Split a list at the second occurrence of the first element

Question

Why should you golf in Haskell? Find out here. In that answer, Zgarb defines a task:

Let's define a function f that splits a list at the second occurrence of the first element, e.g. [0,2,2,3,0,1,0,1] -> ([0,2,2,3],[0,1,0,1]):

Alright then, lets!

You are to take a non-empty list consisting of digits, and output a pair of lists, clearly distinguished, such that the output is the input split before the second occurrence of the first element.

For example, you may output two strings of digits with a non-digit separator. The separator you choose between the elements of each pair and the pairs themselves must be distinct (e.g. space and newline), consistent and not contain digits.

The output must only contain 2 elements however. You may not output empty lists in the output. If you choose to use newlines as a separator between pairs, you may not have a leading newline. You may have trailing newlines no matter what, and you may have trailing whitespace, so long as its sensible.

You may assume:

• The input will be provided in any convenient method, and you may output in any convenient method
• The input will either be a list of digits, or a string
• The input will only contain the integers 0 though to 9
• The first element of the input will always occur at least twice
• The input will always have 3 or more elements in it

Additionally, this is code-golf, so the shortest code in bytes wins

Test cases

[1, 1, 9] -> [[1], [1, 9]]
[4, 9, 4] -> [[4, 9], [4]]
[5, 7, 5, 5] -> [[5, 7], [5, 5]]
[8, 8, 0, 7] -> [[8], [8, 0, 7]]
[7, 1, 5, 7, 4, 2] -> [[7, 1, 5], [7, 4, 2]]
[0, 6, 9, 1, 1, 0, 2] -> [[0, 6, 9, 1, 1], [0, 2]]
[2, 9, 3, 2, 4, 2, 5, 9] -> [[2, 9, 3], [2, 4, 2, 5, 9]]
[0, 2, 2, 3, 0, 1, 0, 1] -> [[0, 2, 2, 3], [0, 1, 0, 1]]
[2, 7, 4, 6, 2, 6, 6, 4, 8, 2] -> [[2, 7, 4, 6], [2, 6, 6, 4, 8, 2]]
[8, 2, 2, 7, 5, 4, 7, 0, 8, 0, 7] -> [[8, 2, 2, 7, 5, 4, 7, 0], [8, 0, 7]]
[8, 7, 8, 9, 4, 2, 9, 4, 5, 7, 5, 1, 9] -> [[8, 7], [8, 9, 4, 2, 9, 4, 5, 7, 5, 1, 9]]
[3, 8, 1, 1, 7, 3, 6, 9, 7, 1, 4, 3, 4] -> [[3, 8, 1, 1, 7], [3, 6, 9, 7, 1, 4, 3, 4]]
[4, 7, 0, 5, 6, 5, 0, 1, 7, 8, 7, 8, 4, 1] -> [[4, 7, 0, 5, 6, 5, 0, 1, 7, 8, 7, 8], [4, 1]]
[2, 1, 8, 0, 3, 2, 2, 5, 7, 9, 4, 3, 5, 1, 9, 6, 9] -> [[2, 1, 8, 0, 3], [2, 2, 5, 7, 9, 4, 3, 5, 1, 9, 6, 9]]
[1, 1, 4, 1, 2, 5, 5, 3, 3, 4, 3, 2, 0, 8, 6, 0, 3] -> [[1], [1, 4, 1, 2, 5, 5, 3, 3, 4, 3, 2, 0, 8, 6, 0, 3]]
[4, 3, 5, 2, 2, 0, 6, 4, 8, 6, 6, 6, 7, 3, 4, 8, 7, 6] -> [[4, 3, 5, 2, 2, 0, 6], [4, 8, 6, 6, 6, 7, 3, 4, 8, 7, 6]]

Answer 1

Haskell, 26 bytes

f(x:y)=([x],0)*>span(/=x)y

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Shortens Zgarb's OG solution

f(x:y)|(a,b)<-span(/=x)y=(x:a,b)

by prepending x to the first element of (a,b) in a pointfree way, that is without explicitly binding (a,b).

It would be nice it we could do (x:)<$>(a,b), but that gives (a,x:b) -- the Functor instance of tuples lets us act on the second element but not the first.

However, Applicative lets us combine tuples as:

(p, f) <*> (a, b) = (p++a, f b)
([x], id) <*> (a, b) = (x:a, b)

It suffices to use *> which ignores f and leaves b unchanged.

