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Change integer value with string in array on some conditions

This is the one of many code challenges from one company that they gave me to solve.
Write a program that outputs sequentially the integers from 1 to 99, but on some conditions:
- when the integer is a multiple of 3 print “Open” instead of the number
- when it is a multiple of 7 print “Source” instead of the number
- when it is a multiple of both 3 and 7 print “OpenSource” instead of the number
I'v wrote this chunk of code:
let array = []
for (let i = 1; i < 100; i++) {
array.push(i);
if (i % 3 === 0) {
array[i] = "open";
}
if (i % 7 === 0) {
array[i] = "source";
}
if (i % 3 === 0 && i % 7 === 0) {
array[i] = "opensource";
}
}
console.log(array);
As you can see from the output, something is messy at index 2-3. All other integers are replaced correctly.
What is wrong with the code?

Array's first index is 0 not 1. So either insert in arr[i - 1] or use the following approach to fix your issue.
Reasoning of my solution - The loop will start from 1 in my case as we want to check from 0 to 100. And after that, I check the logic if the current value(value) is multiple of 3 or 7 or both. And after that, just push the value into the array and it will automatically add the values in respective indexes.
In your scenario - we need to do arr[index - 1] to compensate the difference between current value and index.
let array = []
for (let i = 1; i < 100; i++) {
let value = i;
if (i % 3 === 0) {
value = "open";
}
if (i % 7 === 0) {
value = "source";
}
if (i % 3 === 0 && i % 7 === 0) {
value = "opensource";
}
array.push(value);
}
console.log(array);

If you debug the code, you'll notice, that the first three values are undefined:
let array = []
for (let i = 1; i < 100; i++) {
array.push(i);
if (i < 10) {
console.log(i, i % 3, i % 7, array[i]);
}
if (i % 3 === 0) {
array[i] = "open";
}
if (i % 7 === 0) {
array[i] = "source";
}
if (i % 3 === 0 && i % 7 === 0) {
array[i] = "opensource";
}
}
console.log(array);
So you'll need to fix your indexing logic.

While loop output not correct

I'm busy with a udacity excercise and the following question:
A while loop that:
Loop through the numbers 1 to 20
If the number is divisible by 3, print "Julia"
If the number is divisible by 5, print "James"
If the number is divisible by 3 and 5, print "JuliaJames"
If the number is not divisible by 3 or 5, print the number
I keep submitting the answer but it tells me that my while loop condition is incorrect, Is there anything im doing wrong?
var x = 1;
while (x <= 20) {
if (x/3 === 0) {
console.log("julia" );
} // check divisibility
else if (x/5 === 0) {
console.log("james");
}
else if (x/5 === 0 && x/3 === 0 ) {
console.log("juiliajames");
} // print Julia, James, or JuliaJames
else {
console.log(x);
}
x= x + 1;// increment x
}

You need to use Modulus (%) instead of divide (/). And make x % 5 === 0 && x % 3 === 0 as your first condition.
Change your code like following.
var x = 1;
while (x <= 20) {
if (x % 5 === 0 && x % 3 === 0) {
console.log("juiliajames");
} // print Julia, James, or JuliaJames
else if (x % 3 === 0) {
console.log("julia");
} // check divisibility
else if (x % 5 === 0) {
console.log("james");
} else {
console.log(x);
}
x = x + 1; // increment x
}

var x = 1;
while (x <= 20) {
if(x%3 === 0 && x%5 === 0 ){
console.log("juiliajames" );
} // check divisibility
else if(x%3 === 0){
console.log("juilia");
}
else if (x%5 === 0){
console.log("james");
} // print Julia, James, or JuliaJames
else{
console.log(x);
}
x= x + 1;// increment x
}
If you want to check divisibility, you should use the % operator instead of / operator.

Check x divisible by 5 AND 3 at the begining or it will never been done because if it is divisible by 3 your loop won't go to the else if statment.
To check divisibility use modulo (x/3 === 0 only for x = 0)
var x = 1;
while (x <= 20) {
if(x%5 === 0 && x%3 === 0){
console.log("juiliajames" );
}
else if(x%5 === 0){
console.log("james");
}
else if (x%3 === 0 ){
console.log("juilia");
}
else{
console.log(x);
}
x= x + 1;// increment x
}

