# All together now

Given a list of digits 1 through 9, output whether each digit is grouped together as a single contiguous block. In other words, no two of the same digit are separated by different digits. It's OK if a digit doesn't appear at all. Fewest bytes wins.

Input: A non-empty list of digits 1 through 9. This can be as a decimal number, string, list, or similar sequence.

Output: A consistent Truthy value if all the digits are grouped in contiguous blocks, and a consistent Falsey value if they are not.

True cases:

3
51
44999911
123456789
222222222222222222222


False cases:

818
8884443334
4545
554553
1234567891


var QUESTION_ID=77608,OVERRIDE_USER=20260;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/77608/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.lang>s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
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• Would a list of singleton strings be an acceptable input format? Apr 12, 2016 at 0:22
• Yes, singletons are fine.
– xnor
Apr 12, 2016 at 1:31
• Can anyone tell me what the most efficient algorithm for this problem would be? Or is there a more general problem that this falls under that I can look up?
– user53017
Apr 14, 2016 at 12:36
• @amt528 You can do it in linear time by iterating over each digit and checking that there's no runs of it past the first.
– xnor
Apr 15, 2016 at 2:15
• Could you provide an example of how it's implemented?
– user53017
Apr 15, 2016 at 15:31

# Python 3, 3834 33 bytes

lambda s:s==sorted(s,key=s.index)


This expects a list of digits or singleton strings as argument. Test it on Ideone.

Thanks to @xsot for golfing off 4 bytes!

Thanks to @immibis for golfing off 1 byte!

• If you are allowed to accept a list of strings instead, you can shorten this to lambda s:s==sorted(s,key=s.find)
– xsot
Apr 12, 2016 at 0:19
• Ah, I tried taking a list, but I didn't think of using backticks... I'll ask the OP. Apr 12, 2016 at 0:22
• Am I missing something - why can't you just use s.find? Apr 12, 2016 at 4:31
• @immibis s has to be a list of singleton strings (or I'd have to cast s to list for the comparison), and list.find is not defined... Apr 12, 2016 at 4:33
• @Dennis s.index then? Seems to work for me. Apr 12, 2016 at 4:37

## JavaScript (ES6), 27 bytes

s=>!/(.)(?!\1).*\1/.test(s)


Uses negative lookahead to look for two non-contiguous digits. If at least two such digits exist, then they can be chosen so that the first digit precedes a different digit.

• Or, just use a regex XD. That works too. Apr 12, 2016 at 0:01
• ahem Retina ahem Apr 12, 2016 at 5:58

# 05AB1E, 4 bytes

Code:

Ô¹ÙQ


Explanation:

Ô     # Push connected uniquified input. E.g. 111223345565 would give 1234565.
¹    # Push input again.
Ù   # Uniquify the input. E.g. 111223345565 would give 123456.
Q  # Check if equal, which yields 1 or 0.


Uses CP-1252 encoding.

Try it online!

• You...just beat jelly...I never thought this was possible... Apr 15, 2016 at 8:17

# Jelly, 5 bytes

ĠIFPỊ


Try it online!

### How it works

ĠIFPỊ  Main link. Input: n (list of digits or integer)

Ġ      Group the indices of n by their corresponding values, in ascending order.
For 8884443334, this yields [[7, 8, 9], [4, 5, 6, 10], [1, 2, 3]].
I     Increments; compute the all differences of consecutive numbers.
For 8884443334, this yields [[1, 1], [1, 1, 4], [1, 1]].
F    Flatten the resulting 2D list of increments.
P   Product; multiply all increments.
Ị  Insignificant; check if the product's absolute value is 1 or smaller.

• Five bytes you say? What kind of encoding is that? Apr 12, 2016 at 5:56
• Jelly has its own code page, which encodes each of the 256 characters it understands as a single byte. Apr 12, 2016 at 5:57

# Pyth, 6 5 bytes

1 bytes thanks to FryAmTheEggman

SIxLQ


Inspired by the Python solution here.

Test suite

Explanation:

SIxLQ
xLQ   Map each element in the input to its index in the input. Input is implicit.
SI      Check whether this list is sorted.

• SIxLQ seems to work. Apr 12, 2016 at 1:12
• This is genius. Apr 12, 2016 at 1:52
• The second Q doesn't seem to get parsed properly, it swaps argument order or something so you get all 0s and it always gives true. Here's a test suite. Apr 12, 2016 at 1:55

# MATL, 8 bytes

t!=tXSP=


The output is an array containing only ones for truthy, or an array containing at least one zero for falsey.

Try it online!

