Mastering FizzBuzz in Python: A Step-by-Step Guide

FizzBuzz is one of the most famous coding interview questions. While it may seem simple at first glance, it tests a programmer's ability to handle conditional logic, control flow, and edge cases effectively. In this post, we'll break down a clean and efficient Python solution.

The Problem

The rules of FizzBuzz are straightforward. We need to write a function that prints numbers from 1 to n, but with specific substitutions:

• For multiples of 3, print "Fizz".
• For multiples of 5, print "Buzz".
• For multiples of both 3 and 5, print "FizzBuzz".
• For all other numbers, print the number itself.

The function should return a list of these results.

Example:
fizzbuzz(15) should return:
[1, 2, 'Fizz', 4, 'Buzz', 'Fizz', 7, 8, 'Fizz', 'Buzz', 11, 'Fizz', 13, 14, 'FizzBuzz']

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The Solution

Here is a clear and readable Python implementation:

def fizzbuzz(n):
    """
    Generates the FizzBuzz sequence up to n.
    """
    results = []

    # Loop from 1 to n (inclusive)
    for i in range(1, n + 1):
        # Check for multiples of both 3 and 5 (15) first
        if i % 3 == 0 and i % 5 == 0:
            results.append("FizzBuzz")
        else:
            # Check for multiples of 3
            if i % 3 == 0:
                results.append("Fizz")
            # Check for multiples of 5
            elif i % 5 == 0:
                results.append("Buzz")
            # Otherwise, add the number itself
            else:
                results.append(i)

    return results

# Example usage
print(fizzbuzz(15))

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Code Breakdown

Let's analyze the solution line by line to understand exactly how it works.

1. Function Definition and Initialization

def fizzbuzz(n):
    results = []

We define a function fizzbuzz that accepts an integer n as the upper limit. We also initialize an empty list results to store our output.

2. The Loop

for i in range(1, n + 1):

We use range(1, n + 1) to generate numbers starting from 1 up to n. The + 1 is crucial because the range function in Python is exclusive at the upper bound.

3. The "FizzBuzz" Condition

if i % 3 == 0 and i % 5 == 0:
    results.append("FizzBuzz")

This is the most critical part. We must check if a number is divisible by both 3 and 5 (which means it's a multiple of 15) before checking for 3 or 5 individually. If we checked for 3 first, the number 15 would be labeled "Fizz" instead of "FizzBuzz", which would be incorrect.

4. The "Fizz" and "Buzz" Conditions

else:
    if i % 3 == 0:
        results.append("Fizz")
    elif i % 5 == 0:
        results.append("Buzz")

If the number isn't a multiple of 15, we proceed to check:

• i % 3 == 0: If true, append "Fizz".
• i % 5 == 0: If true, append "Buzz".

Using elif ensures that these conditions are mutually exclusive within the else block of the main condition.

5. The Default Case

else:
    results.append(i)

If none of the above conditions are met (the number isn't divisible by 3 or 5), we simply append the number i to the list.

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Key Takeaways

1. Order Matters: The condition for "FizzBuzz" (divisible by 15) must come before the checks for "Fizz" (3) or "Buzz" (5).
2. Modulo Operator: The % operator is your best friend for checking divisibility. i % n == 0 means i is perfectly divisible by n.
3. Complexity:
   • Time Complexity: O(n), because we iterate through the numbers exactly once.
   • Space Complexity: O(n), because we store a result for every number up to n.

Mastering this logic is a great first step in understanding control flow and algorithmic thinking!