Split string on first occurrence of each character

Question

Related.

Given a printable ASCII string, split it into a list of non-empty strings with a new sub-string beginning every time a character, which has not previously been seen in the same case, occurs.

Examples

"mississippi" → ["m","i","ssissi","ppi"]

"P P & C G" → ["P"," P ","& ","C ","G"]

"AAA" → ["AAA"]

"Adam" → ["A","d","a","m"]

"" → []

---

^{Anecdote: The result will have between 0 and 95 elements. The 95th sub-string will necessarily continue until the end because at that point, all printable ASCII characters have begun a sub-string, so every additional character will have occurred before and thus cannot cause a new sub-string to begin.}

Answer 1

Jelly, 4 bytes

QƤĠị

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Explanation

QƤĠị  Input is a string, say s = "adam"
 Ƥ    For each prefix of s: ["a","ad","ada","adam"]
Q     remove duplicates: ["a","ad","ad","adm"]
  Ġ   Group indices by equal values: [[1],[2,3],[4]]
   ị  Index into s: ["a","da","m"]

The internal representation of the strings, which the TIO link displays, is slightly different.

Answer 2

Retina, 9 bytes

q1,`.
¶$&

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Explanation

Match each character (.), discard repeated matches (q), discard the first match (1,), and insert a linefeed in front of each match ¶$&.

Answer 3

05AB1E, 11 bytes

ÙSk¥sg¸«£õK

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Explanation

Ù             # remove duplicates in input
 S            # split to a list of characters
  k           # get the (first) index of each character in the input
   ¥          # calculate delta's
    sg¸«      # append the length of the input
        £     # split the list into pieces of these sizes
         õK   # remove empty string (for the special case "" -> [])

Answer 4

J, 7 bytes

~:<;.1]

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Explanation

Nub sieve's chance to shine!

~: <;.1 ]
        ]  Input
~:         Nub sieve (1 if the character is the first instance in string)
    ;.1    Split input on 1s in nub sieve
   <       And box each

Answer 5

APL (Dyalog Unicode), 8 bytes^SBCS

(≢¨∪\)⊆⊢

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Answer 6

05AB1E, 8 bytes

Ùvyy¶ì.;

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---

Always will output 1 preceding newline, which is constant and not indicative of a split, the 10-byte alternative that does not output a preceding newline is Ùvyy¶ì.;}¦, you can try that here. According to Adam a preceding or trailing newline is acceptable.

---

Input      = mississippi                               | Stack
-----------#-------------------------------------------+----------------------------------
Ù          # Push unique letters of first input.       | ['misp']
 v         # Iterate through each unique letter.       | []
  yy       # Push 2 copies of the letter (or yD)       | ['m','m']
    ¶      # Push a newline char.                      | ['m','m','\n']
     ì     # Prepended to the letter.                  | ['m','\nm']
      .;   # Replace first instance with '\n + letter' | ['\nmississippi']

After each iteration we get:

['\nmississippi'] > ['\nm\nississippi'] > ['\nm\ni\nssissippi'] > ['\nm\ni\nssissi\nppi']

Which is:

m
i
ssissi
ppi

Answer 7

C,  75   65  63 bytes

Thanks to @Digital Trauma for saving 10 bytes and thanks to both @gastropner and @l4m2 for saving a byte each!

f(char*s){for(int l[128]={};*s;putchar(*s++))l[*s]++||puts(l);}

Prints a leading newline.

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Without a leading newline (71 bytes):

f(char*s){int l[128]={};for(l[*s]=1;*s;putchar(*s++))l[*s]++||puts(l);}

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Answer 8

Perl 6,  58 52  40 bytes

{$/={};.comb.classify({$+=!$/{$_}++}).sort».value».join}

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*.comb.classify({$+=!(%){$_}++}).sort».value».join

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*.classify({$+=!(%){$_}++}).sort».value

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(input is a list of characters, and output is a list of lists of characters)

Expanded:

*                   # parameter for WhateverCode lambda

  .classify(        # classify that list
    {
        $           # anonymous scalar state variable (accumulator)

      +=            # increment it if:

        !           # Bool invert the following
          (
            %       # anonymous hash state variable
          ){ $_ }++ # look to see if the character was seen already
    }
  ).sort\           # sort the Pairs by key (makes the order correct)
  ».value           # get the value from each Pair

---

The output from classify is

{ # Hash
  1 => ['m'],
  2 => ['i'],
  3 => ['s','s','i','s','s','i'],
  4 => ['p','p','i'],
}

And .sort just turns it into:

[
  1 => ['m'],
  2 => ['i'],
  3 => ['s','s','i','s','s','i'],
  4 => ['p','p','i'],
]

».value removes the keys

[
  ['m'],
  ['i'],
  ['s','s','i','s','s','i'],
  ['p','p','i'],
]

Answer 9

Haskell, 39 bytes

foldl(\s c->s++['\n'|all(/=c)s]++[c])""

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Inserts a newline symbol before every character that appears for the first time, resulting in a newline-separated string, with a leading newline. Prepend lines. to produce a list.