My approach works by setting bitwise flags on an integer. If it's divisible by three (value % 3 === 0, where % is 'modulo' which gives an integer remainder from division) The first bit is set and if it's divisible by five the second bit is set. That gives a result that could have three binary values 01, 10 or 11, or in decimal 1, 2 & 3 (The three comes about when both bits are set).
var DIVISABLE_BY_THREE = 1;
var DIVISABLE_BY_FIVE = 2;
var DIVISABLE_BY_THREE_AND_FIVE = 3;
var value = 0;
while(value++ < 20) {
var modulo_3 = (value % 3 === 0) | 0;
var modulo_5 = ((value % 5 === 0) | 0) << 1;
switch(modulo_3 | modulo_5) {
case DIVISABLE_BY_THREE:
console.log("Julia");
break;
case DIVISABLE_BY_FIVE:
console.log("James");
break;
case DIVISABLE_BY_THREE_AND_FIVE:
console.log("JuliaJames");
break;
default:
console.log(value);
break;
}
}

var x = 1;
while (x <= 20) {
var name ="";
if (x % 3 == 0) {
name = name + "julia";
}
if (x % 5 == 0) {
name = name + "james";
}
if(name.length > 0)
console.log(name);
else
console.log(x);
x++;
}

Combination of given number that add up to the target number using DP?

I don't know if I describe the problem correctly. The problem is: there's the given number: 2, 3, 7, and let's say the target number is 10. Use these given numbers only, find the combination that add up to 10. For example: 2 + 2 + 2 + 2 + 2 = 10 or 3 + 7 = 10 or 7 + 3 = 10, etc.
Given that, I came up with the recursive solution (Using Javascript):
function combination(n) {
if(n < 2) {
return 0
}
if(n === 2) {
return 1
}
if(n === 3) {
return 1
}
if(n === 7) {
return 4
}
return combination(n - 2) + combination(n - 3) + combination(n - 7)
}
However, the time of this solution is exponentially growing. So I was trying to solve this by using DP (Dynamic Programming).But I can't think a way of it.
Any ideas ? I would be grateful if someone enlighten me.

You could use a recursive approach by checking the sum and the index. If the values are valid, then proceed.
function combine(array, sum) {
function iter(i, temp) {
var total = temp.reduce((a, b) => a + b, 0);
if (total === sum) {
result.push(temp);
}
if (total >= sum || i >= array.length) {
return;
}
iter(i, temp.concat(array[i]));
iter(i + 1, temp);
}
var result = [];
iter(0, []);
return result;
}
console.log(combine([2, 3, 7], 10));
.as-console-wrapper { max-height: 100% !important; top: 0; }

Thank you guys, I've already figure this out by using iteration, bottom-up.
function combination(n) {
var table = {
'2': 1,
'3': 1,
'7': 4
}
if(n < 2) return 0
if(table[n]) return table[n]
for(let i = 4; i <= n; i++) {
table[i] = (table[i - 2] ? table[i - 2] : 0) + (table[i - 3] ? table[i - 3] : 0) + (table[i - 7] ? table[i - 7] : 0)
}
return table[n]
}

similar to FizzBuzz with a twist

Write a javascript program that displays the numbers from 10 to 100. But for multiples of 4 print "Penny" instead of the number and for multiples of 6 print "Leonard". For numbers which are multiples of both 4 and 6 print "Bazzinga"
I know how to do two parts struggling to print 6 and 4;
function baZzinga (number) {
for (var number = 10; number <= 101; number++)
if(number % 4 == 0) {
console.log("penny");
}
else if (number % 6 == 0) {
console.log("Leonard");
} else if ( not sure what goes here) {
help help help
} else {
console.log(number");
}

You want the and condition first. Try this
var result = document.getElementById("result");
function baZzinga (number) {
for (var number = 10; number <= 101; number++) {
if (number % 4 == 0 && number % 6 == 0) {
result.innerHTML += "Bazinga";
}
else if(number % 4 == 0) {
result.innerHTML += "penny";
}
else if (number % 6 == 0) {
result.innerHTML += "Leonard";
}
else {
result.innerHTML += number;
}
}
}
baZzinga()
<p id="result"></p>
I changed console.log to result.innerHTML because I wanted to demonstrate it in a snippet.

