How do you solve Fizz Buzz on LeetCode?

Fizz Buzz (LeetCode #412) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding divisibility is crucial for this problem.

What is the brute force solution for Fizz Buzz?

The brute force approach for Fizz Buzz is Brute Force, with O(n) time complexity and O(n) space complexity. The brute force approach involves iterating through each number from 1 to n and checking the divisibility conditions for each number. It’s straightforward but not the most efficient way to solve the problem. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Fizz Buzz?

Fizz Buzz uses the following concepts: Math, String, Simulation. The recommended patterns to study are: String manipulation, Conditional statements.

What are the interview tips for Fizz Buzz?

Clarify the problem requirements before coding. Think about edge cases, like the smallest and largest values of n. Explain your thought process as you code to show your understanding.

What are common mistakes when solving Fizz Buzz?

Not checking for both conditions together (3 and 5). Confusing the order of conditions leading to incorrect outputs.

#412

Fizz Buzz

Easy

MathStringSimulationString manipulationConditional statements

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(n)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution is actually the same as the brute force approach in terms of logic, but we can optimize our checks by using a single loop and directly appending results based on the conditions. This is efficient and clear.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty list to store the results.
2Step 2: Loop through each number from 1 to n.
3Step 3: Use a single conditional statement to check if the number is divisible by both 3 and 5, then check for 3, then for 5, and finally append the number if none match.
4Step 4: Return the results list.

solution.py10 lines

1def fizzBuzz(n):
2    result = []
3    for i in range(1, n + 1):
4        output = ''
5        if i % 3 == 0:
6            output += 'Fizz'
7        if i % 5 == 0:
8            output += 'Buzz'
9        result.append(output or str(i))
10    return result

ℹ

Complexity note: The time complexity remains O(n) as we still iterate through all numbers from 1 to n. The space complexity is O(n) since we store n results in the output list.

1Understanding divisibility is crucial for this problem.
2Using string concatenation can simplify the logic.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.