File: fizz-buzz/solution.py

Date: 2026-06-06

Time: 16:52

fizz-buzz/solution.py

Purpose

This file implements LeetCode problem #412 (Fizz Buzz). It owns the single responsibility of generating the FizzBuzz sequence from 1 to n as a list of strings, following LeetCode's Solution class convention.

Key Components

Solution.fizzBuzz(self, n: int) -> list[str] — The sole method. Takes an inclusive upper bound n and returns a list where each element is:

• "FizzBuzz" if the index (1-based) is divisible by both 3 and 5
• "Fizz" if divisible by 3 only
• "Buzz" if divisible by 5 only
• The number as a string otherwise

Patterns

Check order matters. The % 15 check comes first, before % 3 and % 5. This is the standard FizzBuzz idiom — if you checked % 3 first, multiples of 15 would hit "Fizz" and never reach "FizzBuzz". Using % 15 instead of % 3 == 0 and % 5 == 0 is a minor optimization (one modulo vs. two) but mainly a stylistic choice common in LeetCode solutions.

Accumulator pattern. Builds a result list via .append() in a loop rather than using a list comprehension. Straightforward and readable, though a comprehension with a helper or conditional expression would also work.

LeetCode class convention. All solutions in this repo wrap their logic in a Solution class with a specifically-named method matching LeetCode's expected signature.

Dependencies

Imports: None. Pure Python, no standard library or third-party dependencies.

Imported by: The fizz-buzz/test_solution.py file directly. The massive "Imported By" list in the context is misleading — those are test files for *other* problems that happen to share a common test harness pattern importing from their own local solution.py, not from this file specifically.

Flow

1. Initialize empty result list

2. Iterate i from 1 to n inclusive (range(1, n + 1))

3. For each i, evaluate divisibility in priority order: 15 → 3 → 5 → fallback

4. Append the corresponding string

5. Return the accumulated list

Invariants

• Output length is always exactly n (one entry per integer from 1 to n)
• The 1-based indexing is enforced by range(1, n + 1)
• Every 15th element is "FizzBuzz", every 3rd (non-15th) is "Fizz", every 5th (non-15th) is "Buzz"
• For n <= 0, returns an empty list (the range produces nothing)

Error Handling

None. No input validation — if n is negative or zero, the loop simply doesn't execute and an empty list is returned. If n is not an integer, Python's range() will raise TypeError. This is fine for LeetCode where inputs are guaranteed valid.

Topics to Explore

• [file] fizz-buzz/test_solution.py — See how the solution is validated and what edge cases are covered
• [file] fizz-buzz/review.md — The code review notes for this solution, may contain alternative approaches considered
• [file] fizz-buzz/plan.md — The planning document showing how the approach was chosen
• [general] fizzbuzz-variants — LeetCode has follow-ups (Fizz Buzz Multithreaded #1195) that require concurrency; worth comparing the single-threaded baseline
• [function] add-digits/solution.py:Solution.addDigits — Another number-theory problem using modular arithmetic, shows how the repo handles similar math patterns

Beliefs

• fizzbuzz-check-order — The % 15 divisibility check must precede % 3 and % 5 checks; reordering produces incorrect output for multiples of 15
• fizzbuzz-output-length — fizzBuzz(n) always returns a list of exactly n elements for any n >= 1
• fizzbuzz-no-dependencies — The solution uses no imports; it is pure Python with no external dependencies
• fizzbuzz-1-indexed — The output is 1-indexed: result[0] corresponds to integer 1, result[n-1] to integer n