:orphan:

Fizz Buzz
=========

.. highlight:: none

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```

.. highlight:: python

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 :math:`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
----

.. literalinclude:: ../problems/easy/fizz-buzz/fizz_buzz__approach_1.py
    :language: python
    :lines: 7-

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: :math:`O(n)` — single pass through the numbers from 1 to n
| Auxiliary Space: :math:`O(1)`

.. autofunction:: fizz_buzz__approach_1.fizzBuzz