Python One Line FizzBuzz

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 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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