Append and erase

Question

Given one line that consists of only letters, process as following:

You maintain a string that's empty at the beginning.
If the next input character is in the string, remove it from the string.
If the next input character isn't in the string, append it to the string.

Output the final state of the string.

You can safely assume the input consists at least one character (i.e. non-empty), but there's no guarantee that the output isn't empty.

Pseudocode (Feel free to golf this):

str = EMPTY
for each character ch in input
  if ch exists in str
    remove all ch from str
  else
    append ch to str
print str

The input matches the regular expression ^[A-Za-z]+$.

Sample test cases:

ABCDBCCBE -> ADCBE
ABCXYZCABXAYZ -> A
aAABBbAbbB -> aAbB
GG -> (empty)

The input can be given in any applicable way, but it must be treated as a string, and the same for output. The program should not exit with an error.

The shortest program in each language wins!

Extra (Optional): Please explain how your program works. Thank you.

@user202729 No. I changed a little (it does not invalidate any answer) so the input is never empty. — iBug, Commented Nov 28, 2017 at 16:03

Erik the Outgolfer · Accepted Answer · 2017-11-28 17:35:32Z

11

Jelly, 3 bytes

œ^/

Try it online!

Full program.

edited Nov 28, 2017 at 17:35

answered Nov 25, 2017 at 16:59

Erik the Outgolfer

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\$\begingroup\$ Why is œ^/ not enough? \$\endgroup\$
– Jonathan Allan
Commented Nov 27, 2017 at 0:06
\$\begingroup\$ @JonathanAllan The program should not exit with an error. \$\endgroup\$
– Erik the Outgolfer
Commented Nov 27, 2017 at 12:44
1

\$\begingroup\$ the input is never empty Well, now it works. \$\endgroup\$
– user202729
Commented Nov 28, 2017 at 16:10

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Laikoni · Accepted Answer · 2017-11-27 14:12:01Z

10

Haskell, 44 42 bytes

foldl(#)""
s#x|z<-filter(/=x)s=z++[x|z==s]

Try it online! Edit: -2 bytes thanks to Zgarb!

Explanation:

The second line defines a function (#) which takes a string s and a character x and performs either the remove or append. This is achieved by filtering out every occurrence of x in s, resulting in the string z. If x does not occur in s, then z is equal to s and z++[x|z==s] yields the original string with x appended. Otherwise [x|z==s] yields the empty string and only the filtered string is returned.

foldl(#)"" is an anonymous function which takes a string and adds one character after the other the initially empty string "" with the function (#).

edited Nov 27, 2017 at 14:12

answered Nov 25, 2017 at 18:28

Laikoni

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2

\$\begingroup\$ 42 bytes by reusing the filter. \$\endgroup\$
– Zgarb
Commented Nov 25, 2017 at 18:38

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Galen Ivanov · Accepted Answer · 2017-11-25 21:13:25Z

J, 21 19 bytes

#~~:&.|.(2|*)1#.=/~

How it works:

=/~ - makes a table of equality of the characters in the string:

   a =. 'ABCXYZCABXAYZ'
   ]b =: =/~ a 
1 0 0 0 0 0 0 1 0 0 1 0 0
0 1 0 0 0 0 0 0 1 0 0 0 0
0 0 1 0 0 0 1 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 1 0 0 0
0 0 0 0 1 0 0 0 0 0 0 1 0
0 0 0 0 0 1 0 0 0 0 0 0 1
0 0 1 0 0 0 1 0 0 0 0 0 0
1 0 0 0 0 0 0 1 0 0 1 0 0
0 1 0 0 0 0 0 0 1 0 0 0 0
0 0 0 1 0 0 0 0 0 1 0 0 0
1 0 0 0 0 0 0 1 0 0 1 0 0
0 0 0 0 1 0 0 0 0 0 0 1 0
0 0 0 0 0 1 0 0 0 0 0 0 1

1#. - sum of each row by base 1 conversion (how many times the letter occurs)

   ]c =: 1#. b
3 2 2 2 2 2 2 3 2 2 3 2 2

~:&.| - reverse, then apply nub sieve (is the char unique) and reverse again. Thus I find the last occurrences of the characters in the string:

   ]d =. ~:&.|. a
0 0 0 0 0 0 1 0 1 1 1 1 1

* - multiplies the count by 1 for the last position of the character in the sring, by 0 otherwise, computed by the above ~:&.|

   ]e =. c * d
0 0 0 0 0 0 2 0 2 2 3 2 2

2| - modulo 2 (sets to 0 the positions of the chars that have even count):

   ]f =. 2| e 
0 0 0 0 0 0 0 0 0 0 1 0 0

#~ - copy the right argument left arg. times (~ reverses the places of the args)

]f # a A

Try it online!

