This is a quite common interview question because it is simple but also challenging if you have never seen this problem before. As the solution is very straightforward, after this article, you will easily solve this problem when you see it again during the interview.

Problem

Write a function that accepts a single integer, n, and produces an array of strings, answer, where each string corresponds to a number from 1 to n (inclusive). For each number i in this range:

if i is a multiple of both 3 and 5, the corresponding entry in results should be “FizzBuzz”.
if i is only a multiple of 3, the entry should be “Fizz”.
if i is only a multiple of 5, the entry should be “Buzz”.
if i does not meet any of the above criteria, simply use the number itself as a string in results.

A quick walkthrough

Basically the problem is we count from 1 to a positive number of n and store each number in an array, answer, with following conditions:

if the number is a multiple of 3 such as 3, 6, 9 etc, then we store Fizz in the array
if the number is a multiple of 5 such as 5, 10, 15 etc, we store Buzz in the array
if the number is a multiple of both 3 and 5 such as 15, 30, we store FizzBuzz in the array
otherwise we store the number as string in the array

and then finally we return the array.

Examples

Let’s take a look at some examples.

n = 3

It will return [“1”, “2”, “Fizz”].

n = 5

It will return [“1”, “2”, “Fizz”, “4”, “Buzz”].

n = 15

It will return [“1”, “2”, “Fizz”, “4”, “Buzz”, “Fizz”, “7”, “8”, “Fizz”, “Buzz”, “11”, “Fizz”, “13”, “14”, “FizzBuzz”].

Approach

To solve this problem, we are going to make use of modulo.

The modulo operator uses % character and it will return the remainder of two integers left over from the division.

For example, 17 % 3 will return 2 and 15 % 3 will return 0. From that, we can see that if the remainder is 0, the first operand, 15, is divisible by 3. The first operand will be n in our code.

As a result, if the n is divisible by:

3, we will store Fizz
5, we will store Buzz
both 3 and 5, we will store FizzBuzz

Implementation in JavaScript

Here is the complete code in JavaScript.

class solution {
  fizzBuzz(n) {
    const answer = [];
    for (let i = 1; i <= n; i++) {
      if (i % 3 === 0 && i % 5 === 0) {
        answer.push('FizzBuzz');        
      } else if (i % 3 === 0) {
        answer.push('Fizz');
      } else if (i % 5 === 0) {
        answer.push('Buzz');
      } else {
        answer.push(i.toString());
      }
    }
    return answer;
  }
}

Implementation steps

1. Create the class and fizzBuzz method. The method will accept one argument, n.

class solution {
  fizzBuzz(n) {
  
  }
}

2. Declare the array, answer, and we will return it at the end of the algorithm.

class solution {
  fizzBuzz(n) {
    const answer = [];

    return answer;
  }
}

3. We will loop through it starting from 1 until n incrementing by 1 in each loop.

class solution {
  fizzBuzz(n) {
    const answer = [];

    for (let i = 1; i <= n; i++) {

    }

    return answer;
  }
}

4. First, we will check if i can be divisible by both 3 and 5. If so, we store FizzBuzz in the array. This is important to check FizzBuzz first, otherwise, as it will satisfy both 3 and 5, either Fizz or Buzz will get store instead of FizzBuzz.

class solution {
  fizzBuzz(n) {
    const answer = [];
    for (let i = 1; i <= n; i++) {
      if (i % 3 === 0 && i % 5 === 0) {
        answer.push('FizzBuzz');        
      }
    }
    return answer;
  }
}

5. We will also check if it is Fizz (line 9) or Buzz (line 11) and if both are not satisfied, we will store the i as string (line 13).

class solution {
  fizzBuzz(n) {
    const answer = [];

    for (let i = 1; i <= n; i++) {
      if (i % 3 === 0 && i % 5 === 0) {
        answer.push('FizzBuzz');   
      } else if (i % 3 === 0) {
        answer.push('Fizz');
      } else if (i % 5 === 0) {
        answer.push('Buzz');
      } else {
        answer.push(i.toString());
      }
    }

    return answer;
  }
}

Python

Here is the complete code in Python.

class Solution:
  def fizzBuzz(self, n: int) -> List[str]:
    answer = []

    for n in range(1, n + 1):
      if n % 3 == 0 and n % 5 == 0:
        answer.append("FizzBuzz")
      elif n % 3 == 0:
        answer.append("Fizz")
      elif n % 5 == 0:
        answer.append("Buzz")
      else:
        answer.append(str(n))
    
    return answer

Big O

Time complexity

It is a linear time complexity, O(n), because there is no upper limit for n and the loop has to go through until the n.

Space complexity

There is no value increasing in size so it will be O(1) for the space complexity.