The IPv4 address validation algorithm

Leandre — Sun, 07 Feb 2021 16:33:28 +0000

Introduction

We are implementing an IPv4 validation algorithm to check whether a string input is a valid IPv4 address.

What is an IPv4 Address

IPv4 addresses are usually represented in dot-decimal notation, consisting of four decimal numbers, each ranging from 0 to 255, separated by dots(.). Each part represents a group of 8 bits (an octet) of the address.
Example: 172.16.254.1

Implementation

IPv4 has 4 parts that are separated by a dot(.). Each part is a numerical value ranging from 0 to 255.
We don't care about the IP address classes, we only care about a valid IPv4 address

Here is our solution implemented in JavaScript:

function validateIpAddress(ipaddress) {
    if (!ipaddress) return false;

    const ipArray = ipaddress.split('.');
    if (!Array.isArray(ipArray) || ipArray.length !== 4) return false;

    const invalidIPArray = ipArray.filter((octet) => {
        const number = parseInt(octet);
        return isNaN(number) || number < 0 || number > 255;
    });

    if (invalidIPArray.length) return `Wrong octet: ${invalidIPArray}`;

    return true;
}

Breakdown

Step 1: In algorithms, it's always the best approach to check for falsy values to eliminate unwanted errors.

Step 2: Let's split the string argument by dot(.) using split function which returns an array.

Step 3: We check the array length as it must be composed of 4 parts. If the array length different from 4, you guessed it!! It's invalid.

Step 4: Filter the array filter function performing these 2 important tasks:

Parse array value to an integer value
Check and return the value that is not ranging between 0-255, or if the value failed while converting to an integer that returns NaN.

Step 5: Check the returned filtered array length, if the length is not 0, you guessed it!! it's an invalid IP. We return the invalid parts of the given IP address.

filter() method creates an array filled with all array elements that pass a test/condition specified
split() method is used to split a string into an array of substrings and returns the new array.

Conclusion

Feel free to contribute and improve this approach.
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Contributor: Deogratias
Repo

fizz buzz implementation

Leandre — Mon, 01 Feb 2021 11:35:43 +0000

Introduction

We are going to code the famous fizz buzz game. An easy coding interview question that is often used by companies to see if a programmer they want to hire really knows how to code.
It is very famous and we felt like we should consider working on it and trying different approaches to it.
Fizz buzz is a group word game for children to teach them about division. Players take turns to count incrementally, replacing any number divisible by 3 with the word "fizz", any number divisible by 5 with the word "buzz", and any number divisible by 3 and 5.

Example: 1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, Fizz Buzz, 16, 17, Fizz, 19, Buzz, Fizz, 22, 23, Fizz, Buzz, 26, Fizz, 28, 29, Fizz Buzz, 31, 32, Fizz, 34, Buzz, Fizz, ...

Implementation

Let's first try the naive approach:
`
def fizzbuzz_naive_approach():

for i in range(1, 101):

    if i%3 == 0:
        print('fizz')

    if i%5 == 0:
        print('buzz')

    if i%3 and i%5 != 0:
        print(i)

return 0

The problem with this approach is that for multiples of both 3 and 5 it prints fizz on one row and buzz on the row below.

Let's try to improve this:
`
def fizzbuzz_naive_approach2():

for i in range(1, 101):

    if i%3 == 0 and i%5 != 0:
        print('fizz')

    if i%5 == 0 and i%3 != 0:
        print('buzz')

    if i%3 == 0 and i%5 == 0:
        print('fizzbuzz')

    if i%3 != 0 and i%5 != 0:
        print(i)

return 0

The naive approach is improved but still is difficult to read and maintain.

We are going to try and improve the codes implemented above:
`
def fizzbuzz_approach3():

for i in range(1, 101):

    if i%3 == 0 and i%5 == 0:
        print('fizzbuzz')

    elif i%3 == 0:
        print('fizz')

    elif i%5 == 0:
        print('buzz')

    else:
        print(i)

return 0

The implementation above is working, but still has more room for improvement. Wondering how to improve this if it is working?

`
def fizzbuzz_approach4():

for i in range(1 ,101):
    output = ""
    if i%3==0: output += "fizz"
    if i%5==0: output += "buzz"

    if output == "": output = i

    print(output)

return 0

Conclusion

We are now satisfied with the current approach.
Thank you for walking through the different fizz buzz approaches

Bonus

We created a dictionary to help us loop through any multiple we might be interested in, and the good thing about this implementation is that you do not have to change any code in case you add more multiples to the game. You just have to add the dictionary and you are done with any additional multiple.
`
def fizzbuzz_approach5():

multiples_dict = {3:"fizz", 5:"buzz", 7:"bazz"}

for i in range(1 ,101):
    output = ""
    for j in multiples_dict:
        if i%j==0: output += multiples_dict[j]

    if output == "": output = i

    print(output)

return 0

`
Remember there is always room to improve your code
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Link of the code
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DEV Community: Leandre

The IPv4 address validation algorithm

Introduction

What is an IPv4 Address

Implementation

Here is our solution implemented in JavaScript:

Breakdown

Conclusion

fizz buzz implementation

Introduction

Implementation

Conclusion

Bonus

ADeogratias