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@jaysonrowe
Created January 11, 2012 01:39
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FizzBuzz JavaScript solution
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);
}
@KiselevAlexander
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const FizzBuzz = (x) => ((x % 3 === 0 ? 'Fizz' : '') + (x % 5 === 0 ? 'Buzz' : '')) || x;
console.log([...new Array(40)].map((i, k) => [k + 1, FizzBuzz(k + 1)]));

@hari2anand
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My solution, doesn't loop for 100 times, instead gets the results in 27 iterations, try the following,

let i = 0
let j = 1
let limit = 100
//prints zero
console.log(i);
while (true) {
    if (i > limit - 3) break; //can also kept in loop; but I always enjoy infinte loops breaking ;-) and have to subtract with 4 if needed till 99
    i = i + 3;

    //prints for number divisible by 5 that falls before the multiples of 3 lags one more step to reach like 10 needs 9->10
    (i - 2) % 5 === 0 ? console.log(i - 2, " - Buzz") : console.log(i - 2);

    //prints for number divisible by 5 that falls before the multiples of 3 needs lags one step to behind like 5 needs 5<-6
    (i % 5 === 1) ? console.log(i - 1, " - Buzz"): console.log(i - 1);

    //prints if the number is divisble by both 3 and 5
    if (i % 5 === 0) {
        console.log(i, " - FizzBuzz");
        continue
    }

    //prints default for 3 multiples
    console.log(i, " - Fizz")

    j++ //Loop Counter

}

console.log('Total iterations ran: ', j) 

I know the following is not the shortest written code, but it implies arithmetic logic to ensure the code sequence completes in shorter iterations for a given limit/interval :)

let i = 0
let j = 1
let limit = 150
//prints zero
console.log(`${i}`);
while (limit < 3) {
    i++;
    if (i > limit) break
    console.log(`${i}`)
}
while (true) {
    if (i > limit - 3) break; //can also kept in loop; but I always enjoy infinite loops breaking ;-) and have to subtract with 4 if needed till 99
    i = i + 6;

    //prints for number divisible by 5 that falls before the multiples of 3 lags one more step to reach like 10 needs 9->10
    (i - 5 <= limit) ? (i - 5) % 5 === 0 ? console.log(`${i - 5} - Buzz`) : console.log(`${i - 5}`) : console.log();

    //prints for number divisible by 5 that falls before the multiples of 3 needs lags one step to behind like 5 needs 5<-6
    (i - 3 <= limit) ? ((i - 3) % 5 === 1) ? console.log(`${i - 4} - Buzz`) : console.log(`${i - 4}`) : console.log();

    (i - 8 <= limit) ? ((i - 8) % 5 === 0) ? console.log(`${i - 3} - FizzBuzz `) : console.log(`${i - 3} - Fizz `) : console.log();

    (i - 2 <= limit) ? (i - 2) % 5 === 0 ? console.log(`${i - 2} - Buzz`) : console.log(`${i - 2}`) : console.log();

    //prints for number divisible by 5 that falls before the multiples of 3 needs lags one step to behind like 5 needs 5<-6
    ((i-1) <= limit) ? (i % 5 === 1) ? console.log(`${i - 1} - Buzz`) : console.log(`${i - 1}`) : console.log();


    //prints if the number is divisble by both 3 and 5
    if ((i <= limit) && (i % 5 === 0)) {
        console.log(`${i} - FizzBuzz`);
        continue
    }

    //prints default for 3 multiples

    (i <= limit) ? console.log(`${i} - Fizz`) : console.log()

    j++ //Loop Counter

}


console.log(
    `Total iterations ran: ${j}`
);

It takes only 15cycles to print fizz/buzz/fizzbuzz for a 100 iteration :)

@YasirHasn9
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short modern js solution

function fizzBuzz(n){
     for(let i = 0 ; i < n ; ){
            console.log( 
                   (++i % 3 ? "" : "fizz") + (i % 5 ? "" : "buzz") || i
                      )
       }
}

@Mahmud-cse
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Mahmud-cse commented Aug 6, 2021