Community · Accepted Answer · 2020-06-17 09:04:33Z

Brainfuck, 95 bytes

,[<<<[[->+>>>+<<<<]>>>[-<+<->>]<<[[-]<]>[[-]>>[-]>[[-<+>]>]<<[<]<<]<<]<[->>>>[-]<<<]>>>>[->+<]>>[>]>>,]<<<[.<]

Try It Online

How It Works

, Gets first input
[ Starts loop
    <<< Go to start of string
    [ Loop over the string
        [->+>>>+<<<<] Duplicates the current char of the string
        >>>[-<+<->>] Duplicates and subtracts the inputted char from the duplicate of the string char
        <<[[-]<] If the char is different to the input, remove the difference
        > If the char is the same
        [
            [-]>>[-]>[[-<+>]>]<<[<]<< Remove the char from the string and sets the inputted char to 0
        ]
        << Moves to the next char of the string
    ]
    >>>[->+<] adds the inputted char to the string
    >>[>]>>, gets the next input
]
<<<[.<] prints the string

totallyhuman · Accepted Answer · 2017-11-25 17:33:27Z

5

Haskell, 47 bytes

Another one bytes the dust thanks to Bruce Forte.

import Data.List
foldl1(\x y->union(x\\y)$y\\x)

Try it online!

Takes a list of Strings.

Symmetric difference is annoying...

edited Nov 25, 2017 at 17:33

answered Nov 25, 2017 at 17:22

totallyhuman

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\$\begingroup\$ ++ saves 2 bytes over union with this method. \$\endgroup\$
– Ørjan Johansen
Commented Dec 1, 2017 at 10:18

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Martin Ender · Accepted Answer · 2017-11-25 19:48:58Z

3

Retina, 16 bytes

+1`(.)(.*?)\1
$2

Try it online!

edited Nov 25, 2017 at 19:48

answered Nov 25, 2017 at 17:25

Martin Ender

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coltim · Accepted Answer · 2021-01-04 02:39:03Z

3

K (ngn/k), 21 9 bytes

|&2!#'=|:

Try it online!

If a character appears an even number of times in the string, it won't be present in the output. If it appears an odd number of times, only its last instance will remain.

#'=|: count the number of times each distinct character occurs in the string (reversing the order)
|&2! filter down to where the character appears an odd number of times, then reverse

edited Jan 4, 2021 at 2:39

answered Jan 3, 2021 at 17:56

coltim

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Suever · Accepted Answer · 2022-09-04 02:45:14Z

3

MATL, 6 bytes

vi"@X~

Doesn't work in the TIO environment, but works fine on the MATLAB implementation, and thanks to a fresh patch, you may try it on MATL Online

X~ equals setxor, or symmetric difference, which does exactly what the challenge asks. The rest is just looping over the input i"@ and starting with an empty string by concatenating the entire stack which is empty at the start (thanks Luis Mendo).

edited Sep 4, 2022 at 2:45

Suever

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answered Nov 27, 2017 at 21:53

Sanchises

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Community · Accepted Answer · 2020-06-17 09:04:33Z

2

q, 38 bytes

""{$[y in x;except;,][x;y]}/

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 25, 2017 at 20:11

skeevey

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duckmayr · Accepted Answer · 2017-11-26 18:12:44Z

2

R, 92 84 77 bytes

for(i in el(strsplit(scan(,y<-''),y)))y=c(y[y!=i],if(!i%in%y)i);cat(y,sep='')

Try it online!

-15 bytes thanks to djhurio

Explanation

djhurio provided an excellent R answer avoiding a for loop -- as R programmers instinctively do as a rule (myself included). Here's an R answer that utilizes a for loop (and saves a few bytes in the process).

x=scan(,''); -- assign the input into the variable x
y=''; -- create an empty string in a variable called y
for(i in el(strsplit(x,''))) -- for every character i in x
y=c(y[y!=i],if(!i%in%y)i) -- assign to y every element of y that is not equal to i, appending i if i was not already in y
cat(y,sep='') -- print the elements of y with no space between them

