Math & Number Theory

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, otherwise answer[i] == i as a string.

Input: n = 3
Output: ["1","2","Fizz"]

Input: n = 5
Output: ["1","2","Fizz","4","Buzz"]

Input: n = 15
Output: ["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]

Iterate from 1 to n. For each number, build the result string by checking divisibility. Check divisibility by 15 (FizzBuzz) first, or use a StringBuilder approach to concatenate "Fizz" and "Buzz" independently.

For each i from 1 to n, build a string using StringBuilder.
If i % 3 == 0, append "Fizz".
If i % 5 == 0, append "Buzz".
If StringBuilder is still empty, use String.valueOf(i).

class Solution { public List<String> fizzBuzz(int n) { List<String> result = new ArrayList<>(); for (int i = 1; i <= n; i++) { StringBuilder sb = new StringBuilder(); if (i % 3 == 0) sb.append("Fizz"); if (i % 5 == 0) sb.append("Buzz"); if (sb.length() == 0) sb.append(i); result.add(sb.toString()); } return result; } }

Complexity Analysis:

Time complexity: O(n) — one pass through numbers 1 to n.
Space complexity: O(n) — for the output list.