FizzBuzz program in javascript

We will implement a simple algorithm to solve the FizzBuzz problem in Javascript.

Example

In this, we print all the numbers from 1 to N with a specific rule.

1. If the number is divisible by 3 then print ‘Fizz’.
2. If the number is divisible by 5 then print ‘Buzz’.
3. If the number is divisible by both 3 and 5 then print ‘FizzBuzz’.

Input:
10

Output:
1
2
'Fizz'
4
'Buzz'
'Fizz'
7
8
'Fizz'
'Buzz'

Implementation of FizzBuzz in javascript.

• We will loop through all the numbers from 1 to N.
• Then in each iteration we will first check if the number is divisible by both 3 and 5, then print ‘FizzBuzz’.
• Else if it is divisible by only 3 then print ‘Fizz’ or If it is only divisible by 5 then print ‘Buzz’.
• Otherwise just print the number.
• Everything will be written in ES6.

let fizzBuzz = (n) => {
   for(let i = 1; i <= n; 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);
     }
   }
}

Input:
fizzBuzz(15);

Output:
1
2
"Fizz"
4
"Buzz"
"Fizz"
7
8
"Fizz"
"Buzz"
11
"Fizz"
13
14
"FizzBuzz"

Time complexity: O(n).

Space complexity: O(1).

Time and Space complexity

• We are just iterating from 1 to N, so Time complexity is O(n).
• We are using constant space, so Space complexity is O(1).