Note

If you click the TIO link above, you'll find in the header library(methods); this is to deal with the error djhurio experienced regarding the el() function -- the function is provided by the methods package, which in any version of R I've used, is loaded by default, but for whatever reason isn't by TIO. If library(methods) is removed from the header and unlist is substituted for el, I gain four bytes, but so would djhurio, putting our byte counts at 96 88 and 99 respectively.

edited Nov 26, 2017 at 18:12

answered Nov 26, 2017 at 0:42

duckmayr

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\$\begingroup\$ Nice one. Never thought for loop will be shorter. You can make it even more shorter by omitting else statement for(i in el(strsplit(scan(,y<-''),y)))y=c(y[y!=i],if(!i%in%y)i);cat(y,sep=''). \$\endgroup\$
– djhurio
Commented Nov 26, 2017 at 15:56
\$\begingroup\$ @djhurio -- I know, it's almost never the case that in R a for loop will help with anything. Regarding your suggestion: Great idea! The suggestion is now incorporated in the answer. \$\endgroup\$
– duckmayr
Commented Nov 26, 2017 at 16:11
1

\$\begingroup\$ @djhurio -- fair enough; I was too busy looking at the difference introduced by omitted the else statement I didn't see how you had changed the beginning. Editing now. Great work! \$\endgroup\$
– duckmayr
Commented Nov 26, 2017 at 18:10
1

\$\begingroup\$ @djhurio @duckmayr there's a 73 byte solution that is basically taking this solution and using a slightly different approach to extracting characters. I didn't really feel like posting it as a separate answer. Also note that ...[[1]] is longer than el(...) but shorter than unlist(...), provided that ... is a list of length 1. \$\endgroup\$
– Giuseppe
Commented Nov 27, 2017 at 19:52
1

\$\begingroup\$ scratch that, I found a 70 bye answer since 0 is the nul character and gets converted to the empty string. \$\endgroup\$
– Giuseppe
Commented Nov 27, 2017 at 19:56

| Show 2 more comments

djhurio · Accepted Answer · 2017-11-29 22:05:54Z

2

R, 84 bytes

y=el(strsplit(scan(,""),""));cat(unique(y[colSums(outer(y,y,"=="))%%2>0],,T),sep="")

Try it online!

Another solution, but there are better R answers here.

R, 88 bytes

z=table(y<-el(strsplit(scan(,""),"")));cat(setdiff(unique(y,,T),names(z[!z%%2])),sep="")

Try it online!

Thanks to Giuseppe for -7 bytes!

There is a shorter answer by duckmayr.

scan(,"") read input from stdin.
y<-el(strsplit(scan(,""),"")) split input by characters and save as y.
z=table(y<-el(strsplit(scan(,""),""))) compute frequencies of each character and save resulting table as z;
unique(y,,T) take unique characters from the right side.
names(z[!z%%2]) select only even counts and extract names.
setdiff(unique(y,,T),names(z[!z%%2])) remove characters with even count.
cat(setdiff(unique(y,,T),names(z[!z%%2])),sep="") print the output.

edited Nov 29, 2017 at 22:05

answered Nov 25, 2017 at 22:11

djhurio

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\$\begingroup\$ The reason for your error is that el() comes from the methods package, which while typically loaded by default, is not by TIO (discussed in my answer below) \$\endgroup\$
– duckmayr
Commented Nov 26, 2017 at 0:43
\$\begingroup\$ why are you using rev(unique(rev(y)))? Wouldn't just unique(y) work? ooohhh wait I see, you want the unique characters from right to left. In that case unique(y,,T) (setting fromLast=T) will be 88 bytes. \$\endgroup\$
– Giuseppe
Commented Nov 27, 2017 at 20:01

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totallyhuman · Accepted Answer · 2017-11-30 01:06:29Z

2

Python 2, 56 bytes

-2 bytes thanks to xnor. -3 bytes thanks to ovs.

lambda s:reduce(lambda a,c:a.replace(c,'')+c[c in a:],s)

Try it online!

