Some time ago, I need to write function which split string in some specific way. At the first look, task looks trivial, but it was not so easy - especially when I want to decrease code size.


As example we have following input string (it can be arbitrary)

Lorem ipsum dolor sit amet consectetur adipiscing elit sed doeiusmod tempor incididunt ut Duis aute irure dolor in reprehenderit in esse cillum dolor eu fugia ...

We need to splitting it into elements ( groups of adjacent words) using following rules (A-E)

 "Lorem ipsum dolor",  // A: take Three words if each has <6 letters  
 "sit amet",           // B: take Two words if they have <6 letters and third word >=6 letters
 "consectetur",        // C: take One word >=6 letters if next word >=6 letters
 "adipiscing elit",    // D: take Two words when first >=6, second <6 letters
 "sed doeiusmod",      // E: Two words when first<6, second >=6 letters
 "tempor"              // rule C
 "incididunt ut"       // rule D
 "Duis aute irure"     // rule A
 "dolor in"            // rule B
 "reprehenderit in"    // rule D
 "esse cillum"         // rule E
 "dolor eu fugia"      // rule D

So as you can see input string (only alphanumeric characters) is divided to elements (substrings) - each element can have min one and max three words. You have 5 rules (A-E) to divide your string - if you take words one by one from beginning - only one of this rule applied. When you find rule, then you will know how many words move from input to output (1,2 or 3 words) - after that start again: find next rule for the remaining input words.

Boundary conditions: if last words/word not match any rules then just add them as last element (but two long words cannot be newer in one element)

In the output we should get divided string - each element separated by new line or | (you don't need to use double quotes to wrap each element)

Example Input and Output

Here is example input (only alphanumeric ASCII characters):

Lorem ipsum dolor sit amet consectetur adipiscing elit sed doeiusmod tempor incididunt ut Duis aute irure dolor in reprehenderit in esse cillum dolor eu fugia

and its output:

Lorem ipsum dolor|sit amet|consectetur|adipiscing elit|sed doeiusmod|tempor|incididunt ut|Duis aute irure|dolor in|reprehenderit in|esse cillum|dolor eu fugia
  • 2
    \$\begingroup\$ I'm not understanding what's going on with the example. Could you explain how the splitting rules work in general? \$\endgroup\$
    – xnor
    Commented Sep 26, 2019 at 7:08
  • 1
    \$\begingroup\$ @xnor while (word length < 6) {join next word (max 3 words per groups)} else if (word length >= 6) { if (next one length is < 6) {return the pair} else { return group as is } \$\endgroup\$
    – jonatjano
    Commented Sep 26, 2019 at 7:11
  • 1
    \$\begingroup\$ @KamilKiełczewski "you have 6 rules (A-B)". Don't you mean 5 rules (A-E)?.. \$\endgroup\$ Commented Sep 26, 2019 at 7:11
  • 1
    \$\begingroup\$ Can we take the input as a list of words? \$\endgroup\$ Commented Sep 26, 2019 at 7:12
  • 2
    \$\begingroup\$ I'm still pretty confused, but maybe it's just me given the answers and reopen votes. \$\endgroup\$
    – xnor
    Commented Sep 26, 2019 at 17:41

10 Answers 10


Python 2, 92 90 88 87 86 85 bytes

for w in input().split():L=w[:5]<w;x=n+L<3;r+='| '[x]+w;n=n*x-~L
print r[1:]

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-1 byte, thanks to Kevin Cruijssen


Jelly, 23 bytes


A full-program printing the result.

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Given the lengths of three words (a, b, and c) we can write the following mapping for how many word we should take:

a<6?  b<6?  c<6?   words
   1     1     1     3
   1     1     0     2
   1     0     1     2
   1     0     0     2
   0     1     1     2
   0     1     0     2
   0     0     1     1
   0     0     0     1

Treating the comparisons as a single number in binary this is:

bin([a<6,b<6,c<6]):   7   6   5   4   3   2   1   0
             words:   3   2   2   2   2   2   1   1

So we can map like so:

bin([a<6,b<6,c<6]):   7   6   5   4   3   2   1   0
         add eight:  15  14  13  12  11  10   9   8
    divide by five:   3   2   2   2   2   2   1   1

Note that when less than three words remain we want to take all of them, unless there are two left and they are both of length six or more when case C says to take one word. To make this the case we repeat what we have up to length three (with ṁ3 instead of ḣ3) and use that.

a<6?  b<6?         moulded  bin  + 8  div 5 (= words)
   1                111     7    15   3  (i.e. all 1)
   0                000     0     8   2  (i.e. all 1)
   1    1           111     7    15   3  (i.e. all 2)
   1    0           101     5    13   2  (i.e. all 2)
   0    1           010     2    10   2  (i.e. all 2)
   0    0 (i.e. C)  000     0     8   1  (i.e. just 1)

The code then works as follows.