My solution:

function fizzbuzz(num){
    if(num%3==0 && num%5==0){
        console.log('fizzbuzz');
        return;
    }else if(num%3==0){
        console.log('fizz');
    }else if(num%5==0){
        console.log('buzz');
    }
}

for(let i=1;i<=100;i++){
    fizzbuzz(i);
}

@CharlieMaskell
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CharlieMaskell commented Aug 10, 2021

Thought I'd take a crack at it:

var multipliers = [
  [3, "Fizz"],
  [5, "Buzz"]
]

for (let i = 1; i <= 100; i++) {
  let output = "";

  multipliers.forEach(function(item) {
     if (i % item[0] === 0) {
       output += item[1];
     }
  }); 

  console.log(output || i);
}

@GoroJR
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GoroJR commented Aug 15, 2021

/FizzBuzz
Write a program that uses console.log to print all the numbers from 1 to 100,
with two exceptions. For numbers divisible by 3, print "Fizz" instead of the
number, and for numbers divisible by 5 (and not 3), print "Buzz" instead.
When you have that working, modify your program to print "FizzBuzz" for
numbers that are divisible by both 3 and 5 (and still print "Fizz" or "Buzz"
for numbers divisible by only one of those).
(This is actually an interview question that has been claimed to weed out
a significant percentage of programmer candidates. So if you solved it, your
labor market value just went up.)
/
for(let i=1;i<=100;i++, msg="") { if (i%3==0) console.log('Fizz' + i); else if (i%5==0) console.log('Buzz'+ i); if (i%15==0) console.log('FizzBuzz'+ i); else console.log(i)

@LittleDeveloper-CSharp
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LittleDeveloper-CSharp commented Sep 30, 2021

function x(y){ if(y % 3 == 0 && y % 5 == 0) return 'FizzBuzz'; if(y % 3 == 0) return 'Fizz'; if(y % 5 == 0) return 'Buzz'; return y; } for(let i = 1; i <= 100; i++) console.log(x(i));

@AndresCardonaDev
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AndresCardonaDev commented Nov 7, 2021

Hahaha this is funny, i use Fizzbuzz in my interviews, initially to get a quick grasp of basic algorithmic basis, then i elaborate on skills by making a more complex version of the problem creating a counting game generator; anyway here are my short fizzbuzz solutions that i have written so far, seem that the shortests one, is one already accepted as the shortest but... here some guys have achieved two characters shorter https://code.golf/fizz-buzz#javascript 🤯 yet i don't know how 😆 please someone enlighten me

// A short one, an atrocity in terms of readability :D
for(i=0,b='Buzz',f='Fizz';++i<101;)console.log(i%15?i%3?i%5?i:b:f:f+b)
// The shortest i have written, yet i known there are one two characters shorter, no idea how
for(i=0;++i<101;)console.log((i%3?'':'Fizz')+(i%5?'':'Buzz')||i)
// A favorite one because it is made with recursion
f=(i=1)=>i-101?`${(i%3?'':'Fizz')+(i%5?'':'Buzz')||i}\n${f(i+1)}`:'';
console.log(f());
// And well one using array methods to iterate
console.log(Array.from({length:100}, (_,i)=>++i%15 ? i%3 ? i%5 ? i : 'Buzz' : 'Fizz' : 'FizzBuzz').join('\n'));

@Mukunda-Bam
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Mukunda-Bam commented Nov 9, 2021

//for printing in webpage
const n= window.prompt("Enter a number: ");
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);
}

@MaryamFarshbafi
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//Fizz and Buzz
for(i=0; i<=100;i++){
if(i/3==0){
console.log("Fizz");
}
else if( i%5==0 && i/3!==0){
console.log("Buzz");
}

}

@Taimoor0217
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for (let i = 0 ; i < 100 ; i ++){
    let val = "";
    if( i % 3 === 0)
        val += "Fizz"
    if(i%5 ===0)
       val += "Buzz"
    console.log(val)
}

@DaCuteRaccoon
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DaCuteRaccoon commented Mar 31, 2022

Finally! I got it in 64 bytes!