Literally just golfed the pseudocode. :P

edited Nov 30, 2017 at 1:06

answered Nov 25, 2017 at 17:02

totallyhuman

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1

\$\begingroup\$ Save 2 bytes: s=(s+c).replace(c,c[c in s:]). \$\endgroup\$
– xnor
Commented Nov 25, 2017 at 23:33
\$\begingroup\$ @xnor That's some basic golfing executed very cleverly. Thanks! \$\endgroup\$
– totallyhuman
Commented Nov 25, 2017 at 23:34
1

\$\begingroup\$ -1 byte : s=s.replace(c,'')+c[c in s:] \$\endgroup\$
– ovs
Commented Nov 26, 2017 at 9:32
1

\$\begingroup\$ 56 bytes using reduce \$\endgroup\$
– ovs
Commented Nov 26, 2017 at 9:37

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Arnauld · Accepted Answer · 2017-11-25 17:51:09Z

1

JavaScript (ES6), 60 bytes

s=>[...s].map(c=>s=s.match(c)?s.split(c).join``:s+c,s='')&&s

Test cases

let f =

s=>[...s].map(c=>s=s.match(c)?s.split(c).join``:s+c,s='')&&s

console.log(f("ABCDBCCBE"))     // -> ADCBE
console.log(f("ABCXYZCABXAYZ")) // -> A
console.log(f("aAABBbAbbB"))    // -> aAbB
console.log(f("GG"))            // -> (empty)

answered Nov 25, 2017 at 17:51

Arnauld

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\$\begingroup\$ I ported @MartinEnder's Retina answer and it was only 45 bytes... \$\endgroup\$
– Neil
Commented Nov 25, 2017 at 19:44

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

APL+WIN, 19 bytes

Logic similar to Galen's J solution.

(2|+⌿⌽<\⌽c∘.=c)/c←⎕

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 25, 2017 at 20:54

Graham

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Misha Lavrov · Accepted Answer · 2017-11-26 07:17:47Z

1

Wolfram Language (Mathematica), 36 bytes

#//.{a___,x_,b___,x_,c___}:>{a,b,c}&

Try it online!

Takes input and output as a list of characters.

How it works

Uses //. (alias ReplaceRepeated) to find two repeated characters and delete both, until no more repeated characters exist. If the character occurs more than twice, Mathematica will always delete the first two occurrences. So if a character occurs an odd number of times, its last instance will always be the one to survive.

edited Nov 26, 2017 at 7:17

answered Nov 26, 2017 at 0:02

Misha Lavrov

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Prolog 81 byte

a([],O,O).
a([I|J],K,O):-delete(K,I,F),(K=F->append(K,[I],M),a(J,M,O);a(J,F,O)).

Non-obfuscated version:

append_and_eraze([], Output, Output).
append_and_eraze([I | Input], Interim, Output) :-
    delete(Interim, I, Filtered),
    ( Interim = Filtered ->
      append(Interim, [I], Interim1),
      append_and_eraze(Input, Interim1, Output)
    ;
    append_and_eraze(Input, Filtered, Output)
    ).

delete/3 ensures that its third argument unifies with its first argument, with all instances of second argument removed from it.
If those turn out to be the same, we append the element (it wasn't removed).
append/3 as per its name, appends an element to list.
We recur on the elements of the input until we hit the [] (empty list), at which point the intermediate result will unify with desired result.

Test:

?- append_and_eraze(`ABCDBCCBE`, [], X), string_codes(Y, X).
X = [65, 68, 67, 66, 69],
Y = "ADCBE".

?- append_and_eraze(`ABCXYZCABXAYZ`, [], X), string_codes(Y, X).
X = [65],
Y = "A".

?- append_and_eraze(`aAABBbAbbB`, [], X), string_codes(Y, X).
X = [97, 65, 98, 66],
Y = "aAbB".

?- append_and_eraze(`GG`, [], X), string_codes(Y, X).
X = [],
Y = "".

Some Prologs treat strings in double quotes as lists, SWI can be configured to do the same, but for the sake of simplicity, I used string_codes/2 to format output nicely.

Nahuel Fouilleul · Accepted Answer · 2017-11-29 06:36:05Z

1

Perl 5, 28 + 2 (-pF) = 30 bytes

$\=~s/$_//g||($\.=$_)for@F}{

Try it online

edited Nov 29, 2017 at 6:36

answered Nov 26, 2017 at 13:25

Nahuel Fouilleul

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Oliver · Accepted Answer · 2019-03-01 22:22:37Z

1

Japt `-h`, 15 bytes

;£A=øX ?AkX:ApX

Try it online!

answered Mar 1, 2019 at 22:22

Oliver

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\$\begingroup\$ 14 bytes? \$\endgroup\$
– Shaggy
Commented Mar 2, 2019 at 0:25
\$\begingroup\$ @Shaggy That won't work because S.k isn't case sensitive. The workaround would be 15 bytes: tio.run/##y0osKPlf@P/Q4gDbwzsiFOwDsiMMrQK0I/7/… \$\endgroup\$
– Oliver
Commented Mar 2, 2019 at 1:02

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Michael Chatiskatzi · Accepted Answer · 2021-01-03 17:40:57Z

1

Scala, 75..64 49 bytes

-22 bytes thanks to user!