ḲµẈṁ3<6Ḅ+8:5⁸sḢKṄȧƲẎµ¹¿ - Main Link: list of characters
Ḳ                       - split at spaces
                      ¿ - while...
                     ¹  - ...condition: identity (i.e. while there are still words)
 µ                  µ   - ...do: the monadic chain:
  Ẉ                     -   length of each
    3                   -   literal three
   ṁ                    -   mould like ([1,2,3])
      6                 -   literal six
     <                  -   less than? (vectorises)
       Ḅ                -   from binary to integer
         8              -   literal eight
        +               -   add
           5            -   literal five
          :             -   integer divide
            ⁸           -   chain's left argument
             s          -   split into chunks (of that length)
                  Ʋ     -   last four links as a monad (f(x)):
              Ḣ         -     head (alters x too)
               K        -     join with spaces
                Ṅ       -     print & yield
                 ȧ      -     logical AND (with altered x)
                   Ẏ    -   tighten (back to a list of words) 
  • \$\begingroup\$ wow - currently you also put on the leader's yellow jersey 🔥🔥🔥 - btw: nice explanation :) \$\endgroup\$ Commented Oct 1, 2019 at 15:02
  • \$\begingroup\$ Theoretically I am the only one wearing it (since I piped Nick to the post of reaching 23) but I'll very happily share it! \$\endgroup\$ Commented Oct 1, 2019 at 15:37
  • \$\begingroup\$ I don't know that - however you both has different but same size answers - so we have draw here \$\endgroup\$ Commented Oct 1, 2019 at 15:39
  • \$\begingroup\$ Yeah, same byte count, different programs, but a very similar method (now). The timing information is in the revision histories if you're interested, but do note that when looking at revisions of posts that updates within five minutes of another are bundled into one revision stamped at the original time (this feature can cause one to not really be able to tell). For what it's worth I imagine we came up with the similar methods independently too. \$\endgroup\$ Commented Oct 1, 2019 at 15:45

Perl 5, 47 bytes

seems regex can be shorten with this equivalent one

s/(\w{1,5} ){3}|\w{6,} (?=\w{6})|\w+ \w+ /$&

Try it online!

Previous regex

Perl 5, 86 bytes

s/(\w{1,5} ){3}|((\w{1,5} ){2}|\w{6,} )(?=\w{6})|\w{6,} \w{1,5} |\w{1,5} \w{6,} /$&

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Not valid:

Perl 5 (-M5.01 -lnF/(?:\S{1,5}\K\s+){3}|\S{6,}\K\s+(?=\S{6})|\S+\s+\S+\K\s+/), 9 bytes

say for@F

Try it online!

  • \$\begingroup\$ Nice answer. I was about to try and make a Retina answer for this challenge. A trivial port would be 82 bytes. \$\endgroup\$ Commented Sep 26, 2019 at 8:27
  • \$\begingroup\$ however not perfect because the trailing space is not removed \$\endgroup\$ Commented Sep 26, 2019 at 8:29
  • 1
    \$\begingroup\$ removing the trailing space 51 bytes \$\endgroup\$ Commented Sep 26, 2019 at 8:46
  • 3
    \$\begingroup\$ ... which is a made-up language designed specifically for the challenge, meaning that version would be invalid. \$\endgroup\$ Commented Sep 26, 2019 at 11:47
  • 2
    \$\begingroup\$ @pppery Good point, didn't thought about that. I have used large flags before, like this .NET C# answer with two flags, but those are used a static imports. In this case it would indeed be a 'made-up' language (or should I say flag) only designed for this challenge. \$\endgroup\$ Commented Sep 26, 2019 at 15:18

AWK, 65 bytes (thanks manatwork!)


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I didn't even know AWK had a ternary operator

AWK, 97 79 72 bytes (thanks manatwork!)

BEGIN{a[0]="|";a[1]=RS=FS}{printf a[x=n+(L=length($1)>5)<3]$1;n=n*x+L+1}

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I shamelessly stole the algorithm from @TFeld's Python2 solution.

  • 1
    \$\begingroup\$ Compacted it a bit, but only tested with that single input. See if you can get something useful from it: Try it online!. \$\endgroup\$
    – manatwork
    Commented Oct 1, 2019 at 15:02
  • 1
    \$\begingroup\$ Oh, 2 more things: FS's default value is a single space, so you can use it to initialize RS; the challenge says the input will be “only alphanumeric characters”, so you can use printf instead of print, making the assignment to ORS unnecessary. \$\endgroup\$
    – manatwork
    Commented Oct 1, 2019 at 15:30
  • 1
    \$\begingroup\$ Ok, finally spent some time to give a closer look at that array a. A ternary operator would be shorter: Try it online!. \$\endgroup\$
    – manatwork
    Commented Oct 1, 2019 at 15:44
  • \$\begingroup\$ @manatwork I almost feel like you should submit that...I didn't even know AWK had a ternary operator :) \$\endgroup\$ Commented Oct 1, 2019 at 15:51

05AB1E, 28 bytes


Port of @TFeld's Python answer, so make sure to upvote him!

