Fizz Buzz

Problem

https://leetcode.com/problems/fizz-buzz/

Given an integer n, return a string array answer (1-indexed) where:

answer[i] == "FizzBuzz" if i is divisible by 3 and 5.
answer[i] == "Fizz" if i is divisible by 3.
answer[i] == "Buzz" if i is divisible by 5.
answer[i] == i (as a string) if none of the above conditions are true.

Example 1:

Input: n = 3
Output: ["1","2","Fizz"]

Example 2:

Input: n = 5
Output: ["1","2","Fizz","4","Buzz"]

Example 3:

Input: n = 15
Output: ["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]

Constraints:

1 <= n <= 10:sup:`4`

Pattern

Math, String, Simulation

Solution

https://www.youtube.com/watch?v=QPZ0pIK_wsc

There are 4 cases to consider: divisible by 3, divisible by 5, divisible by 3 and 5, and not divisible by 3 or 5. However, observe that the third case is a combination of the first two cases. We can use a string to build up the output for each index 1 to \(n\), appending ‘Fizz’, then appending ‘Buzz’. This saves us on an if statement. Finally we can check if the output string is empty to determine whether we need to print the index.

Code

from typing import List


def fizzBuzz(n: int) -> List[str]:
    """Returns a list of numbers 1 to ``n``, with some numbers replaced by
    'Fizz', 'Buzz', or 'FizzBuzz' depending on their divisibility.
    """
    strings = []
    for i in range(1, n + 1):
        string = ''
        if i % 3 == 0:
            string += 'Fizz'
        if i % 5 == 0:
            string += 'Buzz'
        if len(string) == 0:
            string += str(i)
        strings.append(string)
    return strings

Test

>>> from fizz_buzz__approach_1 import fizzBuzz
>>> fizzBuzz(15)
['1', '2', 'Fizz', '4', 'Buzz', 'Fizz', '7', '8', 'Fizz', 'Buzz', '11', 'Fizz', '13', '14', 'FizzBuzz']

Complexity

Time: \(O(n)\) — single pass through the numbers from 1 to n

Auxiliary Space: \(O(1)\)

fizz_buzz__approach_1.fizzBuzz(n: int) → List[str]: Returns a list of numbers 1 to n, with some numbers replaced by ‘Fizz’, ‘Buzz’, or ‘FizzBuzz’ depending on their divisibility.