_./:("")((?,y)=>if(?toSet y)?filter(y!=)else? +y)

Try it online!

edited Jan 3, 2021 at 17:40

answered Jan 3, 2021 at 0:42

Michael Chatiskatzi

1,4418 silver badges13 bronze badges

1

\$\begingroup\$ Try using /: (foldLeft) here, it may save some bytes (I’m on mobile, so I can’t try it out right now) \$\endgroup\$
– user
Commented Jan 3, 2021 at 1:23
1

\$\begingroup\$ You don't need to print, just return directly: 49 bytes. Also, I think toSet works instead of contains \$\endgroup\$
– user
Commented Jan 3, 2021 at 17:26

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Martin Ender · Accepted Answer · 2017-11-25 17:27:53Z

0

Alice, 9 bytes

/X&@
\io/

Try it online!

Explanation

Basically a port of Erik's answer. Apart from a bit of IP redirection the code is really just:

i&Xo@

which does:

i   Read all input.
&X  Fold symmetric multiset difference over the input.
o   Output the result.
@   Terminate.

answered Nov 25, 2017 at 17:27

Martin Ender

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Erik the Outgolfer · Accepted Answer · 2017-11-25 21:34:48Z

0

APL (Dyalog), 16 bytes

{(,⍨~∩)/⍣(≢⍵)⊖⍵}

Try it online!

If errors were allowed, this would've been 9 bytes:

(,⍨~∩)/∘⊖

answered Nov 25, 2017 at 21:34

Erik the Outgolfer

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\$\begingroup\$ What do you mean by errors? \$\endgroup\$
– FrownyFrog
Commented Nov 25, 2017 at 21:51
\$\begingroup\$ @FrownyFrog The 9-byte version would throw a DOMAIN ERROR if the string is empty, since (,⍨~∩) doesn't have a predefined identity element. \$\endgroup\$
– Erik the Outgolfer
Commented Nov 26, 2017 at 10:29

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chromaticiT · Accepted Answer · 2017-11-25 21:58:22Z

0

Pyth, 15 bytes

Vz?:kN)=-kN=+kN

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answered Nov 25, 2017 at 21:58

chromaticiT

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Justin Mariner · Accepted Answer · 2017-11-25 22:03:46Z

Ruby, 53 bytes

->s{s.reverse.uniq.select{|c|s.count(c)%2>0}.reverse}

Try it online!

Input and output are both an array of chars. Test code calls .chars and .join for convenience.

Explanation

Uses the fact that the letters in the resulting string appear an odd number of times and in the order from right to left.

->s{                # lambda function taking char-array argument
    s.reverse           # reverse the input
    .uniq               # get unique characters
    .select{|c|         # select only those which...
        s.count(c)%2>0      # appear in the input array an odd number of times
    }.reverse           # reverse back and return
}

Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Pyth, 13 bytes

{_xD_Qf%/QT2Q

Takes in input as list of characters. Test it out!

      f     Q            (f)ilter input (Q)
        /QT              On how many times (/) each character (T) appears in the 
                           input (Q)
       %   2             Only allow odd numbers of occurences (when x % 2 = 1)
 _xD_Q                   Sort (D) descending (the first _) by the location (x) of 
                           the last (the second _) inde(x) of the target character
                           in the input (Q)
{                        Remove duplicates

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 26, 2017 at 7:30

Steven H.

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\$\begingroup\$ 10 bytes, but returns 0 instead of an empty string. \$\endgroup\$
– hakr14
Commented Jan 3, 2021 at 1:41

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fergusq · Accepted Answer · 2017-11-26 14:33:54Z

Röda, 34 bytes

{a=[]a-=_ if[_1 in a]else a+=_1;a}

Try it online!

This is a direct translation of the pseudocode. It treats input and output as streams of characters.

Explanation:

{                    /* Anonymous function                   */
    a=[]             /* initialize a                         */
                     /* For each character _1 in the stream: */
    a-=_ if[_1 in a] /*  Remove it from a if a contains it   */
    else a+=_1;      /*  Otherwise append it to a            */
    a                /* Push characters in a to the stream   */
}

FlipTack · Accepted Answer · 2017-11-26 16:25:43Z

0

Python 3, 73 bytes

Not the shortest, but I like this approach.

lambda s:''.join(c*(s.count(c)%2)*(i==s.rfind(c))for i,c in enumerate(s))

Try it online!