I have a few comments on your code -- constructive criticism, I hope! First, you don't need the number parameter in your bazzinga function. Next, the indentation of the code you posted makes it hard to read. Finally, you should almost always use === instead of ==. The === tests for strict equality, whereas == tries to do some type conversions first (and can therefore produce unexpected results). See the official docs.
To answer you question: check for divisibility by 6 AND 8 first. That way, it will override the individual cases. I believe you want something like this:
function bazzinga() {
for (var number = 10; number <= 100; number++) {
if (number % 4 === 0 && number % 6 === 0) {
console.log("Bazzinga");
} else if (number % 4 === 0) {
console.log("Penny");
} else if (number % 6 === 0) {
console.log("Leonard");
}
}
}

Here is a solution using the format you posted:
for (var number = 10; number <= 100; number++) {
if(number % 4 === 0 && number % 6 === 0){
console.log("bazzinga");
} else if(number % 4 === 0) {
console.log("penny");
} else if (number % 6 === 0) {
console.log("Leonard");
} else {
console.log(number);
}
}
Or use the ternary operator to be even more succinct!
for (var i = 10; i <= 100; i++){
var penny = i % 4 === 0;
var leonard = i % 6 === 0;
console.log(penny ? (leonard ? "bazzinga" : "penny"): leonard ? "leonard" : i);
}

function process_num(num) {
return num % 4 == 0 ? num % 6 == 0 ? "Bazzinga" : "Penny" : num % 6 == 0 ? "Leonard" : num;
}
for (x = 10; x <= 100; x++) { console.log( x + ': is ', process_num(x)) }
Nested Ternary operator for conciseness
If it passes outer ternary test it is divisible by 4:
Enter into nested termary one to test if num is also divisible by 6 for the BaZzinga prize!!
If it fails the BaZzinga challenge, we know it previously passed the divisible by 4 test so print "penny"
Failing the outer ternary condition, we know it's not divisible by 4:
Enter nested ternary two to consider if divisible by 6. If so print "Leonard".
If not it's failed both the outer (div by 4) and inner (div by 6) so return the number unchanged.
Now that the logic is contained in the function, we can just create a for loop to iterate over the required numbers printing out the correct values.

FizzBuzz program (details given) in Javascript [closed]

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Can someone please correct this code of mine for FizzBuzz? There seems to be a small mistake. This code below prints all the numbers instead of printing only numbers that are not divisible by 3 or 5.
Write a program that prints the numbers from 1 to 100. But for multiples of three, print "Fizz" instead of the number, and for the multiples of five, print "Buzz". For numbers which are multiples of both three and five, print "FizzBuzz".
function isDivisible(numa, num) {
if (numa % num == 0) {
return true;
} else {
return false;
}
};
function by3(num) {
if (isDivisible(num, 3)) {
console.log("Fizz");
} else {
return false;
}
};
function by5(num) {
if (isDivisible(num, 5)) {
console.log("Buzz");
} else {
return false;
}
};
for (var a=1; a<=100; a++) {
if (by3(a)) {
by3(a);
if (by5(a)) {
by5(a);
console.log("\n");
} else {
console.log("\n");
}
} else if (by5(a)) {
by5(a);
console.log("\n");
} else {
console.log(a+"\n")
}
}

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

/*Write a program that prints the numbers from 1 to 100. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz”*/
var str="",x,y,a;
for (a=1;a<=100;a++)
{
x = a%3 ==0;
y = a%5 ==0;
if(x)
{
str+="fizz"
}
if (y)
{
str+="buzz"
}
if (!(x||y))
{
str+=a;
}
str+="\n"
}
console.log(str);
Your functions return falsy values no matter what, but will print anyway. No need to make this overly complicated.
fiddle: http://jsfiddle.net/ben336/7c9KN/

Was fooling around with FizzBuzz and JavaScript as comparison to C#.
Here's my version, heavily influenced by more rigid languages:
function FizzBuzz(aTarget) {
for (var i = 1; i <= aTarget; i++) {
var result = "";
if (i%3 === 0) result += "Fizz";
if (i%5 === 0) result += "Buzz";
if (result.length ===0) result = i;
console.log(result);
}
}
I like the structure and ease of read.
Now, what Trevor Dixon cleverly did is relay on the false-y values of the language (false , null , undefined , '' (the empty string) , 0 and NaN (Not a Number)) to shorten the code.
Now, the if (result.length ===0) result = i; line is redundant and the code will look like:
function FizzBuzz(aTarget) {
for (var i = 1; i <= aTarget; i++) {
var result = "";
if (i%3 === 0) result += "Fizz";
if (i%5 === 0) result += "Buzz";
console.log(result || i);
}
}
Here we relay on the || operator to say : "if result is false, print the iteration value (i)". Cool trick, and I guess I need to play more with JavaScript in order to assimilate this logic.
You can see other examples (from GitHub) that will range from things like :
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);
}
No variables here, and just check for division by 15,3 & 5 (my above one only divides by 3 & 5, but has an extra variable, so I guess it's down to microbenchmarking for those who care, or style preferences).
To:
for(i=0;i<100;)console.log((++i%3?'':'Fizz')+(i%5?'':'Buzz')||i)
Which does it all in on line, relaying on the fact that 0 is a false value, so you can use that for the if-else shorthanded version (? :), in addition to the || trick we've seen before.
Here's a more readable version of the above, with some variables:
for (var i = 1; i <= 100; i++) {
var f = i % 3 == 0, b = i % 5 == 0;
console.log(f ? b ? "FizzBuzz" : "Fizz" : b ? "Buzz" : i);
}
All in all, you can do it in different ways, and I hope you picked up some nifty tips for use in JavaScript :)