---

Haskell, 55 bytes

(""%)
_%[]=[]
p%s|(a,b)<-span(`elem`s!!0:p)s=a:(a++p)%b

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Repeatedly takes the prefix the first character plus the non-unique characters that follow it.

Answer 10

APL (Dyalog), 9 bytes

Thanks, Erik the Outgolfer for saving 1 byte!

⊢⊂⍨⍳∘≢∊⍳⍨

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Explanation:

⍳⍨: For each character, get the index of its first occurrence. e.g mississippi -> 1 2 3 3 2 3 3 2 9 9 2

⍳∘≢: The range from 1 to the length of the input.

∊: Membership. e.g 1 2 3 4 5 6 7 8 9 10 11∊1 2 3 3 2 3 3 2 9 9 2 -> 1 1 1 0 0 0 0 0 1 0 0

⊢⊂⍨: Partition the input string with new partitions starting at 1s in the vector above

Answer 11

JavaScript (ES6), 37 bytes

Saved 7 bytes: a leading newline was explicitly allowed (Thanks @Shaggy!)

Takes input as an array of characters. Outputs a newline-separated string.

s=>s.map(c=>s[c]=s[c]?c:`
`+c).join``

Test cases

let f =

s=>s.map(c=>s[c]=s[c]?c:`
`+c).join``

;[
  [..."mississippi"],
  [..."P P & C G"],
  [..."AAA"],
  [..."Adam"],
  [...""]    
]
.forEach(s => console.log('{' + f(s) + '}\n'))

Answer 12

Japt, 11 bytes

‰ r@=iRUbY

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Explanation

This was inspired by Magic Octopus Urn's 05AB1E solution.

‰ r@=iRUbY    Implicit: U = input string
‰             Split U into chars, and keep only the first occurrence of each.
   r@          Reduce; for each char Y in this string...
        UbY      Find the first index of Y in U.
      iR         Insert a newline at this index in U.
     =           Set U to the result.
               As reduce returns the result of the last function call, this gives the
               value of U after the final replacement, which is implicitly printed.

Answer 13

Python 2, 81 74 bytes

def f(s):d=sorted(map(s.find,set(s)));print map(lambda a,b:s[a:b],d,d[1:])

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Answer 14

Python 2, 47 bytes

lambda s:reduce(lambda r,c:r+'\n'[c in r:]+c,s)

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Outputs a newline-separated string. Barely beats the program version:

Python 2, 48 bytes

r=''
for c in input():r+='\n'[c in r:]+c
print r

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Answer 15

R, 94 87 bytes

function(s,n=nchar(s),g=substring)g(s,d<-which(!duplicated(g(s,1:n,1:n))),c(d[-1]-1,n))

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Returns a (possibly empty) list of substrings.

Thanks to Michael M for saving 7 bytes!

Answer 16

brainfuck, 66 bytes

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

Formatted:

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

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The leading newline in the output (which is only printed if the input is non-empty) can be removed at the cost of 5 bytes by replacing the body x of the main (outermost) loop with .>,[x].

Answer 17

BQN, 11 6 bytes

⌈`∘⊐⊸⊔

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Explanation

⌈`∘⊐⊸⊔
   ⊐    Classify: replace each letter with a number based on the order of
        their first occurrences
⌈`∘     Scan with maximum: replace each number with the largest number so far
    ⊸⊔  Group the original string into those buckets

A couple of worked examples:

"dodos"            "Mississippi"
⟨ 0 1 0 1 2 ⟩       ⟨ 0 1 2 2 1 2 2 1 3 3 1 ⟩
⟨ 0 1 1 1 2 ⟩       ⟨ 0 1 2 2 2 2 2 2 3 3 3 ⟩
⟨ "d" "odo" "s" ⟩   ⟨ "M" "i" "ssissi" "ppi" ⟩

Answer 18

R, 86 bytes

function(s)strsplit(intToUtf8(Reduce(function(a,b)c(a,9*!b%in%a,b),utf8ToInt(s))),"	")

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Based on the great +1 xnor's answer with some R-specific features.

Takes a string as an input and outputs a vector of strings.

EDIT: added strsplit(...) to output as a vector.

Answer 19

Jelly, 6 bytes

ŒQœṗ⁸Ḋ

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Answer 20

K4, 19 bytes

Solution:

$[#x;(*:'.=x)_;,]x:

Examples:

q)k)$[#x;(*:'.=x)_;,]x:"mississippi"
,"m"
,"i"
"ssissi"
"ppi"
q)k)$[#x;(*:'.=x)_;,]x:"P P & C G"
,"P"
" P "
"& "
"C "
,"G"
q)k)$[#x;(*:'.=x)_;,]x:"AAA"
"AAA"
q)k)$[#x;(*:'.=x)_;,]x:"Adam"
,"A"
,"d"
,"a"
,"m"
q)k)$[#x;(*:'.=x)_;,]x:""
,[""]

Explanation:

8 bytes is just to handle ""...

