Generate all Possibilities of Words

Question

This is not the same as m-ss-ng-lette-s

Challenge:

Given a string with masked characters, generate all possible words, inclusive of non-meaningfull words.

Rules:

• The sequence of the characters in the string does not change.
• The masked character (*) will be one of an English alphabets [a-z].
• There are no line breaks, space, numbers, special characters or punctuations.
• The output can be a list of non-meaningfull words too.
• Any characters in a string can be masked. Not necessarily alternating characters.
• For memory constraint, lets keep the maximum length of input as 10.

Example 1:

Input:

*o*

Output:

aoa
aob
aoc
...
...
...
boa
bob
boc
...
...
...
zoa
zob
zoc
...
...
...
zoz

Example 2:

Input:

*e*c*m*

Output:

aeacama
aeacamb
...
...
...
welcome
...
...
zezczmz

This is code-golf, so shortest answer in bytes wins!

Answer 1

Haskell, 31 bytes

mapM q
q '*'=['a'..'z']
q c=[c]

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Answer 2

Zsh, 25 bytes

eval echo ${1//\*/{a..z}}

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Answer 3

Ruby, 39 bytes

->s{(?a..?z*10).grep /^#{s.tr'*',?.}$/}

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The range is set to terminate after 10 zs, to fit the constraint. Realistically this takes forever to run unless a smaller number is used.

Answer 4

Vyxal j, 5 bytes

×kaVΠ

Try it Online! Input as char lists.

   V  # Replace
×     # Asterisks
 ka   # With the lowercase alphabet
    Π # Take cartesian product

Answer 5

Perl 5 + -n -M5.10.0, 37 bytes

1 byte saved thanks to @Sisyphus!

$"=",";s/\*/{@{[a..z]}}/g;say<"$_\n">

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Explanation

Turns the input into a glob (<"$_\n">) which is then evaluated as a list and printed (say) with newline separators.

Note: This does struggle to produce all the output for multiple *s when running on my phone!

Answer 6

Python 3, 88 86 bytes

def A(x):p=x.find("*");~p and[A(x[:p]+chr(i+97)+x[p+1:])for i in range(26)]or print(x)

Removed 2 bytes thanks to @Adam

Answer 7

Vyxal j, 14 13 bytes

×Oka↔ƛ?×\%V$%

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Could be 8 bytes by using % instead of *

Answer 8

Vyxal j, 9 bytes

×Ẇ⌈yvkaYΠ

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Based on emanresu A's answer, but shaves off another byte.

Answer 9

R, 65 bytes

f=\(s)`if`(grepl("[*]",s[1]),f(sapply(letters,sub,pa="[*]",s)),s)

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Basically the same recursive idea as in Cong Chen's answer.

Answer 10

Julia, 61 bytes

!s=join.(Iterators.product((x>'*' ? x : 'a':'z' for x=s)...))

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(This returns a matrix of all possible words, rather than a list/vector.)

Answer 11

Retina, 30 bytes

/\*/{%`$
25*$(¶$`
,Y0`*`l`\*.*

Try it online! Explanation:

/\*/{

Repeat while there are still any *s to process.

%`$
25*$(¶$`

Add 25 copies of each line.

,Y0`*`l`\*.*

Cyclically transliterate the first * on each line to the lowercase alphabet. The , tells Retina to process all matches and the 0 tells Retina to process the first character of each match.

Answer 12

lin, 49 bytes

"\*".?g.+_.+_"%s"?s.#t ?i len $a.~ `/\".t.~ sf"`'

Try it here! Returns an iterator of strings.

For testing purposes:

"*e*c*m*" ; \outln `' `_
"\*".?g.+_.+_"%s"?s.#t ?i len $a.~ `/\".t.~ sf"`'

Explanation

Prettified code:

"\*".?g.+_.+_ "%s"?s.#t ?i len $a.~ `/\ (.t.~ sf ) `'

Not very short, but surprisingly straightforward.

