Enklactify these strings

Question

Inspired by this comment chain...

I wanna enklact my way out of this challenge but I can't...

@ETHproductions to enklact (v): to implement a lookup table using a subsection consisting of unique elements.

---

Enklacting is a very useful way to compress a lookup table. For example, let's say you had the following list of colors:

red
green
blue
yellow
purple
orange

If you want to be able to take a color as input and return it's index in this list, there's obviously the straightforward way:

["red", "green", "blue", "yellow", "purple", "orange"].index(input())

But there's a way we could do this in way less bytes:

"rgbypo".index(input()[0])

This works because the first (or 0'th) index of each string is unique. This example is obvious, but sometimes it's a little bit harder. What if we wanted to make a lookup table for this list?

Sweet Onion Chicken Teriyaki
Oven Roasted Chicken
Turkey Breast
Italian BMT
Tuna
Black Forest Ham
Meatball Marinara

In this case, we can't do this:

"SOTITBM".index(input()[0])

because there are two different inputs that start with a 'T', namely "Tuna" and "Turkey". We must look at a different index. If you look at the 4th index of each string, you'll notice that they are all unique. So we can do this...

"enklact".index(input()[3])

In this case, the "enklaction string" is "enklact".

That leads us to today's challenge...

Given a list of strings, return any valid enklaction string. Or in other words, given a list of strings, return any new string where each letter is unique, and the string is formed by joining the i'th letter of each string.

If there is no valid enklaction string, your submission must return an empty string or a consistent falsy value instead. As usual, either functions or full programs are allowed, and the input/output formats are permissive (within reason).

Each string will only contain printable ASCII, and this challenge is case sensitive.

This is code-golf, so try to write the shortest program possible in your language of choice!

Test cases

Input:
Programming
Puzzles
Code
Golf

Output (any one of these):
"ozdl"
"gzef"


Input:
the quick
brown fox
jumped over
lazy dogs

Output:
"tbjl"
"hrua"
"eomz"
" wpy"
"qne "
"if o"
"kxvs"

Note that "u dd" and "coog" are not valid.


Input:
Sunday
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday

Output:
""


Input:
AbC
aBc
bbC

Output:
"Aab"


Input:
@#$%^_
Hello_World
How are you?

Output:
"#eo"
"$lw"
"%l "
"^oa"


Input:
a
ab
ac

Output:
""

Answer 1

Python 2, 68 67 61 bytes

lambda a:`{0:j for j in zip(*a)if len(set(j))==len(j)}`[6::5]

Try it online!

Improvements

• From 68 bytes to 67 bytes by Jo King
• From 66 to 65 bytes by Lynn

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If output would not have to be a string:

Python 3, 49 bytes

lambda a:[j for j in zip(*a)if len({*j})==len(j)]

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Answer 2

Python3, 59 bytes

lambda l:{len({*d}):''.join(d)for d in zip(*l)}.get(len(l))

Returns a string with the enklact, None otherwise

Answer 3

Retina, 43 32 bytes

+/^(.).+^\1|^$/ms&m`^.

|""Lm`^.

Try it online! Edit: Saved 11 bytes thanks to @MartinEnder. Explanation:

+

Repeat while the input changes...

/^(.).+^\1|^$/ms&

...only if one line is empty or two lines begin with the same character...

m`^.

...delete the first character of every line. The repeat therefore stops if either a) all the lines begin with different characters, in which case the condition fails and the input does not get changed, or b) at least one line becomes empty, in which case all the characters eventually get deleted, at which point the input stops changing.

|""L`^.

Collect the first character of every line. (If there was no solution, the above loop will have deleted everything and there will be nothing to collect.)

Answer 4

Haskell, 71 bytes

f x|elem""x=""|y<-head<$>x,and[filter(==a)y==[a]|a<-y]=y|1<2=f$tail<$>x

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BMO saved 3 bytes with any null x → elem""x.

Ørjan Johansen saved a byte with sum[1|b<-y,a==b]<2 → filter(==a)[y]==[a].

