return defanged_ip

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the IP address string once to find and replace each period. The input size, n, represents the length of the IP address string. Since the algorithm examines each character in the string exactly once, the time complexity grows linearly with the length of the input string. Therefore, the time complexity is O(n).

Space Complexity

O(N) – The solution constructs a new string by replacing each period in the input IP address with '[.]'. In the worst-case scenario, the IP address consists entirely of periods, so every character needs to be replaced. This results in a new string that could be significantly longer than the original. Therefore, the auxiliary space required to store this new string grows linearly with the length of the original IP address, N.

Edge Cases

Case	How to Handle
Null or empty IP address string	Return an empty string or null, depending on requirements, after checking for null or empty input.
IP address with invalid characters (e.g., letters, symbols other than dots)	Throw an IllegalArgumentException or return an error string indicating invalid input detected.
IP address with more than three digits in a single octet	Throw an IllegalArgumentException or return an error string indicating invalid input, as each octet should not exceed 255.
IP address with fewer or more than four octets	Throw an IllegalArgumentException or return an error string indicating invalid format (must have four octets separated by three dots).
IP address with leading zeros in an octet (e.g., 192.168.001.1)	Consider this a valid octet by trimming the leading zeros, or reject based on specific requirements.
Extremely long IP address string that could cause memory issues (DoS attack)	Check the input string length and reject excessively long inputs to prevent potential resource exhaustion.
IP address with spaces surrounding the dots or numbers	Trim leading/trailing spaces from the entire IP string and from each octet before processing, or reject outright.
IP address where an octet is greater than 255	Throw an IllegalArgumentException or return an error string since octets must be between 0 and 255 inclusive.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What constitutes a valid IP address format? Can you provide a concrete example of an IP address I should expect as input?
Should I validate the input IP address to ensure it follows the standard IPv4 or IPv6 format before defanging?
What should the function return if the input is null, empty, or not a valid IP address?
Are leading or trailing spaces allowed in the input string, and should they be trimmed before processing?
Is it acceptable to modify the input string in place, or should I return a new string with the defanged IP address?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Take the original IP address.
Look at the very first character.
If that character is a period, replace it with "[.]".
If that character is anything else, leave it as it is.
Move on to the next character in the IP address.
Again, check if it's a period; if so, replace it with "[.]", otherwise leave it.
Continue doing this for every single character in the IP address, one after the other.
Once you've gone through the entire IP address and made all the necessary replacements, you're done!

Code Implementation

def defang_i_p_address(ip_address):
    defanged_ip = ''
    for character in ip_address:

        # Need to replace periods with '[.]'
        if character == '.':

            defanged_ip += '[.]'

        # All other characters are kept
        else:

            defanged_ip += character
    return defanged_ip

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Take the IP address as a string.
Go through the string, and whenever you find a period, replace it with '[.]'.
After going through the entire string and making all the replacements, you will have the defanged IP address.
Return the modified string.

Code Implementation

def defang_ip_address(ip_address):
    defanged_ip = ""

    # Iterate through the IP address string.
    for character in ip_address:
        # Replace periods with '[.]'
        if character == '.':

            #Defanged IP requires replacement
            defanged_ip += '[.]'

        else:
            defanged_ip += character

    # Return the defanged IP address.
    return defanged_ip

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty IP address string	Return an empty string or null, depending on requirements, after checking for null or empty input.
IP address with invalid characters (e.g., letters, symbols other than dots)	Throw an IllegalArgumentException or return an error string indicating invalid input detected.
IP address with more than three digits in a single octet	Throw an IllegalArgumentException or return an error string indicating invalid input, as each octet should not exceed 255.
IP address with fewer or more than four octets	Throw an IllegalArgumentException or return an error string indicating invalid format (must have four octets separated by three dots).
IP address with leading zeros in an octet (e.g., 192.168.001.1)	Consider this a valid octet by trimming the leading zeros, or reject based on specific requirements.
Extremely long IP address string that could cause memory issues (DoS attack)	Check the input string length and reject excessively long inputs to prevent potential resource exhaustion.
IP address with spaces surrounding the dots or numbers	Trim leading/trailing spaces from the entire IP string and from each octet before processing, or reject outright.
IP address where an octet is greater than 255	Throw an IllegalArgumentException or return an error string since octets must be between 0 and 255 inclusive.