The shortest code for FizzBuzz

When I go out for drinks with my fellow engineers, we tend to have conversations about technical matters.

Last time we had a good talk about the shortest code for FizzBuzz. FizzBuzz is a common programming challenge often used in job interviews to assess a candidate’s basic programming skills.

“building the short code” is often associated with a concept called “code golf”. Code golf is a type of programming competition where participants strive to write the shortest possible code that solves a given problem.

The requirement of FizzBuzz

The requirements for FizzBuzz are as follows:

1. The program should output numbers from 1 to 100.
2. However, for multiples of 3, it should output “Fizz” instead.
3. For multiples of 5, it should output “Buzz” instead.
4. For numbers that are multiples of both 3 and 5, it should output “FizzBuzz.”

Basic answer

The most common answer is below.

for (let i = 1; i <= 100; i++) {
  if (i % 3 === 0 && i % 5 === 0) {
    console.log("FizzBuzz");
  } else if (i % 3 === 0) {
    console.log("Fizz");
  } else if (i % 5 === 0) {
    console.log("Buzz");
  } else {
    console.log(i);
  }
}

The following code may be better for readability.

for (let i = 1; i <= 100; i++) {
  let msg = "";
  if (i % 3 === 0) {
    msg = "Fizz";
  }
  if (i % 5 === 0) {
    msg += "Buzz";
  }
  console.log(msg || i);
}

The shortest code (NOT RECOMMENDED)

Now let’s consider the shortest code. Ignoring readability, my priority is to create a minimal code, so it is not practical and I do not recommend its use for production. I have considered it only as an exercise for the brain.

Here is my idea.

[...Array(100)].forEach((_, i) =>
  console.log(
    ((i + 1) % 3 ? "" : "Fizz") + ((i + 1) % 5 ? "" : "Buzz") || i + 1
  )
);

The conditional branching part of the if statement has been replaced with a ternary operator. Also, I replaced for looping with [...Array(100)].forEach.

Although less readable, This is the shortest code I could come up with.