((x:), 0) *> (a, b) = (x:a, b)

The 0 could be anything -- it doesn't matter. It would also work to use >> in place of *>.

26 bytes

f(x:y)=([x],y)>>=span(/=x)

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A alternative, this time using the Monad instance and (>>=) :: Monoid a => (a, a0) -> (a0 -> (a, b)) -> (a, b)

27 bytes

f(x:y)=([x],[])<>span(/=x)y

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Using <> to do concatenate elementwise (a, b) <> (c, d) = (a++c, b++d). This is available in Prelude without an import starting in version 8.4.1.

Answer 2

AWK, 17 14 bytes

sub(FS$1,RS$1)

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Thanks to Pedro Maimere for the hint to lop off 3 bytes

The interactions of the rules in the contest allow for pretty trivial AWK solution... Assuming the input can be a blank delimited string of numbers, this will work. If it has to include the brackets and commas (which I wasn't sure about from the linked article about convenient input), then it would be this instead.

sub(", "(a=substr($1,2)),"]\n["a)

And here's one in ><> which might be shrinkable still... It's the first time I've managed to get something to work in that language, so I wouldn't be surprised to learn there some trick I don't know yet.

><>, 41 bytes

i:o&0v
?(0:i<o$v?<=1:+{=&:&:;
 0+1o+19< ^

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I can add a more detail description if anyone is interested, but here's an overview of how it works.

The input it expects is a string of digits. Since the challenge specified that the list was made up of single digit numbers, I chose not to include a delimiter. If that's a requirement, some additional stuff would have to be added...

The first line reads in the first number, uses it to set the register (for comparison as the rest of the list is read in), then pushes a counter that tracks the number of times the first number has been seen. Then it passed control down to the next line.

i:o&0v

That line reads one digit at a time, starting with the second digit, and prints it. There's a conditional code to add a \n when the counter hits 1 (meaning it found the second occurrence of the first digit). The code is interpreted right to left to save characters.

?(0:i<o$v?<=1:+{=&:&:;

The last line is effectively like a "function" call to print a linefeed and tweak the counter so that it will never be called again.

0+1o+19< ^

Answer 3

Vim, 3 bytes/keystrokes

*O<esc>

Jump to next occurrence of word and make a new line.

---

V (vim), 2 bytes

*O

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Since V has implicit escape after O (?), we can use just 2 bytes.

---

Old (general) 8-byter:

Y/<C-r>"<BS><CR>O<esc>

---

Input is each list item on a single line. Output is two lists separated by a blank line.

Uses vim notation for the keystrokes (<C-r> is Ctrl-R, etc.; see :help key-notation).

Explanation:

• Y yank the first line
• /<C-r>"<BS><CR> search for the next occurrence by inserting the yanked text into a search /. Yanking into the search register with "/Y doesn't work (:help quote/). The backspace deletes the line-ending, which is yanked. (Alternately, use y$/<C-r>"<CR>.)
• O<esc> new blank line above.

This generalizes to any type of list as long as each element is a single line.

---

I don't think we can use POSIX vi for this, since I don't think it has <C-r> to insert registers. It doesn't have the search register "/, but that doesn't end up mattering.

Answer 4

Vyxal, 18 7 bytes

Ṙṫ:‟€vp

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Answer 5

K (ngn/k), 11 bytes

{(2#*=x)_x}

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• (...)_x cut the input (x) at...
  • 2#*=x the indices of the first two occurrences of the first value in the input

Answer 6

Factor, 37 35 bytes

[ dup first 1 pick index-from cut ]

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Explanation:

• dup Duplicate the input.

  Stack: (e.g.) { 0 2 2 3 0 1 0 1 } { 0 2 2 3 0 1 0 1 }

• first Get first element.

  Stack: { 0 2 2 3 0 1 0 1 } 0

• 1 Push 1.

  Stack: { 0 2 2 3 0 1 0 1 } 0 1

• pick Put a copy of the object third from the top on top of the stack.

  Stack: { 0 2 2 3 0 1 0 1 } 0 1 { 0 2 2 3 0 1 0 1 }

• index-from Find the index of 0 starting from index 1 in the sequence on top of the stack.

  Stack: { 0 2 2 3 0 1 0 1 } 4

• cut Split a sequence in two at an index.

