Fizz Buzz

Easymathsimulation
Category: Fundamentals
LeetCode →

Approach

Fizz Buzz

Approach

Iterate from 1 to n and apply conditional logic based on divisibility.

Algorithm

  1. Create result list
  2. For each number i from 1 to n:
    • If divisible by 15 (both 3 and 5): add "FizzBuzz"
    • Else if divisible by 3: add "Fizz"
    • Else if divisible by 5: add "Buzz"
    • Else: add number as string
  3. Return result list

Complexity

  • Time: O(n) - single iteration
  • Space: O(n) - result list storage

Key Insights

  • Check divisibility by 15 first (LCM of 3 and 5)
  • Classic interview problem testing basic conditional logic
  • Can be optimized with string concatenation approach

Solution

java
1class Solution {
2    public List<String> fizzBuzz(int n) {
3        List<String> result = new ArrayList<>();
4        
5        for (int i = 1; i <= n; i++) {
6            if (i % 15 == 0) {
7                result.add("FizzBuzz");
8            } else if (i % 3 == 0) {
9                result.add("Fizz");
10            } else if (i % 5 == 0) {
11                result.add("Buzz");
12            } else {
13                result.add(String.valueOf(i));
14            }
15        }
16        
17        return result;
18    }
19}
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