17
\$\begingroup\$

Challenge:

Given a positive integer, output the longest single-digit subsequence that occurs at least twice, AND has boundaries of another digit (or the start/end of the integer).

An example:

Input: 7888885466662716666
The longest subsequence of a single digit would be 88888 (7[88888]5466662716666) with a length of 5. However, this subsequence only occurs once in the integer.
Instead, the result for input 7888885466662716666 should be 6666 (78888854[6666]271[6666]), since it occurs (at least) twice.

Challenge rules:

  • Length of the subsequences takes priority over the amount of times it occurs. (I.e. with input 8888858888866656665666, we output 88888 ([88888]5[88888]66656665666; length 5, occurs twice), and not 666 (88888588888[666]5[666]5[666]; length 3, occurs thrice).
  • If the length of multiple subsequences are equal, we output the one with the largest occurrence-count. I.e. with input 3331113331119111, we output 111 (333[111]333[111]9[111]; length 3, occurs thrice), and not 333 ([333]111[333]1119111; length 3 as well, but occurs twice)
  • If the occurrence-count and length of multiple subsequences are equal, you can output either of them, or all (in any order). I.e. with input 777333777333, the possible outputs are: 777; 333; [777, 333]; or [333, 777].
  • The subsequence must have boundaries of other digits (or the start/end of the integer). I.e. with input 122222233433 the result is 33 (1222222[33]4[33]; length 2, occurs twice) and not 222 (1[222][222]33433, length 3, occurs twice with both invalid).
    • This applies to all numbers that are counted towards the occurrence-counter. I.e. with input 811774177781382 the result is 8 ([8]117741777[8]13[8]2; length 1, occurs thrice) and not 77 (811[77]41[77]781382 / 811[77]417[77]81382; length 2, occurs twice with one invalid) nor 1 (8[1][1]774[1]7778[1]382; length 1, occurs four times with two invalid).
  • You can assume the input won't contain any digits 0 (it will match [1-9]+). (This is to avoid having to deal with test cases like 10002000 that should output 000, where most languages would output 0 by default.)
  • You can assume the input will always contain at least one valid output.
  • I/O are both flexible. Can be a list/array/stream of digits/bytes/characters or as string instead of a single integer.

General rules:

  • This is , so shortest answer in bytes wins.
    Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
  • Standard rules apply for your answer, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
  • Default Loopholes are forbidden.
  • If possible, please add a link with a test for your code.
  • Also, adding an explanation for your answer is highly recommended.

Test cases:

Input:  7888885466662716666 / [7,8,8,8,8,8,5,4,6,6,6,6,2,7,1,6,6,6,6]
Output: 6666                / [6,6,6,6]

Input:  3331113331119111 / [3,3,3,1,1,1,3,3,3,1,1,1,9,1,1,1]
Output: 111              / [1,1,1]

Input:            777333777333                   / [7,7,7,3,3,3,7,7,7,3,3,3]
Possible outputs: 777; 333; [777,333]; [333;777] / [7,7,7]; [3,3,3]; [[7,7,7],[3,3,3]]; [[3,3,3],[7,7,7]]

Input:  122222233433 / [1,2,2,2,2,2,2,3,3,4,3,3]
Output: 33           / [3,3]

Input:  811774177781382 / [8,1,1,7,7,4,1,7,7,7,8,1,3,8,2] 
Output: 8               / [8]

Input:  555153333551 / [5,5,5,1,5,3,3,3,3,5,5,1] 
Output: 1            / [1]

Input:            12321              / [1,2,3,2,1]
Possible outputs: 1; 2; [1,2]; [2,1] / [1]; [2]; [[1],[2]]; [[2],[1]]

Input:  944949949494999494 / [9,4,4,9,4,9,9,4,9,4,9,4,9,9,9,4,9,4]
Output: 4                  / [4]

Input:  8888858888866656665666 / [8,8,8,8,8,5,8,8,8,8,8,6,6,6,5,6,6,6,5,6,6,6]
Output: 88888                  / [8,8,8,8,8]

Input:  1112221112221111               / [1,1,1,2,2,2,1,1,1,2,2,2,1,1,1,1]
Output: 111; 222; [111,222]; [222,111] / [1,1,1]; [2,2,2]; [[1,1,1],[2,2,2]]; [[2,2,2],[1,1,1]]