.fizz and .buzz could be CSS classes, no? In which case:
var n = 0;
var b = document.querySelector("output");
window.setInterval(function () {
n++;
b.classList[n%3 ? "remove" : "add"]("fizz");
b.classList[n%5 ? "remove" : "add"]("buzz");
b.textContent = n;
}, 500);
output.fizz:after {
content: " fizz";
color:red;
}
output.buzz:after {
content: " buzz";
color:blue;
}
output.fizz.buzz:after {
content: " fizzbuzz";
color:magenta;
}
<output>0</output>

With ternary operator it is much simple:
for (var i = 0; i <= 100; i++) {
str = (i % 5 == 0 && i % 3 == 0) ? "FizzBuzz" : (i % 3 == 0 ? "Fizz" : (i % 5 == 0) ? "Buzz" : i);
console.log(str);
}

for(i = 1; i < 101; i++) {
if(i % 3 === 0) {
if(i % 5 === 0) {
console.log("FizzBuzz");
}
else {
console.log("Fizz");
}
}
else if(i % 5 === 0) {
console.log("Buzz");
}
else {
console.log(i)
}
}

In your by3 and by5 functions, you implicitly return undefined if it is applicable and false if it's not applicable, but your if statement is testing as if it returned true or false. Return true explicitly if it is applicable so your if statement picks it up.

As an ES6 generator: http://www.es6fiddle.net/i9lhnt2v/
function* FizzBuzz() {
let index = 0;
while (true) {
let value = ''; index++;
if (index % 3 === 0) value += 'Fizz';
if (index % 5 === 0) value += 'Buzz';
yield value || index;
}
}
let fb = FizzBuzz();
for (let index = 0; index < 100; index++) {
console.log(fb.next().value);
}

Codeacademy sprang a FizzBuzz on me tonight. I had a vague memory that it was "a thing" so I did this. Not the best way, perhaps, but different from the above:
var data = {
Fizz:3,
Buzz:5
};
for (var i=1;i<=100;i++) {
var value = '';
for (var k in data) {
value += i%data[k]?'':k;
}
console.log(value?value:i);
}
It relies on data rather than code. I think that if there is an advantage to this approach, it is that you can go FizzBuzzBing 3 5 7 or further without adding additional logic, provided that you assign the object elements in the order your rules specify. For example:
var data = {
Fizz:3,
Buzz:5,
Bing:7,
Boom:11,
Zing:13
};
for (var i=1;i<=1000;i++) {
var value = '';
for (var k in data) {
value += i%data[k]?'':k;
}
console.log(value?value:i);
}

This is what I wrote:
for (var num = 1; num<101; num = num + 1) {
if (num % 5 == 0 && num % 3 == 0) {
console.log("FizzBuzz");
}
else if (num % 5 == 0) {
console.log("Buzz");
}
else if (num % 3 == 0) {
console.log("Fizz");
}
else {
console.log(num);
}
}

for (var 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);
}
One of the easiest way to FizzBuzz.
Multiple of 3 and 5, at the same time, means multiple of 15.
Second version:
for (var i = 1; i <= 100; 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);
}

In case someone is looking for other solutions: This one is a pure, recursive, and reusable function with optionally customizable parameter values:
const fizzBuzz = (from = 1, till = 100, ruleMap = {
3: "Fizz",
5: "Buzz",
}) => from > till || console.log(
Object.keys(ruleMap)
.filter(number => from % number === 0)
.map(number => ruleMap[number]).join("") || from
) || fizzBuzz(from + 1, till, ruleMap);
// Usage:
fizzBuzz(/*Default values*/);
The from > till is the anchor to break the recursion. Since it returns false until from is higher than till, it goes to the next statement (console.log):
Object.keys returns an array of object properties in the given ruleMap which are 3 and 5 by default in our case.
Then, it iterates through the numbers and returns only those which are divisible by the from (0 as rest).
Then, it iterates through the filtered numbers and outputs the saying according to the rule.
If, however, the filter method returned an empty array ([], no results found), it outputs just the current from value because the join method at the end finally returns just an empty string ("") which is a falsy value.
Since console.log always returns undefined, it goes to the next statement and calls itself again incrementing the from value by 1.