Try it online.


0U                       # Set variable `X` to 0 (it's 1 by default)
#                        # Split the (implicit) input-string on spaces
 v                       # Loop over each word `y`:
      yg                 #  Get the length of the word
        6@               #  And check that it's >= 6
          D              #  Duplicate this
           X+            #  Add variable `X` to it
             3‹          #  And check that it's smaller than 3
               DŠ        #  Duplicate this as well, and triple-swap (a,b,c to c,a,b)
                 X*      #  Multiply the <3 check with variable `X`
                   +     #  Add it to the length >=6 check
                    >U   #  Increase it by 1, and set it as the new variable `X`
  „\n                 è  #  Index the <3 check into the string "\n "
                       y #  And push the current word
 }J                      # After the loop: join the entire stack together
   ¦                     # And remove the leading space
                         # (after which the top of the stack is output implicitly as result)
  • \$\begingroup\$ 🚀 your answer is in the top \$\endgroup\$ Commented Oct 2, 2019 at 6:08

Clean, 206 bytes

import StdEnv,Text
$s=join"|"(map(join" "o map fst)(?[(w,size w<6)\\w<-split" "s]))
?l=case l of[a,b,c:t]|all(snd)[a,b,c]=[[a,b,c]: ?t];[a:t=:[b:_]]|not(snd a||snd b)=[[a]: ?t];[]=[];l=[take 2l: ?(drop 2l)]

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Jelly, 24 bytes


Try it online!

-1 thanks to Jonathan Allan.

  • \$\begingroup\$ can you explain your solution? \$\endgroup\$ Commented Sep 27, 2019 at 21:42
  • \$\begingroup\$ @KamilKiełczewski Eh, I'm a bit tired right now. Might explain tomorrow. \$\endgroup\$ Commented Sep 27, 2019 at 21:43
  • \$\begingroup\$ 🚀 your answer is in the top \$\endgroup\$ Commented Oct 2, 2019 at 6:09

Jelly, 27 23 bytes


Try it online!

A full program that takes as its argument the input string and prints newline-separated groups of words. Takes advantage of the fact that if the current first three words have their length checked to see if <6 and this is then treated as a binary number, the number of words needed will be 1,1,2,2,2,2,2,2,3 for numbers from 0 to 7 respectively.


Ḳ                       | Split at spaces
 µ                  µÐL | Repeat the following until no new results:
  Ẉ                     | - Lengths of lists (i.e. words)
   <6                   | - Less than 6
     ḣ3                 | - First three
       Ḅ                | - Comvert from binary to integer
        +8              | - Add 8
          :5            | - Integer divide by 5
            Ṭ           | - Convert from index to boolean list
             k⁸         | - Split input to this loop iteration at that point
                  ɗ/    | - Reduce using following as a dyad:
               K        |   - Join with spaces
                Ṅ       |   - Output with trailing newline
                 ṛ      |   - Right argument (i.e. rest of list)
  • \$\begingroup\$ do you take idea from Jonathan Allan answer in this version of your answer? \$\endgroup\$ Commented Oct 1, 2019 at 16:00
  • \$\begingroup\$ @KamilKiełczewski only really the idea of outputting the interim results during the loop rather than returning them from the link as a list. \$\endgroup\$ Commented Oct 1, 2019 at 16:11
  • \$\begingroup\$ 🚀 your answer is in the top \$\endgroup\$ Commented Oct 2, 2019 at 6:08

Charcoal, 49 bytes

≔⮌⪪S θWθ«≔⌊⟦Lθ⊕ΣE²›⁶L§θ⁻κ²⟧ι⪫E⁻ι∧⁼鳋⁵L§θ±³⊟θ ¿θ|

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

≔⮌⪪S θ

Split the input into words and reverse it so that we can use Pop to remove words in order.


Repeat while there are still words left.


Estimate the number of words needed as equal to one more than the number of the first two words that are less than 6 letters, but no more than the number of words left.


Adjust the number of words if there are three and the third word is not less than 6 letters, then remove that many words and print them separated with spaces.


Output a separator if there are still more words left.


J, 55 53 bytes


Try it online!

  • \$\begingroup\$ A accept this answer - however can I know why the output is a "table" (literally with borders) ? \$\endgroup\$ Commented Sep 27, 2019 at 5:06
  • 1
    \$\begingroup\$ In J strings of different lengths must be "boxed" if you want to make a list out of them. So I take boxed strings as the input, and since the output requires grouping them, we have a list of boxes each of which contains a list of boxes. The borders you see are just the default way J displays boxed data (it's configurable). The other alternative would be to box the input within my function and return a list of boxed strings, where each string could have multiple words. This felt less consistent to me, though. \$\endgroup\$
    – Jonah
    Commented Sep 27, 2019 at 5:16

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