Input:  911133111339339339339339 / [9,1,1,1,3,3,1,1,1,3,3,9,3,3,9,3,3,9,3,3,9,3,3,9]
Output: 111                      / [1,1,1]
\$\endgroup\$
  • 1
    \$\begingroup\$ Suggested test case: 8888858888866656665666. If I interpreted the challenge correctly, both the Brachylog and 05AB1E solutions fail. \$\endgroup\$ – Mr. Xcoder Sep 17 '18 at 8:23
  • \$\begingroup\$ @Mr.Xcoder Added, thanks. \$\endgroup\$ – Kevin Cruijssen Sep 17 '18 at 8:27
  • \$\begingroup\$ @Arnauld Hmm, it would be one of the winners anyway in my opinion because it occurs as many times as 222 when bounded by other integers. I guess we just shouldn't count the occurrence that is a substring of 1111. Better wait for the OP though, indeed. \$\endgroup\$ – Mr. Xcoder Sep 17 '18 at 8:38
  • 2
    \$\begingroup\$ @Arnauld For 1112221112221111 these are the subsequences and their counts: 1111 (1), 111 (2), 222 (2). Since we only outputs sequences occurring at least twice, the output can be one of: 111, 222, [111,222], [222,111]. (See the fourth rule for some more information.) Basically 1111 will only ever count as 1111, and not as 1 and 111 or 11 and 11. I'll add your test case, but the output is either or both of 111 and 222. \$\endgroup\$ – Kevin Cruijssen Sep 17 '18 at 8:55

18 Answers 18

6
\$\begingroup\$

05AB1E, 14 bytes

γТ1›ÏD€gZQÏ.M

Try it online!

Explanation

γ                # group consecutive equal elements
 Т              # count the occurrence of each group among the list of groups
   1›Ï           # keep only groups with a count greater than 1
      D€gZQÏ     # keep only those with a length equal to the greatest length
            .M   # get the most common item
\$\endgroup\$
  • \$\begingroup\$ @Riley: Unfortunately that would get the first element which is not necessarily the most common one. \$\endgroup\$ – Emigna Sep 18 '18 at 18:54
  • \$\begingroup\$ Oops.. I missed that bullet. \$\endgroup\$ – Riley Sep 18 '18 at 18:56
5
\$\begingroup\$

Jelly, 12 bytes

Œgœ-Q$LÐṀÆṃ'

Try it online!

Previous version – 14 bytes

ŒgŒQ¬TịƲLÐṀÆṃ'

Try it online!

How it works?

Œgœ-Q$LÐṀÆṃ' – Full program. Receives a list of digits as input.
Œg           – Group equal adjacent values.
  œ-Q$       – Multiset difference with itself deduplicate.
      LÐṀ    – Keep those that are maximal by length.
         Æṃ' – Mode. Returns the most common element(s).
-------------------------------------------------------------------------
ŒgŒQ¬TịƲLÐṀÆṃ' – Full program. Receives a list of digits as input.
Œg             – Group equal adjacent values.
  ŒQ           – Distinct sieve. Replace the first occurrences of each value by 1.
                 and the rest by 0. [1,2,3,2,3,2,5]ŒQ -> [1,1,1,0,0,0,1]       
    ¬T         – Negate and find the truthy indices.
      ịƲ       – Then index in the initial list of groups.
               – This discards the groups that only occur once.
        LÐṀ    – Find all those which are maximal by length.
           Æṃ' – And take the mode.
\$\endgroup\$
5
\$\begingroup\$

JavaScript (ES6), 79 73 68 bytes

Takes input as a string. Returns an integer.

s=>[...s,r=q=0].map(o=d=>q=s^d?o[!o[q]|r[q.length]?q:r=q]=s=d:q+d)|r

Try it online!