• "\*".?g regex /\*/g
• .+_.+_ "%s"?s.#t replace * with %s and store as t
• ?i len get number of *s as n
• $a.~ `/\ create length-n "digit" sequences from alphabet
• (.t.~ sf ) `' use t to sprintf each sequence

Answer 13

sh + coreutils + hashcat, 31 bytes

sed s/*/?l/g|xargs hashcat -a 3

Answer 14

05AB1E, 9 bytes

'*A:.»â€S

Input as a list of characters.

Try it online or verify both test cases.

Explanation:

'*A:      '# Replace all "*" in the (implicit) input-list with the lowercase alphabet
    .»     # Left-reduce the list of strings by:
      â    #  Taking the cartesian product
       €S  # Then convert each inner list to a flattened list of characters
           # (after which the result is output implicitly)

Minor note: '* could be W in the legacy version of 05AB1E if the input is guaranteed to always contain a "*", but unfortunately the €S should be €˜J in that case, not saving any bytes.

Answer 15

Charcoal, 25 bytes

⊞υＳＦυ¿№ι*Ｆβ⊞υ⭆ι⎇⁻μ⌕ι*λκ⟦ι

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

⊞υＳ

Start with the input string.

Ｆυ

Check each string.

¿№ι*

If it contains a *, then:

Ｆβ

Loop over the lowercase alphabet.

⊞υ⭆ι⎇⁻μ⌕ι*λκ

Replace the first * in the string with the current letter and push it to the search list.

⟦ι

Otherwise output the string on its own line.

23 bytes using the newer version of Charcoal on ATO:

⊞υωＦＳ≡ι*≔ΣＥυＥβ⁺κμυ≧⁺ιυυ

Attempt This Online! Link is to verbose version of code. Explanation:

⊞υω

Start with the empty string.

ＦＳ

Loop over the characters of the input string.

≡ι*

If the current character is a *, then...

≔ΣＥυＥβ⁺κμυ

... append each lowercase letter to each of the strings so far and join the resulting lists, otherwise...

≧⁺ιυ

... append the current character to each of the strings so far.

υ

Output the final list of strings.

Answer 16

R, 91 bytes

\(a,j=Vectorize(\(x,y)sub('*',y,x,,,T),'y'))`if`(grepl('*',a,,,T),sapply(j(a,letters),f),a)

Recursive solution.

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Answer 17

Factor + spelling, 60 56 bytes

[ 1 group { "*"} ${ ALPHABET } replace [ ] product-map ]

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                              ! "*o*"
1 group                       ! { "*" "o" "*" }
{ "*"} ${ ALPHABET } replace  ! { "abcdefghijklmnopqrstuvwxyz" "o" "abcdefghijklmnopqrstuvwxyz" }
[ ] product-map               ! { "aoa" "boa" ... "zoz" }

Answer 18

Brachylog, 13 bytes

ẹ{"*"∧!Ạ∋|}ᵐc

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This is a generator predicate that will unify its output with each possible combination. The TIO example uses ᶠ - findall to generate all answers, but in a Prolog REPL you could just press ; repeatedly to get each answer.

Explanation

ẹ                Split the string into a list of characters
 {         }ᵐ    Map for each char:
  "*"              If the char is "*"
     ∧!              Cut (i.e. discard the "else" possibility)
        Ạ∋           The output is an element of the alphabet
          |        Else don’t modify the char
             c   Concatenate back to a string

The cut ! is necessary, otherwise the predicate would also generate answers with the asterisk unchanged.

Answer 19

JavaScript, 111 bytes

I feel like it's way too long and I'm missing something big.

s=>(r=s=>s[1]?.at?[...'abcdefghijklmnopqrstuvwxyz'].flatMap(c=>r(s.slice(1)).map(p=>s[0]+c+p)):s)(s.split('*'))

Answer 20

Also Python 3, 86 Bytes:

def f(x):
 for i in range(26):y=x.replace('*',chr(i+97),1);f(y)if'*'in y else print(y)

Answer 21

Julia, 61 bytes

!s=foldl((x,y)->[x.*y...],[x<'a' ? 'a'.+(0:25)' : x for x=s])

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