A Functional version of FizzBuzz
const dot = (a,b) => x => a(b(x));
const id = x => x;
function fizzbuzz(n){
const f = (N, m) => n % N ? id : x => _ => m + x('');
return dot(f(3, 'fizz'), f(5, 'buzz')) (id) (n);
}
for more options in the above replace dot with dots as below
const dots = (...a) => f0 => a.reduceRight((acc, f) => f(acc), f0);
function fizzbuzz(n){
const f = (N, m) => n % N ? id : x => _ => m + x('');
return dots(f(3, 'fizz'), f(5, 'buzz'), f(7, 'bam')) (id) (n);
}
Reference: FizzBuzz in Haskell by Embedding a Domain-Specific Language
by Maciej Piro ́g

for (i=1; i<=100; i++) {
output = "";
if (i%5==0) output = "buzz";
if (i%3==0) output = "fizz" + output;
if (output=="") output = i;
console.log(output);
}

Functional style! JSBin Demo
// create a iterable array with a length of 100
// and map every value to a random number from 1 to a 100
var series = Array.apply(null, Array(100)).map(function() {
return Math.round(Math.random() * 100) + 1;
});
// define the fizzbuzz function which takes an interger as input
// it evaluates the case expressions similar to Haskell's guards
var fizzbuzz = function (item) {
switch (true) {
case item % 15 === 0:
console.log('fizzbuzz');
break;
case item % 3 === 0:
console.log('fizz');
break;
case item % 5 === 0:
console.log('buzz');
break;
default:
console.log(item);
break;
}
};
// map the series values to the fizzbuzz function
series.map(fizzbuzz);

Another solution, avoiding excess divisions and eliminating excess spaces between "Fizz" and "Buzz":
var num = 1;
var FIZZ = 3; // why not make this easily modded?
var BUZZ = 5; // ditto
var UPTO = 100; // ditto
// and easily extended to other effervescent sounds
while (num < UPTO)
{
var flag = false;
if (num % FIZZ == 0) { document.write ("Fizz"); flag = true; }
if (num % BUZZ == 0) { document.write ("Buzz"); flag = true; }
if (flag == false) { document.write (num); }
document.write ("<br>");
num += 1;
}
If you're using using jscript/jsc/.net, use Console.Write(). If you're using using Node.js, use process.stdout.write(). Unfortunately, console.log() appends newlines and ignores backspaces, so it's unusable for this purpose. You could also probably append to a string and print it. (I'm a complete n00b, but I think (ok, hope) I've been reasonably thorough.)
"Whaddya think, sirs?"

check this out!
function fizzBuzz(){
for(var i=1; i<=100; i++){
if(i % 3 ===0 && i % 5===0){
console.log(i+' fizzBuzz');
} else if(i % 3 ===0){
console.log(i+' fizz');
} else if(i % 5 ===0){
console.log(i+' buzz');
} else {
console.log(i);
}
}
}fizzBuzz();

Slightly different implementation.
You can put your own argument into the function. Can be non-sequential numbers like [0, 3, 10, 1, 4]. The default set is only from 1-15.
function fizzbuzz (set) {
var set = set ? set : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
var isValidSet = set.map((element) => {if (typeof element !== 'number') {return false} else return true}).indexOf(false) === -1 ? true : false
var gotFizz = (n) => {if (n % 3 === 0) {return true} else return false}
var gotBuzz = (n) => {if (n % 5 === 0) {return true} else return false}
if (!Array.isArray(set)) return new Error('First argument must an array with "Number" elements')
if (!isValidSet) return new Error('The elements of the first argument must all be "Numbers"')
set.forEach((n) => {
if (gotFizz(n) && gotBuzz(n)) return console.log('fizzbuzz')
if (gotFizz(n)) return console.log('fizz')
if (gotBuzz(n)) return console.log('buzz')
else return console.log(n)
})
}

var num = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
var runLoop = function() {
for (var i = 1; i<=num.length; i++) {
if (i % 5 === 0 && i % 3 === 0) {
console.log("FizzBuzz");
}
else if (i % 5 === 0) {
console.log("Buzz");
}
else if (i % 3 === 0) {
console.log("Fizz");
}
else {
console.log(i);
}
}
};
runLoop();