Commented

s =>                      // s = input string, also used as the current digit
  [ ...s,                 // split s into a list of digit characters
    r =                   // r is the final result
    q =                   // q is the current digit sequence
    0                     // append a final dummy entry to force the processing of the last
  ]                       // sequence
  .map(o =                // o is an object used to keep track of encountered sequences
       d =>               // for each digit d in the array defined above:
    q =                   //   update q:
      s ^ d ?             //     if d is not equal to the current digit:
        o[                //       this statement will ultimately update o[q]
          !o[q] |         //         if q has not been previously seen
          r[q.length] ?   //         or the best result is longer than q:
            q             //           leave r unchanged
          :               //         else:
            r = q         //           set r to q
        ] = s = d         //       reset q to d, set the current digit to d
                          //       and mark q as encountered by setting o[q]
      :                   //     else:
        q + d             //       append d to q
  ) | r                   // end of map(); return r, coerced to an integer
\$\endgroup\$
  • \$\begingroup\$ Maybe I'm saying something incorrect here, but since ...s converts the input to a list of digit characters, isn't it shorter to just take the input as a list of digit characters to begin with, instead of a string? I've allowed flexible I/O. (But I'm assuming it interferes with another part of your code?) \$\endgroup\$ – Kevin Cruijssen Sep 17 '18 at 10:54
  • 2
    \$\begingroup\$ @KevinCruijssen The problem is that I need an extra iteration to process the last sequence. So I'd need to do [...s,0] even if s already is a list. \$\endgroup\$ – Arnauld Sep 17 '18 at 10:57
4
\$\begingroup\$

Retina, 56 bytes

L`(.)\1*
O`
L$m`^(.+)(¶\1)+$
$#2;$1
N`
.+;

N$`
$.&
-1G`

Try it online! Link includes test cases. Explanation:

L`(.)\1*

List all the maximally repeated digit subsequences.

O`

Sort the list into order.

L$m`^(.+)(¶\1)+$
$#2;$1

List all the multiple subsequences with their "count".

N`

Sort in ascending order of count.

.+;

Delete the counts.

N$`
$.&

Sort in ascending order of length. (Where lengths are equal, the previous order due to count is preserved.)

-1G`

Keep the last i.e. longest value.

\$\endgroup\$
4
\$\begingroup\$

R, 102 bytes

function(i)rep(names(sort(-(x=(x=table(rle(i)))[rowSums(x>1)>0,,drop=F])[m<-max(rownames(x)),])[1]),m)

Try it online!

Since there wasn't an R answer yet, I decided to give it a try, and well... it wasn't easy. I don't really know whether it is a good approach, but here it goes.

Inputs and outputs vectors of characters.

\$\endgroup\$
  • \$\begingroup\$ Close to 100 bytes is pretty good for R with this challenge. \$\endgroup\$ – ngm Sep 18 '18 at 18:39
4
\$\begingroup\$

Perl 6, 58 56 bytes

*.comb(/(.)$0*/).Bag.max({.value>1,+.key.comb,.{*}}).key

Try it online!

\$\endgroup\$
3
\$\begingroup\$

Python 2, 123 120 bytes

import re
def f(s):r=re.findall(r'(\d)(\1*)',s);c=r.count;print max((len(a+b)*(c((a,b))>1),c((a,b)),a+b)for a,b in r)[2]

Try it online!

\$\endgroup\$
3
\$\begingroup\$

Powershell, 101 byte

($args|sls '(.)\1*'-a|%{$_.Matches}|group|?{$_.Count-1}|sort @{e={$_.Name.Length,$_.Count}})[-1].Name

Explanied test script:

$f = {

(
    $args|          # for each argument (stings)
    sls '(.)\1*'-a| # searches all
    %{$_.Matches}|  # regex matches
    group|          # group it (Note: Count of each group > 0 by design)
    ?{$_.Count-1}|  # passthru groups with Count not equal 1
    sort @{         # sort all groups by 2 values
        e={$_.Name.Length,$_.Count}
    }
)[-1].Name          # returns name of last group (group with max values)

}

@(
    ,('7888885466662716666', '6666')
    ,('3331113331119111', '111')
    ,('777333777333', '777','333')
    ,('122222233433', '33')
    ,('811774177781382', '8')
    ,('555153333551','1')
    ,('12321', '1','2')
    ,('944949949494999494','4')
    ,('8888858888866656665666','88888')
    ,('1112221112221111','111','222')
) | % {
    $s,$e = $_
    $r = &$f $s
    "$($r-in$e): $r"
}

Output:

True: 6666
True: 111
True: 777
True: 33
True: 8
True: 1
True: 1
True: 4
True: 88888
True: 111
\$\endgroup\$
3
\$\begingroup\$

Python 2, 114 113 bytes

-1 byte thanks to TFeld.

p='';r=[];C=r.count
for c in input():r+=['']*(c!=p);r[-1]+=c;p=c
print max((len(w),C(w),w)for w in r if~-C(w))[2]

Try it online!

