Challenge - Backend Developer

Write a program that prints all the numbers from 1 to 100. However, for multiples of 3, instead of the number, print "Linio". For multiples of 5 print "IT". For numbers which are multiples of both 3 and 5, print "Linianos".

But here's the catch: you can use only one if. No multiple branches, ternary operators or else.

Requirements

• Maximum of 1 if
• You can't use else, else if or ternary
• Unit tests
• Feel free to apply your SOLID knowledge
• You can write the challenge in any language you want. Here at Linio we are big fans of PHP, Kotlin and TypeScript

Learning

• TypeScript: No idea, but i know Flow from facebook and i now use with babeljs, i learned by using this documentation.

Results

Solution was written in following languages:

1. Go
2. JS/TypeScript (Node project).
3. Php 7 standalone with PhpUnitTest.
4. SQL :) [Just for Fun] without Unit Testing, using procedural SQL in MySQL server, can check generated output.
5. Python.

Solution

I wrote the algorithm at the beginning using common javascript, without tests and this is the eloquent solution with program constraints.

const messages          = ["IT", "Linio", "Linianos"];
let isDivisibleByBoth   = 0;
let isDivisibleByThree  = 0;

for(let i=1; i<=100; i++){
    isDivisibleByBoth   = i%3==0 & i%5==0;
    isDivisibleByThree  = i%3==0 & i%5>0;

    if(i%3==0 || i%5==0){
        console.log(messages[(isDivisibleByBoth*2)+isDivisibleByThree]);
        continue;
    }

    console.log(i);
}

This is not one documentable solution, its like not-readable for humans code:

const messages = ["IT", "Linio", "Linianos"];

for(let i=1; i<=100; i++){
    if(i%3==0 || i%5==0){
        console.log(messages[((i%3==0 & i%5==0)*2)+(i%3==0 & i%5>0)]);
        continue;
    }

    console.log(i);
}

Cases that i considered:

• is divisible by both: when number can be divided between three or five.
• is divisible by three: when number just can be divided by three.

I implements one message structure, to get messages with the following equation x(n)*2 + y(n) = Mi, where n is a number between 0 and 100, thus x and y are Real functions.

• 1: true,
• 0: false
• Mi: Message Index
• x(n): n%3==0 & n%5 == 0
• y(n): n%3==0 & n%5>0

Message Index	Message
0	IT
1	Linio
2	Linianos

I will test with following numbers:

• 15: Divisible by both case.
• 6: Divisible by three case.
• 5: Divisible by five case.

n	x	y	result (Mi=2x+y)	Message
15	15%3==0 & 15%5==0 => 1 & 1 => 1	15%3==0 & 15%5>0 => 1 & 0 => 0	(1*2)+0 => 2	Linianos
6	6%3==0 & 6%5==0 => 1 & 0 => 0	6%3==0 & 6%5>0 => 1 & 1 => 1	(0*2)+1 => 1	Linio
5	5%3==0 & 5%5==0 => 0 & 1 => 0	5%3==0 & 5%5>0 => 0 & 0 => 0	(0*2)+0 => 0	IT

I use 2 on equation cause x represent both cases.

Testing

I wrote following function to get message by one number, used to generate output in unit testing, and make assertion tests.

const messages = ["IT", "Linio", "Linianos"];

function getMessageByNumber(n){
    return (n%3==0 || n%5==0)?messages[((n%3==0&n%5==0)*2)+(n%3==0&n%5>0)]:n;
}