for(i=0;++i<101;)console.log((i%3?'':'Fizz')+(i%5?'':'Buzz')||i)

@Guav0
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Guav0 commented Apr 22, 2022

This is my preferred way, fast and easy to read. Normal loop including the higher bound for the FizzBuzz. The ? and : are ternary operators, in (condition) ? truthy : falsy

for (let i = 1; i <= n; i++) {
        console.log((i % 3 ? '' : 'Fizz') + (i % 5 ? '' : 'Buzz') || i)
    }

@Port5k
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Port5k commented May 1, 2022

`for (let x = 1; x <= 100; x++) {

//Filter out the numbers indivisible by either 3 and 5
if (!(x % 3 == 0) && !(x % 5 == 0)) console.log(x);
//Filter out numbers divisible by 3 and 5
else if ((x % 3 == 0) && (x % 5 == 0)) console.log("FizzBuzz");
//Filter out numbers divisible by 3
else if (x % 3 == 0) console.log("Fizz");
//Filter out numbers divisible by 5
else if (x % 5 == 0) console.log("Buzz");

}`

@cynthia-nwannah
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You guys are so amazing. I just started javascript, hope i would reach your level

@Mike-UU07
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const FizzBuzz = (range) => {
for (let i = 0; i <= range; i++) {
if (i === 0) {
console.log(i)
} else if (i % 15 === 0) {
console.log("Fizz Buzz")
} else if (i % 3 === 0) {
console.log("Buzz")
} else if (i % 5 === 0) {
console.log('Fizz')
} else {
console.log(i)
}
}
}

a quick fizzbuzz function where you can choose how many # you want to console.log
you can invoke the function by FizzBuzz() then inside the brackets how many numbers you want

@spirit-x64
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Short and flexible

for(i=0;++i<101;)console.log([3,5].reduce((t,v,j)=>i%v?t+'':t+['Fizz','Buzz'][j],'')||i) // 88 characters
for(i=0;++i<101;)console.log([3,5,7,11].reduce((t,v,j)=>i%v?t+'':t+['Fizz','Buzz','Fuzz','Bizz'][j],'')||i)

@macder
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macder commented Aug 25, 2022

The idea is so that it's readable and maintainable

Clever one liners only come back to haunt you, or the next poor guy working on the code.

Convolution indicates an inadequacy with naming things, not coding prowess. Any developer can forgo variables for expressions, use anonymous functions, and cram everything together without formatting. The responsible ones realize the cost heavily outweighs some weird belief that a few less lines is somehow a benefit.

A declarative approach will give you an edge, not some one liner that causes cognitive overload, making the next change request a fun time...

const isFactorOf = (divisor) => (dividend) => !(dividend % divisor)
const nonZeroIndex = (mapFn) => (index) => mapFn(index + 1)
const makeArrayOf = (fn) => (length) =>
  [...Array(length).keys()].map((index) => fn(index))

const fizzBuzzOrNum = (num) =>
  isFactorOf(15)(num) && "fizzbuzz" ||
  isFactorOf(3)(num) && "fizz" ||
  isFactorOf(5)(num) && "buzz"||
  num

const fizzBuzz = makeArrayOf(nonZeroIndex(fizzBuzzOrNum))

console.log(fizzBuzz(100))

OR the cruel version of the same code, in case you want your future self and co-workers hating you

// only difference from code above is unformatted without variables for expressions to make it look 1337. 

console.log(
  [...Array(100).keys()].map(i=>(i=>!(i%15)&&"FizzBuzz"||!(i%3)&&"Fizz"||!(i%5)&&"Buzz"||i)(i+1))
)

// Gee, thanks for minifying the code so that the build script has less work...

