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!

## 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 stringβand 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.

## Python One-Liners Book: Master the Single Line First!

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