Explanation

f x|elem""x=""                      -- Once any of the strings is empty, return "".
   |y<-head<$>x                     -- Otherwise, let y be all the first letters...
   ,and[                 |a<-y]     -- If, for all a in y,
        filter(==a)y==[a]           -- a occurs just once in y:
                               =y   -- return y.
   |1<2=f$tail<$>x                  -- Else, chop off all the first letters and recurse.

If throwing an error (Prelude.head: empty list) when there's no solution is OK, |elem""x="" can be erased for 61 bytes.

Answer 5

Pyth, 6 bytes

>1{I#C

Test suite.

The output is a singleton list, as allowed by default; the list [ ] (empty list, falsy) is returned in case the string cannot be enklactified.

Explanation

>1{I#C – Full program.
     C – Transpose the input, cropping absences.
    #  – Filter by:
  {I   – Invariant under deduplication.
>1     – Slice to 1. list[:1] in Python.

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Pyth, 5 bytes

This would've been valid if crashing would have counted as a falsy value.

h{I#C

Test suite.

Answer 6

Ruby, 38 bytes

->x,*y{x.zip(*y).find{|z|z==z-[p]|[]}}

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Thanks to G B for pointing out a bug.

Answer 7

Haskell, 76 74 bytes

f t=last$"":(id=<<foldr(zipWith(#))([[]]<$t)t)
x#[s]=[x:s|all(/=x)s]
x#e=e

Try it online! Returns the last valid lookup string or an empty string if no such string exists.

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71 69 bytes

If throwing a consistent exception as falsy value is allowed:

f t=head$id=<<foldr(zipWith(#))([[]]<$t)t
x#[s]=[x:s|all(/=x)s]
x#e=e

Try it online! Throws an empty list exception if no string is found, returns the first valid string otherwise.

-2 bytes thanks to Ørjan Johansen

Answer 8

Jelly, 7 bytes

z0Q€fZḢ

Returns the integer 0 if the strings cannot be enklactified.

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How it works

z0Q€fZḢ  Main link. Argument: A (string array)

z0       Zip/transpose, filling shorter rows with 0.
  Q€     Unique each deduplicate resulting string.
     Z   Zip/transpose, without using a filler.
    f    Filter; keep only string that appear in the results to both sides.
      Ḣ  Head; extract the first string. Returns 0 if the array is empty.

Answer 9

Perl 5, 61 bytes

map$;[$_].=$F[$_],0..@F;$-++}{($_)=grep$-==y///c*!/(.).*\1/,@

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Answer 10

Japt, 9 bytes

Takes input as an array of character arrays, returns an array of characters or undefined

y æ_f eZâ

Try it (Add a new line at the start of the programme with the code qR mq to take input as a newline seperated string, to save you the hassle of creating the arrays.)

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Explanation

y             :Transpose
  æ_          :Pass each array Z through a function and return the first that returns true
    f         :  Filter nulls (used for padding when transposing)
      e       :  Test for equality with
       Zâ     :  Z deduplicated

Answer 11

Stax, 9 8 bytes

åτIⁿs↓µg

Run and debug it

Explanation (unpacked):

M{c0-u=}j Full program, implicit input
          e.g. ["Programming", "Puzzles", "Code", "Golf"]
M         Transpose
                ["PPCG", "ruoo", "ozdl", "gzef", "rl\0\0", "ae\0\0", "ms\0\0", "m\0\0\0", "i\0\0\0", "n\0\0\0", "g\0\0\0"]
 {     }j Find first matching element:
            e.g. "PPCG"
  c0-       Copy and remove zero bytes (padding)
                 "PPCG" "PPCG"
     u      Unique
                 "PPCG" "PCG"
      =     Check if equal:
                 1
          First matching here: "ozdl". If none is found, the stack is empty
          Implicit output if anything on stack

Answer 12

R, 127 bytes

function(S,s=sapply(S,substring,x<-1:max(nchar(S)+1),x))cat(rbind(s[!apply(s,1,anyDuplicated)&!rowSums(s==""),],"")[1,],sep="")

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sapply typically returns a matrix when all the length(FUN(X[[i]])) are equal, except when length(FUN(X[[i]]))==1, in which case it returns a vector. In order to use the matrix operations, then, we have to substring one further than we need to guarantee a matrix, which is why x extends to max(nchar(S)+1).