@ihtisham-Ullah
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function fizzBuzz(n) {
for(let i=1; i<=n; i++)
{

if((i%3==0) && (i%5==0))
{
console.log("FizzBuzz");
}

if((i%3==0) && (i%5!==0))
{
console.log("Fizz");
}

if((i%5==0) && (i%3!==0))
{
console.log("Buzz");
}

if((i%3!==0) && (i%5!==0))
{
console.log(i);
}

}

}

@lpmagdiel
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const fizzBuzz = n => (n%15? (n%3? (n%5? (n%7? '' : 'woff') :'buzz') : 'fizz') : 'fizzbuzz') || n;

@NayanaKalhane
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public class Solution {

public static void main(String[] args) throws IOException {
    int n = 0;
    
    for (int i = 1; i <= n; i++){
        if(i % 3 == 0 && i % 5 == 0)
        System.out.println("FizzBuzz");
    else if (i % 3 == 0 && i % 5 != 0)
        System.out.println("Fizz");
    else if (i % 5 == 0 && i % 3 != 0)
        System.out.println("Buzz");
    else
        System.out.println(i);  
    }
}

}

@paschalyn1
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function fizzBuzz(arr){
let len = arr.length;
for(let i = 0; i < len; i++){
//multiples of 3 and 5
if(arr[i] % 3 === 0 && arr[i] % 5 === 0){
arr[i] = "FizzBuzz";
}
//multiples of 3 but not multiples of 5
if(arr[i] % 3 === 0 && arr[i] % 5 !== 0){
arr[i] = "Fizz";
}
//multiples of 5 but not multiples of 3
if(arr[i] % 3 !== 0 && arr[i] % 5 === 0){
arr[i] = "Buzz";
}
}
return arr;
}

//call the function torender result.
console.log(fizzBuzz([1, 2, 3,4,5,6,7,8,9,10,11,12,13,14,15]))

@ahmadad62
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ahmadad62 commented Oct 6, 2022

Switch Case Solution
`

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;
}
}`

@Metteus96
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for (num = 1; num < 101; num++) {
resultado = "";
if (num % 3 === 0) {
resultado = "Fizz";
}
if (num % 5 === 0) {
resultado = resultado + "Buzz";
}
if (resultado === "") {
resultado = num;
};
console.log(resultado);
};

@Khris234
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Khris234 commented Dec 7, 2022

omo..!!!....HEADache don dey worry me...

@Iwumezie-pamela
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Iwumezie-pamela commented Dec 21, 2022

for (let i=1; i <=100; i++){
   if(i  %  3 === 0  && 1 % 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)
  }
}

@ayodejii
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short modern js solution

function fizzBuzz(n){
     for(let i = 0 ; i < n ; ){
            console.log( 
                   (++i % 3 ? "" : "fizz") + (i % 5 ? "" : "buzz") || i
                      )
       }
}

using short synthax is fun and games until you want to read the code.

@0xadada
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0xadada commented Jan 22, 2023

short modern js solution

function fizzBuzz(n){
     for(let i = 0 ; i < n ; ){
            console.log( 
                   (++i % 3 ? "" : "fizz") + (i % 5 ? "" : "buzz") || i
                      )
       }
}

using short synthax is fun and games until you want to read the code.

We're all here for fun and games, none of this is production code bruh

@nmors
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nmors commented Jan 23, 2023

We're all here for fun and games, none of this is production code bruh

Agreed. If you need to adopt FizzBuzz for production use, you are in the wrong place. FizzBuzz enterprise edition is recommended for these use cases https://github.com/EnterpriseQualityCoding/FizzBuzzEnterpriseEdition

@b0yblake
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let i = 1;
while(i <= 100) {
console.log((${i%3===0?'Fizz':''}${i%5===0?'Buzz':''}) || i)
i++;
}``

@b0yblake
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b0yblake commented May 31, 2023

Using currying is highly technical but for some reason, I think is not more reliable than one-line code (It divides so many fn to solved). I just think we can use HOC or using closures to limit the context.

@essenmitsosse
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essenmitsosse commented May 31, 2023

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)

@Megafry
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Megafry commented Jun 1, 2023

@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().map((_,i) => {
    i++
    console.log((`${i%3===0?'Fizz':''}${i%5===0?'Buzz':''}`) || i)
})

@essenmitsosse
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@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.

@mortocks
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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())

@nehalkadyan
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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.

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

@Wlad2000
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Wlad2000 commented Feb 1, 2024

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);

@GTRshaoran
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//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); }

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