4 Ways to Solve FizzBuzz in Python

FizzBuzz is a challenge that involves writing code that labels numbers divisible by three as “Fizz,” four as “Buzz” and numbers divisible by both as “FizzBuzz.” Here’s how to solve it in Python.

Written by Emmett Boudreau
Published on Oct. 05, 2022
Image: Shutterstock / Built In

The technical interview is hard to master and can be a nerve-racking experience. Not only do you need to know what you are talking about, but you also have to prove it to the person interviewing you. Fortunately, most fears of failure in this regard are exaggerated, and often, the interview will boil down to only a few potentially difficult questions.

4 Methods for Solving FizzBuzz in Python

1. Conditional statements.
2. String concatenation.
3. Itertools.
4. Lambda.

One very common problem that programmers are asked to solve in technical interviews and take-home assignments is the FizzBuzz problem. FizzBuzz is a word game designed for children to teach them about division. In the game, each number divisible by three will be returned with a `Fizz` and any number divisible by four will return a `Buzz`. I was never a big fan of the test, but it can help weed out weaker applicants.

While the test is pretty easy to pass so long as you know the right operators, there are a variety of different ways to solve it. However, some solutions might prove to be more impressive than others, and I think this is something to keep in mind when working on this problem for a real interview. In addition to demonstrating these alternative methods of solving FizzBuzz, we are going to time each solution and compare the respective results.

How to Solve FizzBuzz in Python

1. Conditional Statements

The most popular and well-known solution to this problem involves using conditional statements. For every number in n, we are going to need to check if that number is divisible by four or three. If the number is divisible by three, it will print `Fizz`; if the number is divisible by four, it will print `Buzz`. The key here is simply knowing what operators to use to check for divisibility. In Python, we can use the modulus operator, `%`.

In computing, the modulo operation is meant to return the signed remainder of division. If a number is divisible by another, the remainder will always be zero, so we can use that to our advantage whenever we make our FizzBuzz function. We will structure condition blocks like this, where `num` is the iteration in a list of numbers.

``````    if num % 3 == 0:
print('Fizz')
``````

We can now build an iterative loop following the same principle, except we’ll be adding `Fizz` and `Buzz`:

``````for num in range(1,101):
string = ""
if num % 3 == 0:
string = string + "Fizz"
if num % 4 == 0:
string = string + "Buzz"
if num % 4 != 0 and num % 3 != 0:
string = string + str(num)
print(string)``````

2. String Concatenation

Though incredibly similar to its regular conditional loop counterpart, the string concatenation method is another really great way to solve this problem. Of course, this method is also all but too similar to the conditional method. The significant difference here is that the conditionals are simply going to be affecting a small sequence of characters put into the string data-type.

``````for num in range(1,21):
string = “”
if num % 3 == 0:
string = string + “Fizz”
if num % 5 == 0:
string = string + “Buzz”
if num % 5 != 0 and num % 3 != 0:
string = string + str(num)
print(string)
``````

3. Itertools

Another way we could approach this problem — as well as other iteration problems — is to use the standard library tool, itertools. This will create a loop with better performance than most other iteration methods. Itertools can be thought of as an iteration library that is built to mirror several other extremely performant libraries from other languages, except using pythonic methods for solving problems.

Itertools will need to be imported, however, it is in the standard library. This means pip won’t be necessary, but itertools is still considered a project dependency. We are going to utilize three different methods from this module:

• `cycle()`: Cycle is a function takes a basic data-type and creates an iterator out of it. This function is useful and makes building custom iterators incredibly easy in Python.
• `count()`: Count is another generator that iterates a range. This iterator is often called an “infinite iterator,” which basically means that the count() function could essentially loop on and on forever.
• `islice()`: The islice function is short for “iteration slice.” We can use this iterator to cut out particular elements in a data structure and iterate them.

Combining these methods will allow us to create a new function where we can solve the FizzBuzz problem without using the typical iteration methods in Python that we might be used to.

``````import itertools as its
def fizz_buzz(n):
fizzes = its.cycle([""] * 2 + ["Fizz"])
buzzes = its.cycle([""] * 4 + ["Buzz"])
fizzes_buzzes = (fizz + buzz for fizz, buzz in zip(fizzes, buzzes))
result = (word or n for word, n in zip(fizzes_buzzes, its.count(1)))
for i in its.islice(result, 100):
print(i)
``````

The benefits of using this methodology is that the itertools library’s methods of iteration are typically going to be a lot faster than the pythonic methods of iteration. While itertools is still pythonic, it is likely that the speed of iterative looping is going to improve when using this library over the typical for loop in Python. Needless to say, creating a faster algorithm than any other applicant could certainly put you on the map for getting the job. This is a valuable module and application of the module for programmers who are still searching for employment.

4. Lambda

Another method we could use to solve this problem is even more Pythonic of a solution, and it makes use of Python’s bridge to scientific computing, lambda. There are a lot of standard functions that can be used with these lambda expressions, and they certainly come in handy. One of the most frequently used methods in this regard is the `map()` method. This method is going to take an expression that we can create using lambda as well as an iterative data structure.

``````print(*map(lambda i: 'Fizz'*(not i%3)+'Buzz'*(not i%5) or i, range(1,101)),sep='\n')
``````

For this example, I used the range generator, and the “not” keywords in order to reverse the polarity of the modulus operators usage.

What’s the Best Way to Solve FizzBuzz in Python?

With all of these new ways to solve the problem, you might be wondering which one you should use. Of course, there are going to be trade-offs between the solutions, but in order to really make a great impression, we could narrow our decision down to using either the lambda method or the itertools method.

The lambda method has the advantage of being incredibly concise. However, depending on what code the map() method uses for iteration, it might trail behind the itertools method in terms of speed due to its less efficient iteration. The only way to figure out whether or not this is the case is to run some tests and compare our interpreter return times. So, that is going to be the mission between comparing these two heaps of code. In order to facilitate this comparison, I am going to be using the IPython magic in-line command, `%timeit`. Let’s start by trying it out on the itertools method. Since I wrote this as a function earlier, I can simply time the function call:

``````%timeit fizz_buzz(101)
``````

We will do the same with the lambda method:

``````%timeit print(*map(lambda i: 'Fizz'*(not i%3)+'Buzz'*(not i%5) or i, range(1,101)),sep='\n')
``````

Just as I predicted, the itertools method came in just a little faster, while the lambda method lagged slightly behind losing less than a millisecond off of the overall interpretation time. The answer here is somewhat of a mixed bag because the concise nature of the lambda expression and `map()` function in tandem make the lambda method appear to be a lot more impressive. But the compile time of the itertools method is most certainly impressive because of its speed.

As is often the case in programming, there are multiple ways to do one thing, and as is also often the case, some ways are significantly better than others. There are certainly some trade-offs depending on what methodology you select, but this is what defines your own style as a programmer. I believe regardless  of the decision that is made, using these faster methods will almost certainly make any aspiring programmer look a lot more proficient in their take home assignment. Furthermore, any aspiring programmer could certainly learn a lot more about programming and the language they are programming in by trying out different methods of doing the same thing.