Loops through the string, keeping only those characters where:

(s.count(c)%2) == 0 - The character appears an even number of times.
(i==s.rfind(c)) - The current index is the last appearance of the character in question.

edited Nov 26, 2017 at 16:25

answered Nov 26, 2017 at 16:17

FlipTack

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idrougge · Accepted Answer · 2017-11-27 18:35:14Z

0

REXX, 102 bytes

a=arg(1)
s=''
do while a>''
  b=right(a,1)
  if countstr(b,a)//2 then s=b||s
  a=changestr(b,a,'')
  end
say s

Try it online!

How it works: Take the rightmost letter, see if the number of occurrences is even or odd (which also doubles as a truth value) and if odd, add it to the output string. Then remove all occurrences of the letter from the input string. Repeat until input is depleted.

answered Nov 27, 2017 at 18:35

idrougge

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Xcali · Accepted Answer · 2017-11-27 23:52:38Z

0

Perl 5, 22 + 1 (`-p`) = 23 bytes

s/(.)(.*?)\1/$2/&&redo

Try it online!

answered Nov 27, 2017 at 23:52

Xcali

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Jakob · Accepted Answer · 2017-11-28 01:45:03Z

Java 8, 93 bytes

A lambda from String to String. Just an implementation of the pseudocode in the question.

s->{String o="";for(char c:s.toCharArray())o=o.indexOf(c)<0?o+c:o.replace(c+"","");return o;}

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Java 8, 182 bytes

Here's another lambda of the same type that uses streams! It's probably more efficient.

s->s.join("",s.chars().mapToObj(c->(char)c+"").filter(c->s.replaceAll("[^"+c+"]","").length()%2>0).distinct().sorted((c,d)->s.lastIndexOf(c)-s.lastIndexOf(d)).toArray(String[]::new))

Try It Online

Ungolfed

s ->
    s.join(
        "",
        s.chars()
            .mapToObj(c -> (char) c + "")
            .filter(c -> s.replaceAll("[^" + c + "]", "").length() % 2 < 0)
            .distinct()
            .sorted((c, d) -> s.lastIndexOf(c) - s.lastIndexOf(d))
            .toArray(String[]::new)
    )

Giuseppe · Accepted Answer · 2017-11-29 22:30:26Z

R, 70 bytes

function(s){for(i in utf8ToInt(s))F=c(F[F!=i],i*!i%in%F);intToUtf8(F)}

Try it online!

I was encouraged by djhurio to post this solution; djhurio's answer can be found here.

This uses the same idea as duckmayr's answer, but it leverages a numeric approach by converting the string to its codepoints rather than splitting it into characters, and is a function rather than a full program so it can return the new string rather than printing to stdout.

function(s) {
 for(i in utf8ToInt(s))           # convert string to codepoints and iterate over it
  F=c(F[F!=i],                    # remove duplicates and append
      i*!i%in%F)                  # 0 if in F, i otherwise
 intToUtf8(F)                     # collapse from codepoints to string
}

One important observation is that F is initialized to FALSE or 0 and utf8ToInt(0)=="", so this will succeed for the empty string as well as correctly collapsing the codepoints.

Append and erase

36 Answers 36

Jelly, 3 bytes

Haskell, 44 42 bytes

J, 21 19 bytes

Brainfuck, 95 bytes

How It Works

Haskell, 47 bytes

Retina, 16 bytes

K (ngn/k), 21 9 bytes

MATL, 6 bytes

q, 38 bytes

R, 92 84 77 bytes

Explanation

Note

R, 84 bytes

R, 88 bytes

Python 2, 56 bytes

JavaScript (ES6), 60 bytes

Test cases

APL+WIN, 19 bytes

Wolfram Language (Mathematica), 36 bytes

How it works

Prolog 81 byte

Perl 5, 28 + 2 (-pF) = 30 bytes

Japt -h, 15 bytes

Scala, 75..64 49 bytes

Alice, 9 bytes

Explanation

APL (Dyalog), 16 bytes

Pyth, 15 bytes

Ruby, 53 bytes

Explanation

Pyth, 13 bytes

Röda, 34 bytes

Python 3, 73 bytes

REXX, 102 bytes

Perl 5, 22 + 1 (-p) = 23 bytes

Java 8, 93 bytes

Java 8, 182 bytes

Ungolfed

R, 70 bytes

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Japt `-h`, 15 bytes

Perl 5, 22 + 1 (`-p`) = 23 bytes

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