\$\endgroup\$
3
\$\begingroup\$

Haskell, 72 bytes

import Data.Lists
g!x|y<-countElem x g=(y>1,1<$x,y)
(argmax=<<(!)).group

How it works

(argmax=<<(!)).group       -- expands to: f i = argmax (group i !) (group i)
    group                  -- split the input list into subsequences of equal digits
                           -- e.g. "1112211" -> ["111","22","11"]

                           -- find the element of this list where the function !
                           -- returns the maximum value. First parameter to !
                           -- is the grouped input list, second parameter the
                           -- the element to look at 

g!x|
    y<-countElem x g       -- let y be the number of occurrences of x in g
  = (  ,   ,  )            -- return a triple of
     y>1                   -- a boolean y>1  (remember: True > False)  
        1<$x               -- length of x (to be exact: all elements in x
                           -- replaced by 1. This sorts the same way as the
                           -- length of x)
             y             -- y
                           -- a triples sorts lexicographical
\$\endgroup\$
  • \$\begingroup\$ Do you not need to use Haskell + lists as language because Data.Lists is not part of base? \$\endgroup\$ – ბიმო Sep 18 '18 at 22:02
  • \$\begingroup\$ @BWO: don't know. I've always used a plain "Haskell", even when I imported an exotic library (e.g. Gloss for graphical output or Matrix). I use "Haskell + something" if I don't want to include the byte count for the imports. I think we had this topic on meta, but I cannot find it anymore. If I remember correctly, we had no general definition of "standard library". What should be the reference for Haskell? The Haskell Report, GHC's base, Haskell Plattform, something else? \$\endgroup\$ – nimi Sep 18 '18 at 22:34
  • \$\begingroup\$ IMO it should be as with C/JavaScript/.. that (if it matters) we need to use Haskell (GHC) or Haskell (Hugs) etc. because the implementation specifies a language on PPCG. So for a GHC answer that would include base and for all the other ones I wouldn't know :D \$\endgroup\$ – ბიმო Sep 18 '18 at 23:39
  • \$\begingroup\$ Do you perhaps have a TIO link so it can be tested? Or is the Data.Lists library not available on TIO or another online Haskell compiler? \$\endgroup\$ – Kevin Cruijssen Sep 19 '18 at 10:23
  • 1
    \$\begingroup\$ @KevinCruijssen: yes Data.Lists is missing on TIO. You can test it with this version. \$\endgroup\$ – nimi Sep 19 '18 at 16:45
3
\$\begingroup\$

R, 85 bytes

function(x,R=rle(x),a=ave(R$v,R,FUN=length))rep(R$v[o<-order(a<2,-R$l,-a)[1]],R$l[o])

Try it online!

  • Input : a vector of separated integer digits e.g. c(1,8,8...)

  • Output : a vector of separated integer digits

Unrolled code with explanation :

function(x){                # x is a vector of digits : e.g. c(1,1,8,8,1,1)

R = rle(x)                  # Get the sequences of consecutive repeating digits
                            # doing run length encoding on x, i.e. : R is a list
                            # with the digits (R$values) and the number of their
                            # consecutive occurrencies (R$lengths)
                            # N.B. you can use R$v for R$values and R$l for R$lenghts

a=ave(R$v,R,FUN=length)     # Group R$v by R$l AND R$v, count the occurrencies 
                            # for each group and "unroll" the value of each 
                            # group to the original R$v length.
                            # Here basically we count the occurrencies of the same 
                            # sequence.

o<-order(a<2,-R$l,-a)[1]    # Get the indexes used to order by a < 2 then by -R$l and
                            # finally by -a; store the first index in "o".
                            # Here basically we use order to select the first sequence 
                            # repeated at least twice, in case of ties the sequence 
                            # with the greatest length and in case of ties the most 
                            # repeated sequence.

rep(R$v[o],R$v[o])          # Using the index "o", we reconstruct the sequence repeating
                            # R$l[o] times R$v[o]
}

Alternative version accepting vector of integer or character digits :

R, 88 bytes

function(x,R=rle(x),a=ave(R$v,R,FUN=length))rep(R$v[o<-tail(order(a>1,R$l,a),1)],R$l[o])

Try it online!

  • Input : a vector of separated characters or digits e.g. c("1","8","8"...) or c(1,8,8...)

