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 <= 104This approach iterates through numbers from 1 to n and applies conditional logic using modulus operations to determine if a number should be represented as "Fizz", "Buzz", or "FizzBuzz". If none of these conditions are met, the number itself is returned as a string.
The function fizzBuzz takes an integer n and generates an array of strings. It checks each number from 1 to n for divisibility by 3 and 5. The use of strdup efficiently allocates memory for the strings returned.
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Time Complexity: O(n) as we iterate through each number from 1 to n once.
Space Complexity: O(n) for the output array.
This approach uses a hash map to store possible outputs, simplifying conditional checks. By mapping integers to their respective Fizz or Buzz values, we consolidate decision logic, reducing redundancy in the code.
This C solution defines a constant array storing potential word outputs for Fizz, Buzz, and FizzBuzz. Indexes based on divisibility determine which string to select, reducing the number of branch conditions.
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Java
Python
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Time Complexity: O(n), since each element is evaluated once.
Space Complexity: O(n) due to storage for result strings.
| Approach | Complexity |
|---|---|
| Simple Iteration with Condition Checks | Time Complexity: |
| Hash Map Optimized Approach | Time Complexity: |
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