Fix my FizzBuzz

Question

FizzBuzz is where a range of positive integers is taken, and numbers divisible by 3 are replaced with "Fizz", divisible by 5 with "Buzz" and divisible by 15 with "FizzBuzz".

I've got some FizzBuzz output here. Unfortunately, it's been shuffled:

4
Fizz
Fizz
Buzz
2

A bit of logic shows us that this is 2, Fizz, 4, Buzz, Fizz shuffled.

Your challenge is to take a range of FizzBuzz, shuffled, and put it in any order such that it's valid FizzBuzz. For example, given Fizz, Fizz, Buzz, 11 we can work out that the correct order is Fizz, Buzz, 11, Fizz (corresponding to 9, 10, 11, 12)

Input will always be rearrangeable into valid FizzBuzz.

You may choose any values that cannot appear as integers to represent Fizz, Buzz and FizzBuzz.

This is code-golf, shortest wins!

Testcases

1, 2, Fizz, 4, Buzz -> 1, 2, Fizz, 4, Buzz
Fizz, 4, Buzz, 1, 2 -> 1, 2, Fizz, 4, Buzz
Fizz, Fizz, Buzz, 11 -> Fizz, Buzz, 11, Fizz
Fizz, Buzz, 8, Buzz, 7, Fizz -> Buzz, Fizz, 7, 8, Fizz, Buzz
Buzz, Fizz -> either way
29, 31, FizzBuzz -> 29, FizzBuzz, 31
28, Fizz, FizzBuzz, 29 -> Fizz, 28, 29, FizzBuzz
8, 7 -> 7, 8

Answer 1

Vyxal, 21 15 bytes

ṖÞ∞ƛ₍₃₅T∨;ÞS$↔h

Try it Online!

Why use clever mathematics when you can just generate infinite FizzBuzz?

-6 thanks to Steffan and emanresuA

Explained

ṖÞ∞ƛ₍₃₅T∨;ÞS$↔h
Ṗ                # All permutations of the input
 Þ∞              # An infinite list of positive integers starting at 1
   ƛ     ;       # To each number n:
    ₍₃₅          #   [n % 3 == 0, n % 5 == 0]
       T         #   Truthy indices of that
        ∨        #   Logical or that with the number. This gets either the indices of where Fizz or Buzz would usually be or n
         ÞS      # Sublists of the infinite fizzbuzz
           $↔    # Remove sublists which don't have everything from the input
             h   # Get the first item of that

Answer 2

J, 54 bytes

(-:&(/:~)"1#])#]\1((,~{~0=])2-@#.0=5 3|])@+#i.@+9>.>./

Try it online!

_1, _2, _3 represent Fizz, Buzz, FizzBuzz.

• Generate all integers from 1 to <list length> + max(9, <max value>)
• Fizz Buzz them
• Sliding window across them of size <list len>
• Find ones which set equal the input

Answer 3

JavaScript (ES6), 82 bytes

Saved 2 bytes thanks to a suggestion by @emanresuA

Expects and returns -1 for Fizz, -2 for Buzz, -3 for FizzBuzz.

f=(a,i=k=0)=>[...b=a.sort().map(_=>-!(++i%3)-2*!(i%5)||i)].sort()+''==a?b:f(a,++k)

Try it online!

The underlying algorithm is the same as in my answer to the twin challenge.

Answer 4

Charcoal, 55 bytes

ＷＳ⊞υι⮌⊟ΦＥ⁷ＥＥυ⁻⁺⌈ＥυΣ⁺ψνκμ∨⁺⎇﹪λ³ωFizz⎇﹪λ⁵ωBuzzＩλ⬤ι⁼№ιλ№υλ

Try it online! Link is to verbose version of code. Explanation:

ＷＳ⊞υι

Input the shuffled range.

⌈ＥυΣ⁺ψν

Calculate the largest integer in the range.

Ｅυ⁻⁺...κμ

Calculate a descending range of integers starting with that integer, but also offset by the outer loop index (see below).

Ｅ...∨⁺⎇﹪λ³ωFizz⎇﹪λ⁵ωBuzzＩλ

Calculate the FizzBuzz output for that range.

⮌⊟ΦＥ⁷...⬤ι⁼№ιλ№υλ

Calculate 7 ranges sliding up by 1 each time, and output one which matches the input, reversing it to put it back into order. (7 is needed for the edge case of Buzz Fizz.)

37 bytes for a boring numeric version:

Ｉ⮌⊟ΦＥ⁷ＥＥθ⁻⁺⌈θκμ∨±⁺¬﹪λ³⊗¬﹪λ⁵λ⬤ι⁼№ιλ№θλ

Try it online! Link is to verbose version of code. Uses -1 for Fizz, -2 for Buzz and -3 for Fizzbuzz.

Answer 5

05AB1E, 28 bytes

∞εD35SÖƶ(O‚0Kθ}Ig„üÿ.V.Δ{I{Q

Not too happy about this.. I have the feeling this can be a lot shorter.. :/

Try it online or verify all test cases.

Explanation:

∞               # Push the infinite positive list: [1,2,3,...]
 ε              # Map each value to:
  D             #  Duplicate the current value
   35S          #  Push pair [3,5]
      Ö         #  Check if the value is divisible by 3 and/or 5
       ƶ        #  Multiply the check by their 1-based index
        (       #  Negate them
         O      #  Sum them (-1 if only divisible by 3; -2 if only 5; -3 if both;
                #  0 if neither)
          ‚     #  Pair it with the current value
           0K   #  Remove all 0s
             θ  #  Pop and keep the last value
 }              # Close the map
  Ig            # Push the length of the input-list
    „üÿ         # Push string "ü<length>"
       .V       # Evaluate and execute it as 05AB1E code,
                # `ün` created overlapping lists of size `n`
         .Δ     # Then find the first which is truthy for:
           {    #  If it's sorted
              Q #  is equal to
            I{  #  the sorted input-list

Answer 6

Python 3.8 (pre-release), 149 bytes

lambda a,e=enumerate:{p for p in permutations(a)if all(j in(l:=k+i-n,-(l%3<1)-2*(l%5<1))for k,j in e(p)for n,i in e(p)if i>0)}
from itertools import*

Try it online!

Uses -1 for Fizz, -2 for Buzz and -3 for FizzBuzz.

Returns all possible options.

The idea is to try all permutations and then, for each permutation, find the first integer in the list and check if the rest of the list agrees with that one.

Answer 7

Python3, 276 bytes:

def v(r,l,k=[]):
 if any([r,l])==0:yield k;return
 for i,a in enumerate(l):
  if a==r[0]or(type(a)==str and r[0]%{'Fizz':3,'Buzz':5,'FizzBuzz':15}[a]==0):yield from v(r[1:],l[:i]+l[i+1:],k+[a])
def f(l):
 c=0
 while(c:=c+1):
  if(V:=next(v([*range(c,c+len(l))],l),0)):return V

Try it online!