Logs 'Fizz' for multiples of three, 'Buzz', for multiples of 5, and 'Fizz Buzz' for multiples of 3 & 5

A short review;

• You don't need the array, you could just use i
• Part of the challenge is to realize that you only have to check for 3 and 5
• i is a good short variable, I would go for num -> n or even count
• I only use fat arrow syntax => for inline functions
• Function names ideally follow <verb>Thing syntax so fizzBuzz -> logFizzBuzz

This is my counter proposal;

function logFizzBuzz(count){ for (let i = 1; i <= count; i++) { let out = ''; if (i % 3 === 0){ out += 'Fizz'; } if (i % 5 === 0){ out += 'Buzz'; } console.log(out || i); } }; logFizzBuzz(45);

There are a couple of things that are worrisome in this code. The one that caught me first was the shadowing of the num variable... You have 2 different variables called num.

In the populateArray function you should have a different name for it, not num.

This got me thinking, and the name num is a poor name choice. It should be count or similar, because it's not an actual number you are computing FizzBuzz on, but rather the count of values....

In addition, your loop for populating the array is

for (let i = 1; i < num + 1; i++) { fizzBuzzNumbers.push(i); } 

Loops typically are 0-based in computing, and when you need a 1-based loop like you do now, instead of having let i = 1; i < num + 1; i++ it is more common to have let i = 1; i <= num; i++ - it is probably inconsequential on performance, but having the additional + 1 in each check of the loop is messy. Note that <= x is the same as < x + 1

Finally, your if-else-if-else-if ... chain has too many comparisons. Your first comparison compares for both %5 and %3 so for your subsequent checks you don't need to check for the negative match on things.

You have:

if (fizzBuzzNumbers[i] % 3 === 0 && fizzBuzzNumbers[i] % 5 === 0) { console.log('FizzBuzz'); } else if (fizzBuzzNumbers[i] % 3 === 0 && fizzBuzzNumbers[i] % 5 !== 0) { console.log('Fizz'); } else if (fizzBuzzNumbers[i] % 3 !== 0 && fizzBuzzNumbers[i] % 5 === 0) { console.log('Buzz'); } else { console.log(fizzBuzzNumbers[i]); } 

But that could be just:

if (fizzBuzzNumbers[i] % 3 === 0 && fizzBuzzNumbers[i] % 5 === 0) { console.log('FizzBuzz'); } else if (fizzBuzzNumbers[i] % 3 === 0) { console.log('Fizz'); } else if (fizzBuzzNumbers[i] % 5 === 0) { console.log('Buzz'); } else { console.log(fizzBuzzNumbers[i]); } 

Technically, your code would read simpler if it was also in a forEach loop, instead of a for(...) loop. Consider the outer loop:

fizzBuzzNumbers.forEach((value) => { if (value % 3 === 0 && value % 5 === 0) { console.log('FizzBuzz'); } else if (value % 3 === 0) { console.log('Fizz'); } else if (value % 5 === 0) { console.log('Buzz'); } else { console.log(value); } }); 

Aside from fizzBuzzNumbers[i] always being identical to i (as others pointed out), I think this is a fantastic first attempt. I.e. if num was 3, fizzBuzzNumbers would be [1,2,3] and i would go from 1-to-2-to 3 without even needing to reference the array at all beyond length, which is just your input num. Your array could actually be ['big', 'bad', 'wolf'] and the output would be the same for an input of 3.

One resource that I think is worth a watch after trying this problem is Tom Scott's FizzBuzz: One Simple Interview Question on YouTube. He breaks down some common ways that people approach the problem, as well as why some approaches are better than others.

Your solution is good. One big change your code could use is to simplify into a single loop. Generating an array counts as the first loop, of course. Replacing it with a for loop or a while loop works well for this application. You already have a for loop, so let's reconstruct your fizzBuzz function to harness that.

Remove the array and the populateArray function

Removing the fizzBuzzNumbers array and the accompanying function will simplify the rest of the function drastically.

const fizzBuzz = (num) => { for (let i = 0; i < fizzBuzzNumbers.length; i++) { if (fizzBuzzNumbers[i] % 3 === 0 && fizzBuzzNumbers[i] % 5 === 0) { console.log("FizzBuzz"); } else if (fizzBuzzNumbers[i] % 3 === 0 && fizzBuzzNumbers[i] % 5 !== 0) { console.log("Fizz"); } else if (fizzBuzzNumbers[i] % 3 !== 0 && fizzBuzzNumbers[i] % 5 === 0) { console.log("Buzz"); } else { console.log(fizzBuzzNumbers[i]); } } }; 

Refactor the old array references

So, we want to change fizzBuzzNumbers.length to num to use the num parameter. I'll go ahead and change num to max for clearer understanding of its intent. I'll change the initialization of the for loop to i = 1 to start at 1 as intended and the condition to i <= max.

const fizzBuzz = (max) => { for (let i = 1; i <= max; i++) { if (fizzBuzzNumbers[i] % 3 === 0 && fizzBuzzNumbers[i] % 5 === 0) { console.log("FizzBuzz"); } else if (fizzBuzzNumbers[i] % 3 === 0 && fizzBuzzNumbers[i] % 5 !== 0) { console.log("Fizz"); } else if (fizzBuzzNumbers[i] % 3 !== 0 && fizzBuzzNumbers[i] % 5 === 0) { console.log("Buzz"); } else { console.log(fizzBuzzNumbers[i]); } } }; 

Additionally, all references to fizzBuzzNumbers[i] can simply be changed to i instead:

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

Nit-picky suggestions

This code works just as your original code does, but there are a few nit-picky things that could be adjusted.

If a number is divisible by both 3 and 5, you can reduce the condition to i % 15 === 0:

const fizzBuzz = (max) => { for (let i = 1; i <= max; i++) { if (i % 15 === 0) { console.log("FizzBuzz"); } else if (i % 3 === 0 && i % 5 !== 0) { console.log("Fizz"); } else if (i % 3 !== 0 && i % 5 === 0) { console.log("Buzz"); } else { console.log(i); } } }; 

The negation checks when a number is divisible by either 5 or 3 (exclusive) can be removed (they are implied):

const fizzBuzz = (max) => { for (let i = 1; i <= max; i++) { if (i % 15 === 0) { console.log("FizzBuzz"); } else if (i % 3 === 0/* && i % 5 !== 0 */) { console.log("Fizz"); } else if (/* i % 3 !== 0 && */i % 5 === 0) { console.log("Buzz"); } else { console.log(i); } } }; 

Keep practicing! Make sure to keep your code readable even if it's just for yourself (variable names, code indentation, comments, etc).