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412. Fizz Buzz

easy

Return a list of strings: 'Fizz' for multiples of 3, 'Buzz' for multiples of 5, 'FizzBuzz' for multiples of both, otherwise the number as a string. The legendary screening question — easy to write, easy to over-engineer.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

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.

Constraints

  • 1 <= n <= 10^4

Examples

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"]

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Output

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Hints

Progressive — try the first before opening the next.

Hint 1

Order of checks matters: test 15 (i.e., both 3 AND 5) FIRST. If you check 3 alone first, multiples of 15 would print 'Fizz' instead of 'FizzBuzz'.

Hint 2

Equivalently: build the string by concatenating 'Fizz' if i % 3 == 0 and 'Buzz' if i % 5 == 0; if empty, use str(i). One pass, no nested if.

Hint 3

Avoid the modulo entirely with two counters mod 3 and mod 5: reset on hit. Micro-optimization but interviewers occasionally ask.

Hint 4

Loop 1..n, append result to a list, return.

Solution approach

Reveal approach

Loop i from 1 to n inclusive. For each i, determine the string by concatenation: piece = ''; if i % 3 == 0 then piece += 'Fizz'; if i % 5 == 0 then piece += 'Buzz'; if piece is empty, piece = str(i). Append piece to the result list. Return the list. O(n) time, O(n) space for the output. The concatenation pattern is one line shorter than the if/elif/elif/else cascade and naturally handles multiples of 15 — order doesn't matter because both checks fire. The modulo-free counter variant maintains f = 0 and b = 0; each iteration increments both; on f == 3 reset and append 'Fizz', on b == 5 reset and append 'Buzz'. Same O(n) but no division.

Complexity

Time
O(n)
Space
O(n)

Related patterns

  • math
  • simulation

Related problems

Asked at

Companies reported asking this problem (sourced from public Glassdoor, Blind, and Levels.fyi interview posts).

  • Amazon
  • Microsoft
  • Apple

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