The FizzBuzz problem is a common exercise posed in code interviews to test your proficiency in writing simple Python code.

**Problem**: Print all numbers from 1-100 to the shell with three exceptions:

- For each number divisible by three you print
`"Fizz"`

, - For each number divisible by five you print
`"Buzz"`

, and - For each number divisible by three and five you print
`"FizzBuzz"`

.

**Example**: The first 15 numbers of the FizzBuzz sequence are the following.

1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz ...

How to write a Python one-liner that solves this problem?

Here’s an interactive overview:

**Exercise**: Do both one-liners produce the same results? Run the code to check!

Let’s dive into those one-liners to gain a deeper understanding and improve your Python skills!

Table of Contents

## FizzBuzz One-Liner 1: Generator Expression + String Concatenation + Short Circuiting

The following one-liner solves the problem in an elegant way using a fine understanding of more advanced Python features (source).

print('\n'.join('Fizz' * (i%3==0) + 'Buzz' * (i%5==0) or str(i) for i in range(1,101)))

The one-liner creates a string using the `join`

function with the newline character as a delimiter. Here’s a short explanation of the function:

The `string.join(iterable)`

method concatenates all the string elements in the `iterable`

(such as a list, string, or tuple) and returns the result as a new string. The `string`

on which you call it is the delimiter string—and it separates the individual elements. For example, `'-'.join(['hello', 'world'])`

returns the joined string `'hello-world'`

.

So, what’s the iterable, you pass into the `join()`

function? It’s a generator expression of the form: `expression for variable in context`

. You go over all integer values in the context 1 to 100 using the `range()`

function. So, you obtain the remaining `expression for i in range(1, 101)`

. What’s the expression part?

It consists of three elements:

`'Fizz' * (i%3==0)`

— The modulo expression`i%3==0`

returns`True`

only if the integer`i`

is divisible by 3, otherwise it returns`False`

. So, you multiply the string`'Fizz'`

either with`True`

(=1) or with`False`

(=0). As a result, you obtain the empty string`''`

in all cases except if the integer`i`

is divisible by 3—in which case you obtain the string`'Fizz'`

.`'Buzz' * (i%5==0)`

— The modulo expression`i%5==0`

returns`True`

only if the integer`i`

is divisible by 5, otherwise it returns`False`

. So, you multiply the string`'Buzz'`

either with`True`

(=1) or with`False`

(=0). As a result, you obtain the empty string`''`

in all cases except if the integer`i`

is divisible by 5—in which case you obtain the string`'Buzz'`

.- You use string concatenation to glue together the previously obtained strings. In most cases, this will be the empty string. If
`i`

is divisible by 3, you obtain the string`'Fizz'`

. If`i`

is divisible by 5, you obtain the string`'Buzz'`

. And if`i`

is divisible by 3 and 5, you obtain the string`'FizzBuzz'`

. `or str(i)`

— In the case you obtained a non-empty string in`{'Fizz', 'Buzz', 'FizzBuzz'}`

in the previous step, the`or`

operation simply returns this string. This is called short circuiting—and it’s used in many programming languages such as Python to improve efficiency of logical operations.- But if the string is empty, it is interpreted as a logical
`False`

. Thus, Python returns the second operand of the`or`

operation. The second operand simply is the string representation of the integer`i`

.

A very interesting implementation of the FizzBuzz problem indeed!

## FizzBuzz One-Liner 2: Slicing

An alternative is given in the following nice one-liner (source):

for i in range(1, 101): print('FizzBuzz'[i*i%3*4:8--i**4%5] or i)

Wow—what a short and concise one-liner solution! But how does it work?

- You iterate over all values from
`i=1`

to`i=100`

and print a string. So far so good. - You use the
`or`

operation and slicing to determine the string`'FizzBuzz'[start:end] or i`

generates the output. - You use the property of short circuiting in Python: If
`'FizzBuzz'[start:end]`

is empty, the integer`i`

is returned, otherwise, the non-empty string is returned. - You carve out a substring from
`'FizzBuzz'`

using slicing as follows.

Slicing is a concept to carve out a substring from a given string. Use slicing notation `s[start:stop:step]`

to access every `step`

-th element starting from index `start`

(included) and ending in index `stop`

(excluded). All three arguments are optional, so you can skip them to use the default values (`start=0`

, `stop=len(lst)`

, `step=1`

). For example, the expression `s[2:4]`

from string `'hello'`

carves out the slice `'ll'`

and the expression `s[:3:2]`

carves out the slice `'hl'`

.

In the example, you have the slicing operation `'FizzBuzz'[i*i%3*4:8--i**4%5]`

.

`start = i*i%3*4`

— Note that the multiplication`*`

and modulo operation`%`

have the same priority, so they are evaluated from left to right. If integer`i`

is divisible by 3,`i*i`

is also divisible by 3, and the start index is 0. In all other cases, the start index is 4. Thus, the slice either starts with`'Fizz'`

or`'Buzz'`

.`stop = 8--i**4%5`

— This is 4 in all cases except if the number`i`

is divisible by 5, in which case this is 8.

So, there are four cases:

- The number is divisible only by 3:
`start=0`

,`stop=4`

–>`'Fizz'`

- The number is divisible only by 5:
`start=4`

,`stop=8`

–>`'Buzz'`

- The number is divisible by both 3 and 5:
`start=0`

,`stop=8`

–>`'FizzBuzz'`

- The number is divisible by neither 3 nor 5:
`start = 4`

,`stop=4`

–>`''`

Phew! This was a hard nut to crack, wasn’t it?

## Python One-Liner 3: Map + Lambda

You can find detailed tutorials on the map and lambda functions here:

Those two functions can be used to solve the FizzBuzz problem (source):

print(list(map(lambda i: "Fizz"*(i%3==0)+"Buzz"*(i%5==0) or str(i), range(1,101))))

It’s similar to Method 1 and by now you’re able to figure it out. Think of the different values the integer `i`

can take.

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While working as a researcher in distributed systems, Dr. Christian Mayer found his love for teaching computer science students.

To help students reach higher levels of Python success, he founded the programming education website Finxter.com. He’s author of the popular programming book Python One-Liners (NoStarch 2020), coauthor of the Coffee Break Python series of self-published books, computer science enthusiast, freelancer, and owner of one of the top 10 largest Python blogs worldwide.

His passions are writing, reading, and coding. But his greatest passion is to serve aspiring coders through Finxter and help them to boost their skills. You can join his free email academy here.