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This 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.
Time Complexity: O(n) as we iterate through each number from 1 to n once.
Space Complexity: O(n) for the output array.
1def fizzBuzz(n):
2 result = []
3 for i in range(1, n + 1):
4 if i % 3 == 0 and i % 5 == 0:
5 result.append("FizzBuzz")
6 elif i % 3 == 0:
7 result.append("Fizz")
8 elif i % 5 == 0:
9 result.append("Buzz")
10 else:
11 result.append(str(i))
12 return result
13The Python solution utilizes a list to track results. A for loop checks divisibility and either appends Fizz, Buzz, or the number itself.
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.
Time Complexity: O(n), since each element is evaluated once.
Space Complexity: O(n) due to storage for result strings.
1#
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.