FizzBuzz

Alright, let's start with perhaps the most famous technical interview challenge ever.

Legend has it that this challenge was created as a way for employers to see if recent Computer Science graduates were comfortable with programming fundamentals. Evidently, CS programs were teaching students a lot of theory, but not helping them build problem-solving skills.

Here's how the challenge goes:

/*
  Write a function that takes a number as a parameter.

  • If that number is a multiple of 3, it should return "Fizz".
  • If that number is a multiple of 5, it should return "Buzz".
  • If that number is a multiple of both 3 AND 5, it should return "FizzBuzz".
  • Otherwise, it should return `undefined`.
*/

fizzbuzz(2);    // undefined
fizzbuzz(3);    // "Fizz"
fizzbuzz(5);    // "Buzz"
fizzbuzz(15);   // "FizzBuzz"
fizzbuzz(40);   // "Buzz"
fizzbuzz(45);   // "FizzBuzz"

Difficulty

This is by far the least challenging problem you'll tackle in this section. I wanted to start off with something relatively straightforward.

That said, it also relies on JS problem-solving skills, which isn't something we've tackled directly in the course.

Playground

To tackle this challenge, you'll provide the definition for the fizzbuzz function below.

You can use the output in the Result pane to see if your solution works or not.

Code Playground

Requirements:

• If `n` is a multiple of 3, it should return "Fizz".

• If `n` is a multiple of 5, it should return "Buzz".

• If `n` is a multiple of both 3 AND 5, it should return "FizzBuzz".

• Otherwise, it should return `undefined`.

function fizzbuzz(n) {

// TODO

}

export default fizzbuzz;

My attempt

Attempt code (success)

Here's what I wound up with in the video:

function fizzbuzz(n) {
  if (n % 3 === 0 && n % 5 === 0) {
    return 'FizzBuzz';
  } else if (n % 3 === 0) {
    return 'Fizz';
  } else if (n % 5 === 0) {
    return 'Buzz';
  }
}

Thinking about it, I think we could be even clearer. I sorta like the idea of doing something like this:

function fizzbuzz(n) {
  const isDivisibleByThree = n % 3 === 0;
  const isDivisibleByFive = n % 5 === 0;

  if (isDivisibleByThree && isDivisibleByFive) {
    return 'FizzBuzz';
  } else if (isDivisibleByThree) {
    return 'Fizz';
  } else if (isDivisibleByFive) {
    return 'Buzz';
  }
}

As I mentioned in the video, there are definitely things we could do which would be more clever. For example, we could reduce duplication by building up a string:

function fizzbuzz(n) {
  let output = '';

  if (n % 3 === 0) {
    output += 'Fizz';
  }

  if (n % 5 === 0) {
    output += 'Buzz';
  }

  return output || undefined;
}

Personally, though, I feel like this is a step in the wrong direction. I have to burn more calories to translate this code into the original requirements.