  Stack: { 0 2 2 3 } { 0 1 0 1 }

Answer 7

Jelly, 7 bytes

œṡḢ©®;Ɱ

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œṡḢ©®;Ɱ  Main Link
œṡ       Split at the first occurence of
  Ḣ      The first element (pops the element)
   ©     (also copies that element to the register)
      Ɱ  For each block
    ®;   Prepend the register

As pointed out by Nick Kennedy in the comments (full credit to them), Ḣ;Ɱœṡ@¥ also works and is slightly more functionally pure (though Ḣ still modifies the list itself, so it's not entirely pure):

Ḣ;Ɱœṡ@¥  Main Link
Ḣ        Cut off and return the first element
      ¥  Last two as a dyad (for chaining; this makes the right argument the modified list for both inner dyads rather than applying consecutively)
 ;Ɱ      Prepend the first element to each of
   œṡ@   The modified list, split at the first occurrence of the first element

Answer 8

Python 3.8, 41 bytes

lambda l:[l[:(i:=l.index(l[0],1))],l[i:]]

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Answer 9

Retina 0.8.2, 18 17 bytes

^((.).*?)\2
$1¶$2

Try it online! Link is to test suite that double-spaces the output for convenience. Takes input as a string of digits. Explanation: Simply finds the earliest next match of the first character and inserts a newline before it. Edit: Saved 1 byte thanks to @Jakque.

Answer 10

JavaScript (ES6), 31 bytes

Thanks Razetime for -1 byte.

a=>a.match(/^(.).*?(?=\1)|.+/g)

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Answer 11

J, ¹⁷ 16 bytes

(]{.,<@;@}.)<;.1

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In a sentence:

Cut on first element and then meld together the tail elements.

Consider f 0 2 2 3 0 1 0 1

• <;.1 Cut using the first element as the fret:

  ┌───────┬───┬───┐
  │0 2 2 3│0 1│0 1│
  └───────┴───┴───┘
  

• {.,<@;@}. First element of that result {. catted with tail of that result }., razed ; and then reboxed <@.

  ┌───────┬───────┐
  │0 2 2 3│0 1 0 1│
  └───────┴───────┘
  

Answer 12

BQN, 12 11 bytes^SBCS

⊢⊔˜·∨`·»⊑=«

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Explanation:

⊢⊔˜·∨`·»⊑=«   # tacit function which can take input as either a list or a string
          «   # the input list shifted left
         =    # equality comparison with
        ⊑     # the first element of the input
      ·»      # shift the result right
   ·∨`        # 'or' scan
⊢⊔˜           # group the input according to those values

Answer 13

R, 38 37 36 bytes

Edit: -1 byte thanks to pajonk, as well as outputting the right-way-around now, and then -1 byte thanks to digEmAll

function(l)by(l,cumsum(l==l[1])>1,c)

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Answer 14

Zsh, 35 bytes

shift>$1
ls
for x
(rm $x&&od;<<<$x)

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Explanation:

• >$1: create a file named the first input item
• shift: shift the input array, removing the first item
• ls: list the directory. Since we created $1, this prints the first item
• for x: for each item $x in the input array: (
  • rm $x: try to remove the item. If the item $x is the same as the first input, and the item has not already been removed, this will succeed
  • &&: if that succeeds:
    • od: print 0000000 as a separator
  • <<<$x: print the item

With some loose interpretation of what is allowed as a separator, we could have:

Zsh, 33 bytes

shift>$1
ls
for x
rm -v $x||<<<$x

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• rm -v is verbose; if the removal is successful, a message removing 'x' is printed which is arguably a separator, as well as printing the item
• ||: if that fails:
  • <<<$x: print the item. We only need to do this if removal fails, because rm -v prints the item already

Answer 15

Red, 49 46 bytes

func[x][reduce[take/part x find next x x/1 x]]

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Answer 16

Scala, 35 30 bytes

Saved 5 bytes thanks to @cubic lettuce!

x=>x splitAt x.indexOf(x(0),1)

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Hopefully, I'm not FGITW'ing this.

Answer 17

JavaScript (ES6), 34 bytes

a=>[a,a.splice(a.indexOf(a[0],1))]

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Answer 18

Haskell, 56 bytes

f(h:t)=(h#t)[h]
(h#(x:y))a|x==h=(a,x:y)|0<1=(h#y)$a++[x]

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Because why not?

flawr's suggestion, 33 bytes

f(x:y)|(a,b)<-break(==x)y=(x:a,b)

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This is based on Zgarb's solution (f(x:y)|(a,b)<-span(/=x)y=(x:a,b)).