Just want to share my way to solve this
for (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);
}
}

var limit = prompt("Enter the number limit");
var n = parseInt(limit);
var series = 0;
for(i=1;i<n;i++){
series = series+" " +check();
}
function check() {
var result;
if (i%3==0 && i%5==0) { // check whether the number is divisible by both 3 and 5
result = "fizzbuzz "; // if so, return fizzbuzz
return result;
}
else if (i%3==0) { // check whether the number is divisible by 3
result = "fizz "; // if so, return fizz
return result;
}
else if (i%5==0) { // check whether the number is divisible by 5
result = "buzz "; // if so, return buzz
return result;
}
else return i; // if all the above conditions fail, then return the number as it is
}
alert(series);

Thats How i did it :
Not the best code but that did the trick
var numbers = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
for(var i = 0 ; i <= 19 ; i++){
var fizz = numbers[i] % 3 === 0;
var buzz = numbers[i] % 5 === 0;
var fizzBuzz = numbers[i] % 5 === 0 && numbers[i] % 3 === 0;
if(fizzBuzz){
console.log("FizzBuzz");
} else if(fizz){
console.log("Fizz");
} else if(buzz){
console.log("Buzz");
} else {
console.log(numbers[i]);
}
}

As much as this is easy logic it can be a daunting task for beginners. Below is my solution to the FizzBuzz problem:
let i = 1;
while(i<=100){
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);
}
i++;
}

considering performance and readability, please find my take on this problem
way 1: instead of doing a math modules operation in an if loop, which results in performing 3 times taking it a step above reduces the overhead
function fizzBuzz(n) {
let count =0;
let x = 0;
let y = 0;
while(n!==count)
{
count++;
x = count%3;
y = count%5;
if(x === 0 && y ===0)
{
console.log("fizzbuzz");
}
else if(x === 0)
{
console.log("fizz");
}
else if(y === 0)
{
console.log("buzz");
}
else
{
console.log(count);
}
}
}
fizzBuzz(15);
way 2: condensing the solution
function fizzBuzz(n) {
let x = 0;
let y = 0;
for (var i = 1; i <= n; i++) {
var result = "";
x = i%3;
y = i%5;
if (x === 0 && y === 0) result += "fizzbuzz";
else if (x === 0) result += "fizz";
else if (y === 0) result += "buzz";
console.log(result || i);
}
}
fizzBuzz(5)

Here's my favorite solution. Succinct, functional & fast.
const oneToOneHundred = Array.from({ length: 100 }, (_, i) => i + 1);
const fizzBuzz = (n) => {
if (n % 15 === 0) return 'FizzBuzz';
if (n % 3 === 0) return 'Fizz';
if (n % 5 === 0) return 'Buzz';
return n;
};
console.log(oneToOneHundred.map((i) => fizzBuzz(i)).join('\n'));

function fizzBuzz(n) {
for (let i = 1; i < n + 1; 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);
}
}
}
fizzBuzz(15);

Different functional style -- naive
fbRule = function(x,y,f,b,z){return function(z){return (z % (x*y) == 0 ? f+b: (z % x == 0 ? f : (z % y == 0 ? b: z))) }}
range = function(n){return Array.apply(null, Array(n)).map(function (_, i) {return i+1;});}
range(100).map(fbRule(3,5, "fizz", "buzz"))
or, to incorporate structures as in above example: ie [[3, "fizz"],[5, "buzz"], ...]
fbRule = function(fbArr,z){
return function(z){
var ed = fbArr.reduce(function(sum, unit){return z%unit[0] === 0 ? sum.concat(unit[1]) : sum }, [] )
return ed.length>0 ? ed.join("") : z
}
}
range = function(n){return Array.apply(null, Array(n)).map(function (_, i) {return i+1;});}
range(100).map(fbRule([[3, "fizz"],[5, "buzz"]]))
OR, use ramda [from https://codereview.stackexchange.com/questions/108449/fizzbuzz-in-javascript-using-ramda ]
var divisibleBy = R.curry(R.compose(R.equals(0), R.flip(R.modulo)))
var fizzbuzz = R.map(R.cond([
[R.both(divisibleBy(3), divisibleBy(5)), R.always('FizzBuzz')],
[divisibleBy(3), R.aklways('Fizz')],
[divisibleBy(5), R.always('Buzz')],
[R.T, R.identity]
]));
console.log(fizzbuzz(R.range(1,101)))