We then filter for rows that have no duplicates and no empty strings. Since we can only return a single string, we take the first one, except that when there are no rows that meet the criteria, we would throw an error, so we append an extra row of "" to the end.

Then we print out the enklactified string or the empty string.

Answer 13

R, 116 107 95 bytes

R+pryr

pryr::f(for(i in 1:min(nchar(v)))`if`(anyDuplicated(t<-substr(v,i,i)),0,{cat(t,sep="")
break}))

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base R

function(v)for(i in 1:min(nchar(v)))`if`(anyDuplicated(t<-substr(v,i,i)),0,{cat(t,sep="")
v=0})

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These two variants saving 9 bytes thanks to Giuseppe.

Explanation:

This effectively truncates all stings in the vector v to the length of the shortest and iterates through subsequent indexes. Then checks whether there are any duplicates within chosen letters and if not, pastes them together and prints with cat. If all indexes return duplicated results, this prints empty string.
All wraped in an anonymous pryr function with break to halt the loop, or base R function zeroing the vector to break the loop.

Answer 14

05AB1E, 7 bytes

øʒDÙQ}н

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Explanation

ø        # Zip the input array
 ʒ   }   # Filter by ...
  DÙQ    # ... an entry is equal to itself deduplicated
      н  # Take the first element

Answer 15

Python 3, 75 bytes

def f(t):c=[s.pop(0)for s in t];return all(t)and(f(t),c)[len(t)==len({*c})]

Operates on character lists instead of strings. Returns False if no valid enklaction string exists.

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Answer 16

Wolfram Language (Mathematica), 54 bytes

#&@@Select[PadRight@#,#~FreeQ~0&&Union@#==Sort@#&]&

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Takes a list of list of characters as input, returns a list of characters. Contains U+F3C7, corresponding to "Transpose" operator.

Returns #1 and throws a bunch of ignorable errors when there is no suitable string.

Explanation:

PadRight@#

Pad the input so that each "string" (list of characters) is the same length. This appends integer 0s (not string "0"s). Then transpose.

Select[ ... ,#~FreeQ~0&&Union@#==Sort@#&]

Select the strings that have no integer 0s in them and have all unique characters.

#&@@

Get the first one.

Answer 17

Charcoal, 23 21 bytes

-2 bytes thanks to @Neil!

§ΦＥ⌊ＥθＬι⭆θ§λι⬤ι⁼¹№ιλ⁰

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Answer 18

C (gcc), 161 bytes

f(s,i)char**s;{char**t,a[255],*u=a;for(i=0;memset(a,0,255),u&&~i;i+=!!~i&&u)for(t=s;(u=u?*t++:0)&&~(i=u[i]?i:-1)&&!a[u[i]]++;);while(~i&&(u=*s++))putchar(u[i]);}

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Each character position is tested for duplication and skipped if a duplicate is detected; this is continued until the shortest string ends. It's ASCII only, sadly: DBCS/UTF-8 strings break this function badly!

Answer 19

Japt, 12 bytes

Returns undefined for non-enklactic strings.

y ·æ_¬n ä¦ e

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Explanation:

y ·æ_¬n ä¦ e
y             // Split the input at newlines and transpose
  ·           // Join on newlines 
   æ_         // Return the first item that returns truthy when ran through:
     ¬n       //   Sort
        ä¦    //   Reduce with !=
           e  //   All items are truthy (unique)

Answer 20

JavaScript (ES6), 66 bytes

Returns either a string or undefined if no solution exists.

f=(a,i=0)=>a.every(s=>(o[k+=c=s[i],c]^=1)&&c,o=k=[])?k:c&&f(a,i+1)

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Commented

f = (           // f = recursive function taking:
  a,            //   a[] = input array
  i = 0         //   i   = pointer
) =>            //
  a.every(s =>  // for each string s in a[]:
    (o[k +=     //   append to the key string k
      c = s[i], //   the character c at position i in s
      c] ^= 1   //   and toggle o[c] (undefined -> 1 -> 0)
    ) && c,     //   additionally, make sure that c is defined
    o = k = []  //   start with o = k = empty array
  ) ?           // every() is true if all characters were distinct and defined:
    k           //   in which case we return k
  :             // else:
    c &&        //   provided that every() didn't fail because of an undefined character,
    f(a, i + 1) //   try again at the next position