Answer 19

PowerShell Core, 60 57 bytes

$a,$b=$args
$r=(,$a),$y
$b|%{$r[$r[1]-or$_-eq$a]+=,$_}
$r

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It takes the input as an array of ints, returns two arrays

Another approach for 59 bytes

param($a)$a[0..(($i=$a|% i*f $a[0] 1)-1)],$a[$i..$a.Length]

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Takes the input as a string, returns two arrays

Answer 20

Red, 67 bytes

func[b][collect[keep/only take/part b find next b b/1 keep/only b]]

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Answer 21

Python 2, 45 bytes

i=input()
a=i.index(i[0],1)
print i[:a],i[a:]

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Answer 22

Wolfram Language (Mathematica), 26 bytes

g[b_,a___,b_,c___]=b.a|b.c

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-10 bytes thanks to @att

Answer 23

05AB1E, 8 7 bytes

ćk>Ig‚£

Try it online or verify all test cases.

7 bytes alternative provided by @ovs:

ćk>°RÅ¡

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Explanation:

ć       # Extract head of the (implicit) input-list; pop and push remainder-list and
        # first item separated to the stack
 k      # Get the first 0-based index of this item in the remainder-list
  >     # Increase it by 1 to make it a 1-based index
   Ig   # Get the length of the input-list
     ‚  # Pair them together
      £ # And split the (implicit) input-list into parts of that size
        # (after which the result is output implicitly as result)

ćk>     # Same as above
   °    # Pop and push 10 to the power this 1-based index
    R   # Reverse it, so we have a 1 with some leading 0s
     Å¡ # Split the (implicit) input-list at the truthy indices (or singular index in
        # this case: the 1)
        # (after which the result is output implicitly as result)

Answer 24

Vyxal ṡ, 5 bytes

hẆḢḣf

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Explanation:

h      # Get the first element of the list
 Ẇ     # Split list on head, without removing it from the list
  Ḣḣ   # Get the first element from the resulting list
    f  # Flatten the rest of the resulting list
       # 'ṡ' flag - print both lists, separated by a space

Answer 25

Japt, 8 bytes

Feels like there should be a shorter way for this, but can't seem to find it right now.

Selecting the right-hand element ensures we don't select the first letter itself, the check on the incremented variable ensures we only split once.

óϦUΪT°
ó        // Split the input string between char pairs where it's not true that
 Ï       // the right-hand element
  ¦      // is different than
   UÎ    // the first char of the input
     ª   // or
      T° // T, initially equals zero, plus plus.

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Answer 26

JavaScript (V8), 54 bytes

x=>[x[s="slice"](0,i=x[s](1).indexOf(x[0])+1),x[s](i)]

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Answer 27

MATLAB/Octave, 63 bytes

function y=f(x)
l=find(x==x(1));y={x(1:l(2)-1),x(l(2):end)};end

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Outputs cell aray with 2 cells, which hold appropriate vectors. I've chosen such output because rules say 2 lists must be distinguished, not necessarily be separate variables. And outputting 2 variables turned out to give a little longer code.

Ungolfed/explained:

function y = f(x)
l = find( x==x(1) );   % indices of elements equal to first element
l2 = l(2);             % index of second occurence
y = { x(1:(l2-1)),...  % vector containing elements before 2nd occurence
      x(l2:end) };     % vector containing elements from 2nd occurence
end

---

Interestingly, it's also possible to create anonymous function that does the same, but it's 2 bytes longer:

@(x){x(1:find(x(2:end)==x(1),1)),x(find(x(2:end)==x(1),1)+1:end)}

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It is possible to shorten it more as flawr noticed, resulting in 42 bytes but it's an Octave-only solution, not working for MATLAB:

@(x){x(1:(l=find(x==x(1))(2))-1),x(l:end)}

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Answer 28

Husk, 12 bytes

Fȯ:;:←¹↕≠←¹t

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no split at index builtin, but ↕ (span) helps a bit.

it's similar to xnor(and Zgarb)'s answer, but argument destructuring and functors don't exits, so it just uses a fold.

Answer 29

Vyxal, 8 bytes

ḣ$£¥€ƛ¥p

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This is possible in 8 bytes Imao

Answer 30

Stax, 10 bytes

τÄ∩T╕(û▒(Ç

Run and debug it

annoyingly long, but i guess it works.

Stax, 12 bytes(regex)

êt┴≈∟·M╤\+6)

Run and debug it

Stax, 22 bytes(tsh's regex)

"^(.).*?(?=\1)|.+$"|Fm

Run and debug it