$[#x;(*:'.=x)_;,]x: / the solution
                 x: / save input as variable x
$[  ;         ; ]   / $[condition;true;false]
  #x                / length of x ("" has length 0, i.e. false)
             _      / cut right at left indices
     (      )       / do together
          =x        / group x into key/value (char!indices)
         .          / return value (get indices)
      *:'           / first (*:) each
               ,    / enlist, "" => [""]

Answer 21

Perl, 30 bytes

Includes +1 for p

Give input without trailing newline on STDIN. Output is also without trailing newline:

echo -n adam | perl -pE 's%.%$v{$&}+++!pos?$&:$/.$&%eg'; echo

If you don't care about leading and trailing newlines this 25 (+3 for -p because the code contains ') also works:

#!/usr/bin/perl -p
s%%$/x!$v{$'&~v0}++%eg

Answer 22

JavaScript, 61 54 52 bytes

Takes input as an array of characters.

s=>s.map(x=>a[x]?a[y]+=x:a[x]=a[++y]=x,a=[],y=-1)&&a

---

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o.innerText=JSON.stringify((f=
s=>s.map(x=>a[x]?a[y]+=x:a[x]=a[++y]=x,a=[],y=-1)&&a
)([...i.value=""]));oninput=_=>o.innerText=JSON.stringify(f([...i.value]))

<input id=i><pre id=o></pre>

Answer 23

Stax, 8 bytes

ç↓‼►▐NVh

Run and debug online

The ascii representation of the same program is this.

c{[Ii=}(m

For each character, it splits when the index of the current character is the current position.

c            copy the input
 {    }(     split the string when the result of the enclosed block is truthy
  [          duplicate the input string under the top of the stack
   I         get the character index of the current character
    i=       is it equal to the iteration index?
        m    print each substring

Answer 24

Java 8, 193 169 155 151 bytes

s->{for(int l=s.length(),i=0,j;i<l;i++)if(s.indexOf(s.charAt(i))==i){for(j=i;++j<l&&s.indexOf(s.charAt(j))!=j;);System.out.println(s.substring(i,j));}}

-14 bytes thanks to @raznagul (for something obvious I somehow missed myself..)
-3 bytes thanks to @O.O.Balance (again for something obvious I somehow missed myself.. :S)

Explanation:

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s->{                    // Method with String parameter and no return-type
  for(int l=s.length(), //  The length of the input-String
          i=0,j;        //  Index integers
      i<l;i++)          //  Loop `i` from 0 to `l` (exclusive)
    if(s.indexOf(s.charAt(i))==i){
                        //   If the character at index `i` hasn't occurred yet:
      for(j=i;++j<l     //    Inner loop `j` from `i` to `l` (exclusive),
          &&s.indexOf(s.charAt(j))!=j;);
                        //     as long as the character at index `j` has already occurred
      System.out.println(//    Print:
        s.substring(i,j));}}
                        //     The substring of the input from index `i` to `j` (exclusive)

Answer 25

><>, 22 17 14 bytes

-1 byte thanks to Emigna

i:::a$1g?!o1po

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Prints a leading and trailing newline.

It keeps track of which letters have already appeared by putting a copy of the character at that corresponding spot on the second row, and printing a newline if the value fetched from that position was not 1. Ends in an error when it tries to print -1

Answer 26

Perl 5 + -pl, 22 bytes

s@.@$/x!$h{$&}++.$&@ge

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Answer 27

R, 83 82 bytes

function(s)if(s>"")substring(s,i<-which(!duplicated(utf8ToInt(s))),c(i[-1],1e6)-1)

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Different approach to int 21h Glory to Ukraine's answer.

Note, though, that int 21h Glory to Ukraine could still beat this by reverting to the (allowed) output format of a single newline-separated string.

(Edit: after scrolling down after posting, I noticed that this approach is similar an earlier answer by Giuseppe, but saves some bytes by using utf8ToInt instead of substring to split-up the input vector).

Answer 28

Haskell, 62 bytes

r#c|any(elem c)r=init r++[last r++[c]]|1<2=r++[[c]]
foldl(#)[]

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Answer 29

JavaScript (ES6), 68 bytes

s=>s.map(c=>o[c]?t+=c:(t&&m.push(t),t=o[c]=c),t='',o=m=[])&&[...m,t]

Takes input as a list of characters.

Test cases:

let f=
s=>s.map(c=>o[c]?t+=c:(t&&m.push(t),t=o[c]=c),t='',o=m=[])&&[...m,t]

console.log(f([...'mississippi']));
console.log(f([...'P P & C G']));
console.log(f([...'AAA']));
console.log(f([...'Adam']));
console.log(f([...'']));

Answer 30

PHP, 317 bytes

function SplitOnFirstUnique($s){
    $len = strlen($s); 
    $output = [];
    $newstring = '';
    for ($i=0; $i < $len ; $i++) { 
        $newstring = $newstring.$s[$i];
        if(!in_array($s[$i] , $output  )){
            $output[] = $newstring;
            $newstring = '';
        }
    }
    return $output;
}

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