Answer 21

Husk, 9 bytes

ḟS=UḞz:∞ø

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Explanation

fS=UḞz:∞ø
    Ḟz:∞ø  Transpose the input dropping characters of longer strings
    Ḟ        Fold right
     z:      Zip with prepend
       ∞ø    Infinite list of empty lists
ḟS=U       Find the first string without duplicates, returns an empty string if none
ḟ            Return first value satisfying predicate
  =          Equal
 S U         to itself with duplicates removed

Answer 22

Perl 6, 27 bytes

{([Z] $_).first:{.Set==$_}}

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After seeing that the ruby version has been upvoted, I have copied the approach and used character lists, instead of strings. I don't like it.

My older, and more correct submission is as below:

Perl 6, 38 bytes

Split, zip, check uniqueness, join.

{[~] ([Z] @_>>.comb).first:{.Set==$_}}

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Answer 23

Retina, 81 56 bytes

m`$
$.=*£¢
N$`.
$.%`
¶

~`(.*?¢)+
L`.{$#1}
A`(.).*\1|£|¢

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-25 bytes thanks to @Neil

---

Transposing rectangular text in Retina is surprisingly difficult.

Answer 24

C (gcc), 121 113 110 bytes

i;d;f(s)char**s;{char**_=s,x[255]={0},y[99]={0};for(d=i=0;*_;)d+=x[y[i++]=*(*_++)++]++;d=d?*x?0:f(s):puts(y);}

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Ungolfed

void enklactify(char *strings[]) {
    int quit = 0;
    while (!quit) {
        char **arg = strings;      // current row
        int exists[255] = {0};     // which characters exist in the column
        char output[99] = {0};     // will hold output string
        int index = 0;             // where to insert in output
        int duplicates = 0;        // have we found any duplicates?
        while (*arg != NULL) {
            char *word = *arg;     // current word
            char first = *word;    // first letter of current word
            if (exists[first])
                duplicates = 1;    // we found a duplicate
            exists[first] = 1;     // mark it as in our string
            output[index] = first; // add it to our output
            index++;
            (*arg)++;              // advances the current word to the next
                                   // character by reference (chops first char)
            arg++;                 // advance to next whole argument
        }

        if (!duplicates) {         // This is a valid solution
            puts(output);
            quit = 1;
        }

        if (exists[0]) {           // We hit the null terminator of one of the
                                   // input strings, so we failed
            quit = 1;
        }
    }
}

-3 thanks to ceilingcat

This abuses the ability to just add 1 to a string pointer in C to get the 'tail' of the string. The main byte savers are:

• d+=x[y[i++]=*(*_)++]++ which appends the first character of the first string of _ to y, advances the first string of _ to remove its first character, adds the x entry at that character to d, and increments said x entry
• q=d?*x:puts(y) which prints y if d is non-zero while setting q to a non-zero value, or sets q to non-zero if the first element of x is non-zero (if we were at the end of one of the strings then said element would be non-zero)

Edit: Shaved bytes by switching from while loop to recursive tail call and removing brackets from for loop.

Answer 25

Pyth, 13 bytes

e+kf{IT@.TQ.t

Try it here

e+kf{IT@.TQ.t
           .tQ   Transpose the (implicit) input with padding.
        .TQ      Transpose the input without padding.
       @         Take the strings in both.
   f{IT          Find the ones that have no duplicates.
e+k              Get the last, or an empty string.

Answer 26

Red, 139 bytes

func[b][m: length? first b foreach a b[m: min m length? a]repeat n m[c: copy[]foreach a b[append c a/(n)]if c = unique c[return rejoin c]]]

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Explanation:

Takes the input as a block (list) of strings. Returns the enklaction string or none otherwise.

f: func[b][
    m: length? first b                   ; the minimal length of the first string  
    foreach a b[m: min m length? a]      ; find the minimal length of all strings
    repeat n m[                          ; limit the search to the minimal length
        c: copy[]                        ; an empty block  
        foreach a b[append c a/(n)]      ; for each string append the current char
        if c = unique c[return rejoin c] ; if all chars are unique, return the block
    ]  
]