  • Output : a vector of separated characters if the input was a vector of characters, a vector of digits if the input was a vector of digits

\$\endgroup\$
  • \$\begingroup\$ Can you add an explanation? I don't understand how it's working. \$\endgroup\$ – JayCe Sep 20 '18 at 13:43
  • \$\begingroup\$ @JayCe: done! (I've added details that you well know, just for non-R users ;) ) \$\endgroup\$ – digEmAll Sep 20 '18 at 17:29
  • \$\begingroup\$ ty! It makes sense now. \$\endgroup\$ – JayCe Sep 20 '18 at 18:18
2
\$\begingroup\$

Red, 256 250 bytes

func[s][p: func[b][sort parse b[collect[any keep[copy a skip thru any a]]]]first
last sort/compare collect[foreach d p p s[if 1 < k: length? to-block d[keep/only
reduce[form unique d k]]]]func[x y][(reduce[length? x/1 x/2])< reduce[length? y/1 y/2]]]

Try it online!

Really, realy long solution this time... (sigh)

Takes the input as a string.

Explanation:

f: func [ s ] [
    p: func [ b ] [                        ; groups and sorts the adjacent repeating items
        sort parse b [ 
            collect [                      
                any keep[
                    copy a skip thru any a ; gather any item, optionally followed by itself  
                ]
            ]
        ]
    ]
    t: copy []
    foreach d p p s [                     ; p p s transforms the input string into a block of sorted blocks of repeating digits
        if 1 < k: length? to-block d [    ; filters only the blocks that occur more than once
            insert/only t reduce [ form unique d k ] ; stores the digits and the number of occurences
                                          ; "8888858888866656665666" -> [["5" 3] ["666" 3] ["88888" 2]]
        ]
    ]
    first last sort/compare t func [ x y ] ; takes the first element (the digits) of the last block of the sorted block of items
        [ (reduce [ length? x/1 x/2 ]) < reduce [ length? y/1 y/2 ] ] ; direct comparison of the blocks
]
\$\endgroup\$
2
\$\begingroup\$

Java (JDK 10), 213 bytes

s->{int l=99,X[][]=new int[10][l],d,D=0,m=0,M=0;for(var x:s.split("(?<=(.))(?!\\1)"))X[x.charAt(0)-48][x.length()]++;for(;M<1&&l-->1;)for(d=0;d++<9;)if((m=X[d][l])>1&m>M){M=m;D=d;}for(;l-->0;)System.out.print(D);}

Try it online!

Explanation (outdated)

s->{                                    // Lambda for Consumer<String>
 int l=99,                              //  Length of token, max is 99.
     X[][]=new int[10][l],              //  Array containing the occurrences per token
     d,                                 //  digit value
     D=0,                               //  digit holder for best sequence candidate
     m=0,                               //  holder of the current candidate
     M=0;                               //  best candidate for the current length of token.
 for(var x:s.split("(?<=(.))(?!\\1)"))  //  Tokenize the string into digit-repeating sequences
  X[x.charAt(0)-48][x.length()]++;      //   Add one occurrence for the token
 for(;M<1&&l-->1;)                      //  While no value has been found and for each length, descending. Do not decrease length if a value has been found.
  for(d=0;d++<9;)                       //   for each digit
   if((m=X[d][l])>1&m>M){               //    if the current occurrence count is at least 2 and that count is the current greatest for the length
    M=m;D=d;                            //     mark it as the current best
   }                                    //
 for(;l-->0;)System.out.print(D);       //  Output the best-fitting subsequence.
}                                       // 

Credits

\$\endgroup\$
  • 1
    \$\begingroup\$ I'm afraid there is a small flaw in your j*o>M check. If I understand correctly it takes the max length * occurrence-count. But for a test case like 1113311133933933933933 for example, the 111 would be (3 * 2 = 6), and the 33 would be (2 * 6 = 12). So it outputs 33 having the highest occurrence, instead of 111 being the longest occurring at least twice. Also, var r="";for(;O-->0;)r+=D;return r; can be golfed to for(;O-->0;)System.out.print(D); in Java 10, or even shorter in Java 11: return(D+"").repeat(O);. \$\endgroup\$ – Kevin Cruijssen Sep 18 '18 at 11:48
  • \$\begingroup\$ @KevinCruijssen I think I fixed it. \$\endgroup\$ – Olivier Grégoire Sep 18 '18 at 12:23
  • 1
    \$\begingroup\$ That indeed looks better, and nice way of golfing bytes at the same time. You just forgot to update your explanation. And you can golf 1 more byte changing int X[][]=new int[10][99],d,l=99, to int l=99,X[][]=new int[10][l],d,. \$\endgroup\$ – Kevin Cruijssen Sep 18 '18 at 12:46
  • 1
    \$\begingroup\$ @KevinCruijssen Thanks! I also golfed one more byte by writing d++<9 instead of ++d<10. Sorry for the rest: I'm rather tired today =_= \$\endgroup\$ – Olivier Grégoire Sep 18 '18 at 13:15
2
\$\begingroup\$

