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S1Q1 · Describe Number Based on Divisibility

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Function: describe_number(num: int) -> str

Core idea: check divisibility by 15 first, then 3, then 5, then default to "Normal".

def describe_number(num: int) -> str:
    if num % 15 == 0:
        return "FizzBuzz"
    elif num % 3 == 0:
        return "Fizz"
    elif num % 5 == 0:
        return "Buzz"
    return "Normal"

Key rules: - Check % 15 before % 3 and % 5 -otherwise FizzBuzz cases get caught early - Exactly one of the four strings is returned - The classic FizzBuzz problem -appears constantly in interviews

Problem Statement

Problem

Write a function describe_number(num) that returns "FizzBuzz" if divisible by both 3 and 5, "Fizz" if by 3 only, "Buzz" if by 5 only, and "Normal" otherwise.

Examples:

Input 
9
Output 
"Fizz"
Input 
10
Output 
"Buzz"
Input 
15
Output 
"FizzBuzz"
Input 
7
Output 
"Normal"

Understanding the problem

num % 3 == 0 % 5 == 0 Result
9 ✅ ❌ "Fizz"
10 ❌ ✅ "Buzz"
15 ✅ ✅ "FizzBuzz"
7 ❌ ❌ "Normal"

Why check % 15 first?

15 is divisible by both 3 and 5. If you check % 3 first, 15 would return "Fizz" before reaching the % 5 check. Checking % 15 first ensures the combined case is caught correctly.

Solution approaches

def describe_number(num: int) -> str:
    if num % 15 == 0:
        return "FizzBuzz"
    elif num % 3 == 0:
        return "Fizz"
    elif num % 5 == 0:
        return "Buzz"
    else:
        return "Normal"

def describe_number(num: int) -> str:
    result = ""
    if num % 3 == 0:
        result += "Fizz"
    if num % 5 == 0:
        result += "Buzz"
    return result if result else "Normal"

Build the result by concatenating -no need to check % 15 separately. Both strings get appended in the right order automatically when both conditions are true.

def describe_number(num: int) -> str:
    mapping = {
        (True, True):   "FizzBuzz",
        (True, False):  "Fizz",
        (False, True):  "Buzz",
        (False, False): "Normal",
    }
    return mapping[(num % 3 == 0, num % 5 == 0)]

Map the pair of boolean conditions to the result. No if-else at all.

describe_number = lambda num: (
    "FizzBuzz" if num % 15 == 0 else
    "Fizz"     if num % 3  == 0 else
    "Buzz"     if num % 5  == 0 else
    "Normal"
)

Chained ternary as a lambda -all logic in one expression.

Key takeaways

01

Check the combined case first

Always check % 15 (or % 3 and % 5) before the individual checks. Getting this order wrong is the most common FizzBuzz mistake in interviews.

02

String concatenation avoids % 15

The concatenation approach naturally handles the combined case -both strings get appended when both conditions are true, producing "FizzBuzz" automatically.

03

Boolean tuple as dict key

(num % 3 == 0, num % 5 == 0) produces a tuple of booleans mapping cleanly to all four outcomes. A neat pattern for multi-condition dispatch.