Answer 27

APL+WIN, 35 33 bytes

2 bytes saved thanks to Adám

Prompts for the lines of text as a character matrix:

⊃((↑⍴¨a)=+/¨((a⍳¨a)=⍳¨⍴¨a))/a←,⌿⎕

Try it online! Courtesy of Dyalog Classic

Explanation:

a←,⌿⎕ prompts for input and creates a nested vector of the input matrix columns

((a⍳¨a)=⍳¨⍴¨a) creates a binary vector for each nested element with a 1 for each unique element

((↑⍴¨a)=+/¨ sums each binary vector and compares to number of characters in each element

(...)/a←⊂[1]⎕ selects only those elements where number of uniques = column length

⊃ converts nested vector back to a matrix of each valid enklaction string 

Answer 28

Röda, 80 77 bytes

f a{a|seq 0,#_|try{{|i|a|[_[i:i+1]]|orderedUniq|concat|[_]if[#_1=#a]}_|head}}

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-1 byte thanks to Cows quack

Explanation:

f a{
  a|         /* Push the strings in a to the stream */
             /* For each string (_): */
  seq 0,#_|     /* Push a range from 0 to the length of _ to the stream */
  try{       /* Ignore errors during the following block */
    {|i|        /* For each i in the stream: */
      a|           /* Push strings in a to the stream */
      [_[i:i+1]]|  /* For each string, push the ith character to the stream */
      orderedUniq| /* Remove duplicate characters */
      concat|      /* Join the characters into a string */
      [_]if        /* Push the string to the stream if */
      [#_1=#a]     /* Its length is the length of a */
    }_|
    head        /* Take the first string in the stream and return it */
  }
}

The try keyword is used to dismiss errors that occur if i is larger than the length of the smallest string in a, or there is no answer and head causes an error.

Answer 29

Java 10, 106 bytes

a->{for(int i=0;;i++){var r="";for(var s:a)r+=s[i];if(r.length()==r.chars().distinct().count())return r;}}

Will give an error instead of returning an empty String if no solution can be found. Input is a character-matrix.

Try it online.

Explanation:

a->{                  // Method with character-matrix parameter and String return-type
  for(int i=0;;i++){  //  Loop `i` upwards
    var r="";         //   Result-String, starting empty
    for(var s:a)      //   Loop over the character-arrays of the input
      r+=s[i];        //    And append every `i`'th character to `r`
    if(r.length()==r.chars().distinct().count())
                      //   If `r` only contains unique characters
      return r;}}     //    Return `r` as result

Answer 30

Clojure, 59 bytes

#(for[s(apply map list %):when(=(count(set s))(count %))]s)

Returns a list of lists of characters.

Answer 31

Scala, 97 bytes

l=>l.flatMap(h=>(0 to l.map(_.size).min-1).find(i=>l.map(_(i)).toSet.size==l.size)map(h))mkString

input is List[String] and returns an empty String if no solution found.

Try it online.

Explanation:

l                                    // List("Programming", "Puzzles", "Code", "Golf")
  .flatMap(                          //
    h =>                             // "Programming"
      (0 to l.map(_.size).min-1)     // Range(0, 3) range from 0 to size of smallest word ("Code".size - 1 = 3)
        .find(                       // returns an Option[Int]: if for a given index, all words have a different char, return Some(i)
          i =>                       // index in words, let's say i = 2
            l.map(_(i)).toSet.size   // Set('o', 'z', 'd', 'l').size = 4; checks the nbr of unique chars for the given index
              == l.size              // and compare it to the nbr of words (4)
        )                            // 
        map(h)                      // Option(i).map(i => h(i)) which gives Some(2).map(h) = Some('o') and for another index: None.map(h) = None; this is why the begining is l.flatMap and not l.map.
  )mkString                         // joins the List[Char] into a String

Answer 32

C# (.NET Core), 167 bytes (149 + 18 for using System.Linq;)

p=>{int l=p.Count(),m=p.Min(s=>s.Length);for(int i=-1;i++<m;){var d=p.Select(s=>s[i]).Distinct();if(d.Count()==l)return string.Join("",d);}return"";}

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