Ruby, 68 67 bytes

->a{(b=a.chunk &:+@).max_by{|x|[(c=b.count x)<2?0:x[1].size,c]}[1]}

Try it online!

Inputs and outputs arrays of chars.

The approach is pretty straightforward: we identify the runs of consecutive digits (chunk using unary + as identity function) and take the maximum - first by the size of the run (reset to zero if its occurrence count is < 2), then by the count itself.

\$\endgroup\$
2
\$\begingroup\$

PCRE, 152 bytes

(\d)(?<!(?=\1)..)(?=(\1*)(?!\1).*(?!\1).\1\2(?!\1))(?!(?:(?=\2((\3?+)(\d)(\5*)))){1,592}?(?=\2\3.*(?!\5).\5\6(?!\5))(?:\1(?=\1*\4\5(\7?+\5)))*+(?!\1))\2

See it in action on: https://regex101.com/r/0U0dEp/1 (just look at the first match in each test case)

This is just for fun, since regex isn't a real programming language in and of itself, and the solution is limited :P

Because a zero-width group such as (?:)+ only matches once and doesn't repeat indefinitely, and because PCRE internally makes copies of groups quantified with limits, I've had to use a magic number in there ("{1,592}"), which means we can only look up to 592 contiguous sets of digits ahead to find a competing set that could be longer than the one currently under inspection. More info on this concept here.

\$\endgroup\$
1
\$\begingroup\$

Perl 5, 88 bytes

my($m,%s);++$i%2*$s{$_}++&&($n=$s{$_}/9+length)>$m&&($a=$_,$m=$n)for pop=~/((.)\2*)/g;$a

Try it online!

Slightly ungolfed, with tests:

sub f {
  my($m,%s);
  my($i,$n,$a);           #not needed in golfed version
  ++$i % 2  *  $s{$_}++
  && ($n=$s{$_}/9+length) > $m
  && ($a=$_, $m=$n)
    for pop=~/((.)\2*)/g; #i.e. 7888885466662716666 => 7 88888 5 4 6666 2 7 1 6666
  $a
}
for(map[/\d+/g],split/\n/,join"",<DATA>){ #tests
  my($i,@e)=@$_;
  printf "%-6s   input %-24s   expected %-10s   got %s\n",
    (grep f($i) eq $_, @e) ? "Ok" : "Not ok", $i, join('|',@e), f($i);
}
__DATA__
Input:  7888885466662716666     Output: 6666
Input:  3331113331119111        Output: 111
Input:  777333777333            Output: 777|333
Input:  122222233433            Output: 33
Input:  811774177781382         Output: 8
Input:  555153333551            Output: 1
Input:  12321                   Output: 1|2
Input:  944949949494999494      Output: 4
Input:  8888858888866656665666  Output: 88888
Input:  1112221112221111        Output: 111|222
\$\endgroup\$
1
\$\begingroup\$

Wolfram Language (Mathematica), 67 bytes

#&@@@MaximalBy[Select[Tally@Split@#,Last@#>1&],{Length@#,#2}&@@#&]&

Pure function. Takes a list of digits as input and returns a list of subsequences (in no particular order) as output. Not sure if the "must appear at least twice" clause can be handled more cleanly. Try it online!

\$\endgroup\$
  • 1
    \$\begingroup\$ Could you perhaps add a TIO link for it? \$\endgroup\$ – Kevin Cruijssen Sep 18 '18 at 6:22
  • \$\begingroup\$ If you really insist... \$\endgroup\$ – LegionMammal978 Sep 18 '18 at 10:17
1
\$\begingroup\$

Japt -h, 12 bytes

Input & output are strings.

ò¦ ñÊf@ÓZè¶X

Try it

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.