Introduction

In this blog post, we will explore how to solve the LeetCode problem "412" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 <= 104

Explanation

Here's the solution to the FizzBuzz problem:

• Iterate and Check: Iterate from 1 to n. For each number, check its divisibility by 3 and 5.
• Conditional Logic: Apply conditional logic to determine if "Fizz", "Buzz", "FizzBuzz", or the number itself should be added to the result array.

• String Conversion: Convert the number to a string when no divisibility conditions are met.

• Runtime & Space Complexity: O(n) runtime, O(n) storage.

Code

    def fizzBuzz(n: int) -> list[str]:
    """
    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.
    """
    answer = []
    for i in range(1, n + 1):
        if i % 3 == 0 and i % 5 == 0:
            answer.append("FizzBuzz")
        elif i % 3 == 0:
            answer.append("Fizz")
        elif i % 5 == 0:
            answer.append("Buzz")
        else:
            answer.append(str(i))
    return answer