🏷️ Category: Algorithms / Logic | 🎯 Difficulty:
Easy| 🔥 Frequency: ★★★★★
📋 Problem Statement
For numbers 1 to n:
- Print “FizzBuzz” if divisible by both 3 and 5
- Print “Fizz” if divisible by 3
- Print “Buzz” if divisible by 5
- Otherwise print the number
📥 Input → n = 15
📤 Output → [1, 2, "Fizz", 4, "Buzz", "Fizz", 7, 8, "Fizz",
"Buzz", 11, "Fizz", 13, 14, "FizzBuzz"]
🧠 What the Interviewer is Testing
| Concept | Why It Matters |
|---|---|
| Modulo operator | Basic arithmetic |
| Condition order | Logic trap awareness |
| Clean code | Readability |
⚠️ Classic trap: Check divisible by 15 first, then 3, then 5. If you check 3 first, “FizzBuzz” never prints correctly.
💡 Approach
Check order matters:
1️⃣ n % 15 == 0 → "FizzBuzz" ← MUST be first
2️⃣ n % 3 == 0 → "Fizz"
3️⃣ n % 5 == 0 → "Buzz"
4️⃣ else → str(n)
✅ Optimal Solution
def fizzbuzz(n: int) -> list:
result = []
for i in range(1, n + 1):
if i % 15 == 0:
result.append("FizzBuzz")
elif i % 3 == 0:
result.append("Fizz")
elif i % 5 == 0:
result.append("Buzz")
else:
result.append(str(i))
return result
# ── Test Cases ─────────────────────────────────
print(fizzbuzz(15))
# ✅ ['1','2','Fizz','4','Buzz','Fizz','7','8',
# 'Fizz','Buzz','11','Fizz','13','14','FizzBuzz']
🔄 Pythonic One-liner Version
def fizzbuzz_clean(n: int) -> list:
return [
"FizzBuzz" if i % 15 == 0 else
"Fizz" if i % 3 == 0 else
"Buzz" if i % 5 == 0 else
str(i)
for i in range(1, n + 1)
]
📊 Complexity Analysis
┌─────────────┬────────────────────────────────────┐
│ Time O(n) │ Single loop through 1 to n │
│ Space O(n) │ Output list of n elements │
└─────────────┴────────────────────────────────────┘
🎤 How to Explain in the Interview
“I check divisibility by 15 first — that handles the FizzBuzz case. Then 3 for Fizz, 5 for Buzz, otherwise the number itself. The order is critical — checking 3 or 5 before 15 would break the FizzBuzz output.”
🔁 Common Follow-up Questions
Q1 → Make it extensible for any rules?
def fizzbuzz_generic(n: int, rules: dict) -> list:
# rules = {3: "Fizz", 5: "Buzz", 7: "Boom"}
result = []
for i in range(1, n + 1):
word = ''.join(v for k, v in sorted(rules.items()) if i % k == 0)
result.append(word if word else str(i))
return result
📁 Set 1 of 4 — Python Interview Prep Series — devprepguide.com