FizzBuzz JavaScript solution

FizzBuzz.js

for (var i=1; i <= 20; i++)
{
    if (i % 15 == 0)
        console.log("FizzBuzz");
    else if (i % 3 == 0)
        console.log("Fizz");
    else if (i % 5 == 0)
        console.log("Buzz");
    else
        console.log(i);
}

If you want fun and games:

new Array(100)
    .fill()
    .map((_,i) => `${i%3===0?'Fizz':''}${i%5===0?'Buzz':''}`)
    .forEach(console.log)

or a little more testable

const fizzbuzz = (_,i) => `${i%3===0?'Fizz':''}${i%5===0?'Buzz':''}`

new Array(10).fill().map(fizzbuzz).forEach(console.log)

@essenmitsosse

The use of forEach is not necessary here since map already traverses the array.
But I like how you used fill

new Array(100).fill().forEach((_,i) => {
    i++
    console.log((`${i%3===0?'Fizz':''}${i%5===0?'Buzz':''}`) || i)
})

@Megafry If you want fizzbuzz to be side-effect free, it still makes sense to separate the two. Also, I usually wouldn't expect map to have a side effect — that's what forEach is for. It is an additional traversal, but in most cases, the gain in clarity should outweigh the performance hit.

Not the smallest / most elegant solution but when that pesky product manager wants to add "boop" on 7 it won't require a refactor

const fizzBuzz = (duration = 100) => Array(duration).fill().map((_, i) => {
  i++
  const subs = new Map([[15, "Fizzbuzz"], [3, "Fizz"], [5, "Buzz"] ])
  const match = Array.from(subs.keys()).find(k => i % k === 0)
  return match ? `${subs.get(match)}` : `${i}`;
})

console.log(fizzBuzz())

for (var i = 1; i <= 100; i++) {
switch (true) {
case i % 5 == 0 && i % 3 == 0:
console.log("FizzBuzz");
break;
case i % 5 == 0 && i % 3 != 0:
console.log("Buzz");
break;
case i % 3 == 0 && i % 5 != 0:
console.log("Fizz");
break;
default:
console.log(i);
break;
}
}

Thanks man, was looking for switch case implementation on this one.

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

const fizzBuzz = (num, rules) => {
  const result = [];
  for (let i = 1; i <= num; i++) {
    result.push(applyRules(i, rules));
  }
  return result;
};

const applyRules = (num, rules) => {
  const output = rules.reduce((result, rule) => {
    if (rule.condition(num)) {
      result += rule.output;
    }
    return result;
  }, '');

  return output || num;
};
 // Example usage:
  const isMultipleOf = (n) => (num) => num % n === 0;
  const rules = [
    { condition: isMultipleOf(3), output: 'Fizz' },
    { condition: isMultipleOf(5), output: 'Buzz' },
  ];
    const resultArray = fizzBuzz(15, rules);
    console.log(resultArray);

//para imprimir en la página web const n= window.prompt("Ingrese un número: "); for(i=1; i<=n; i++) if(i%3==0 && i%5==0){ console.log("FizzBuzz\n"); } else if(i%3==0 && i%5!=0){ console.log("Fizz\n"); } else if(i%3!=0 && i%5==0){ console.log("Buzz\n"); } else{ console.log(i\n); }

Which way would be "better regarded" as a best answer, by using for loops or by using .forEach() and .map()? Or it really does not matter as long as the program executes what is expected of it? Cheers

@iuliaGrig The answer is: It depends.

In this example, it's so simple that it doesn't matter for most intents and purposes. For more complex examples a rule of thumb could be:

• .forEach and .map are generally more readable (even though plenty of people already might disagree). .forEach should be used if there are side-effects to the function (like console.log), .map must be used if you want to create a new list (.forEach doesn't return anything)
• for-loops are generally more performant, even though the difference only really matters for very large lists

A v8 optimized version (I'd be quite surprised if it can be made faster by keeping the same signature):

function fizzBuzz(limit = 100) {
  var result = new Array(limit),
    f = "Fizz",
    b = "Buzz",
    fb = "FizzBuzz",
    i = 0;

  for (; i + 15 <= limit; i += 15) {
    result[i] = i + 1;
    result[i + 1] = i + 2;
    result[i + 2] = f;
    result[i + 3] = i + 4;
    result[i + 4] = b;
    result[i + 5] = f;
    result[i + 6] = i + 7;
    result[i + 7] = i + 8;
    result[i + 8] = f;
    result[i + 9] = b;
    result[i + 10] = i + 11;
    result[i + 11] = f;
    result[i + 12] = i + 13;
    result[i + 13] = i + 14;
    result[i + 14] = fb;
  }

  var pattern = [0, 0, f, 0, b, f, 0, 0, f, b, 0, f, 0, 0, fb];
  for (; i < limit; i++) result[i] = pattern[i % 15] || i + 1;

  return result;
};