Fizz Buzz

LeetCode 412 | Difficulty: Easy

Easy

Problem Description

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.

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"]

Constraints:

• 1 <= n <= 10^4

Topics: Math, String, Simulation

---

Approach

Mathematical

Look for mathematical patterns or formulas. Consider: modular arithmetic, GCD/LCM, prime factorization, combinatorics, or geometric properties.

When to use

Problems with clear mathematical structure, counting, number properties.

String Processing

Consider character frequency counts, two-pointer approaches, or building strings efficiently. For pattern matching, think about KMP or rolling hash. For palindromes, expand from center or use DP.

When to use

Anagram detection, palindrome checking, string transformation, pattern matching.

---

Solutions

Solution 1: C# (Best: 316 ms)

Metric	Value
Runtime	316 ms
Memory	N/A
Date	2018-02-21

Solution

public class Solution {
    public IList<string> FizzBuzz(int n) {
        
        List<string> result = new List<string>();
        for(int i=1;i<=n;i++)
        {
            if(i%3==0 && i%5 ==0)
            {
                result.Add("FizzBuzz");
            }
            else if(i%3 == 0)
            {
                result.Add("Fizz");
            }
            else if(i%5 == 0)
            {
                result.Add("Buzz");
            } 
            else 
            {
                result.Add(i.ToString());
            }
        }
        
        return result;
        
    }
}

---

Complexity Analysis

Approach	Time	Space
Solution	$O(n)$	$O(1) to O(n)$

---

Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.