FizzBuzz

problemsolving
Published

December 15, 2024

What is FizzBuzz?

The FizzBuzz challenge is straightforward: write a program that prints the numbers from 1 to 100. However, for multiples of three, print “Fizz” instead of the number. For multiples of five, print “Buzz.” For numbers that are multiples of both three and five, print “FizzBuzz.”

The Classic Approach (using if, else if, and else)

This is the most common and easily understood solution. We use nested if statements to check the divisibility of the number by 3 and 5.

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);
  }
}

This code iterates through numbers 1 to 100. The modulo operator (%) gives us the remainder after division. If the remainder is 0, the number is divisible.

A More Concise Approach (using a ternary operator)

We can make the code more compact using JavaScript’s ternary operator. While potentially less readable for beginners, it demonstrates a more concise coding style.

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

This version builds the output string incrementally. If output remains empty after checking for “Fizz” and “Buzz”, it defaults to printing the number itself.

Beyond the Basics: Improving Readability and Efficiency

While the above solutions work perfectly, consider these points for larger-scale applications:

  • Function Decomposition: For better organization and reusability, encapsulate the FizzBuzz logic within a function.
function fizzBuzz(n) {
  for (let i = 1; i <= n; i++) {
    // ... (same FizzBuzz logic as before) ...
  }
}

fizzBuzz(100);
  • Error Handling: Add checks to handle invalid input (e.g., non-numeric input).