FizzBuzz is one of the most well-known katas. It originated as a game to teach children about division, and is now sometimes used in programming job interviews as a minimum standard.
Iterating through the numbers between 1 and n:
- If the number divisible by three, print "Fizz"
- If the number is divisible by five, print "Buzz"
- If the number is divisible by both three and five, print "FizzBuzz"
- If the number is not divisible by three or five, print the number
1 2 fizz 4 buzz fizz 7 8 fizz buzz
For our solution, we are going to return a newline-separated string, rather than printing individual values
Attempt 1: iterate
The most simple solution is what interviewers are looking for, as it demonstrates coding ability at the minimum standard.
- Iterate through numbers
1 to n
- For each number, check what it is divisible by, create a string accordingly
- Return all the strings
For our first method, we're going to use some python features (but not many, this approach is almost pseudo-code)
Things to note:
range(0, 10)will print all the numbers from 0-9, but not 10.We need to add a wee
+1to make sure we're iterating through all numbers
In : list(range(1, 10)) Out: [1, 2, 3, 4, 5, 6, 7, 8, 9] In : list(range(1, 10+1)) Out: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
- You can append to a string using
In : string = 'a' In : string += 'bcd' In : string Out: 'abcd'
- You can use a boolean expression to return a 'truthy' value, which I though was pretty cool
In : line = '' In : i = 1 In : (line or i) Out: 1
class Basic: def fizzbuzz(n=100): every =  for i in range(1, n+1): line = '' if i % 3 == 0: line += 'fizz' if i % 5 == 0: line += 'buzz' every.append(line or str(i)) return '\n'.join(every)
Let's see if it works!
In : print(fizzbuzz.Basic.fizzbuzz(10)) 1 2 fizz 4 buzz fizz 7 8 fizz buzz
Noice, we've implemented our basic solution and got a job! But maybe there's a better way?
Attempt 2: Map
This attempt is pretty similar, but it uses
map to execute the same function over every number in the sequence.
class Map: def fizzbuzz(n=100): return '\n'.join(list(map(Map.rep, range(1, n)))) def rep(n): line = '' if n % 3 == 0: line += 'fizz' if n % 5 == 0: line += 'buzz' return line or str(n)
Attempt 3: Lazy/Functional
After a lot of tinkering, I finally made a functional/lazy solution! I'm not going to go into detail around this solution - I'll save that for a standalone post, as it will get quite involved.
from itertools import cycle, count, islice class Lazy: def fizzbuzz(n=100): return '\n'.join( word or str(i) for word, i in Lazy.sequence(n) ) def sequence(n): 'Return a lazy fizzbuzz sequence of the required size' return islice( Lazy.cycle_with_count(Lazy.fb_seq()), n ) def fb_seq(): return map(''.join, zip( cycle(['', '', 'Fizz']), cycle(['', '', '', '', 'Buzz']) )) def cycle_with_count(seq, start=0): return zip(seq, count(start))
Python 3.6.5 |Anaconda, Inc.| (default, Apr 26 2018, 08:42:37) Type 'copyright', 'credits' or 'license' for more information IPython 6.4.0 -- An enhanced Interactive Python. Type '?' for help. In : import fizzbuzz In : %timeit fizzbuzz.Basic.fizzbuzz(100) 32.4 µs ± 1.8 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each) In : %timeit fizzbuzz.Map.fizzbuzz(100) 34.4 µs ± 1.85 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each) In : %timeit fizzbuzz.Lazy.fizzbuzz(100) 34.6 µs ± 1.33 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
The fizzbuzz challenge ended up being more interesting that I initially expected. I haven't encountered another Kata yet where the most basic solution performs neck-and-neck with solutions that require a more in-depth knowledge.
'Til next time