### Explanation

Consider the input 22331, which satisfies the condition. Testing if each character equals each other gives the 2D array

1 1 0 0 0
1 1 0 0 0
0 0 1 1 0
0 0 1 1 0
0 0 0 0 1


The final result should be truthy if the rows of that array (considered as atomic) are in (lexicographical) decreasing order. For comparison, input 22321 gives the array

1 1 0 1 0
1 1 0 1 0
0 0 1 0 0
1 1 0 1 0
0 0 0 0 1


in which the rows are not sorted.

t!   % Take string input. Duplicate and tranpose
=    % Test for equality, element-wise with broadcast: gives a 2D array that
% contains 0 or 1, for all pairs of characters in the input
t    % Duplicate
XS   % Sort rows (as atomic) in increasing order
P    % Flip vertically to obtain decreasing order
=    % Test for equality, element-wise


# R, 66484643 38 bytes

function(s)!any(duplicated(rle(s)$v))  This is a function that accepts the input as a vector of digits and returns a boolean. To call it, assign it to a variable. Not the shortest but I thought it was a fun approach. We run length encode the input and extract the values. If the list of values contains duplicates then return FALSE, otherwise return TRUE. Verify all test cases online Saved 20 bytes thanks to MickyT, 3 thanks to Albert Masclans, and 5 thanks to mnel! • 26 by switching to latest R version and using table. Aug 21, 2022 at 18:36 ## Retina, 17 bytes M(.)(?!\1).+\1 0  Try it online! (Slightly modified to run all test cases at once.) The first regex matches digits which are separated by other digits, so we get a 0 for valid inputs and anywhere between 1 and 9 for invalid inputs (due to the greediness of the the .+, we can't get more than n-1 matches for n different digits). To invert the truthiness of the result, we count the number of 0s, which is 1 for valid inputs and 0 for invalid ones. • I made a shorter one, but it's close enough to yours that it should be a comment instead. Use AntiGrep instead of Match, then remove the last line: A(.)(?!\1).+\1 for 15 bytes. Also works for multiple inputs. Truthy is the input, falsy is nothing. One does not simply out-golf Martin at his own language. :) Apr 22, 2016 at 19:11 • @mbomb007 I think I actually considered that, but unfortunately, the challenge asks for a consistent truthy (and falsy) value, so printing the input as truthy isn't allowed. Apr 22, 2016 at 19:43 # J, 8 bytes -:]/:i.~  Test it with J.js. ### How it works -:]/:i.~ Monadic verb. Argument: y (list of digits) i.~ Find the index in y of each d in y. ]/: Sort y by the indices. -: Match; compare the reordering with the original y.  • :] :i :-1 Apr 13, 2016 at 2:02 • Not sure if joke or golfing suggestion... Apr 13, 2016 at 2:19 ## Java, 161 156 bytes Because Java... Shamelessly stealing borrowing the regex from this answer because I started out trying to do this with arrays and math manipulation, but it got hideously complex, and regex is as good a tool as any for this problem. import java.util.regex.*;public class a{public static void main(String[] a){System.out.println(!Pattern.compile("(.)(?!\\1).*\\1").matcher(a[0]).find());}}  Ungolfed: import java.util.regex.*; public class a { public static void main(String[] args) { System.out.println(!Pattern.compile("(.)(?!\\1).*\\1").matcher(args[0]).find()); }  Laid out like a sensible Java person: import java.util.regex.Matcher; import java.util.regex.Pattern; public class { public static void main(String[] args) { Pattern p = Pattern.compile("(.)(?!\\1).*\\1"); Matcher m = p.matcher(args[0]); System.out.println(!m.find()); } }  • like a sensible Java person That would be, not using Java ever. – cat Apr 13, 2016 at 20:07 • Other solutions are just providing a function, would make it a lot shorter. Something like s->s.match("(.)(?!\\1).*\\1") Apr 13, 2016 at 21:47 • But then we couldn't revel in the verboseness of the answer. Apr 14, 2016 at 0:54 # Pyth, 7 bytes {IeMrz8  # Ruby, 23 bytes Anonymous function. Accepts a string. Regex strat. ->n{/(.)(?!\1).*\1/!~n}  Regex breakdown /(.)(?!\1).*\1/ (.) # Match a character and save it to group 1 (?!\1) # Negative lookahead, match if next character isn't # the same character from group 1 .* # Any number of matches \1 # The same sequence as group 1  !~ means if there are no matches of the regex within the string, return true, and otherwise return false. # Mathematica, 26 bytes 0<##&@@Sort[#&@@@Split@#]&  # MATL, 13 11 bytes u"G@=fd2<vA  Thanks to Luis Mendo for saving two bytes! Try it Online! Explanation  % Grab the input implicitly u % Find the unique characters " % For each of the unique characters G % Grab the input again @= % Determine which chars equal the current char f % Find the locations of these characters d % Compute the difference between the locations 2< % Find all index differences < 2 (indicating consecutive chars) v % Vertically concatenate all stack contents A % Ensure that they are all true % Implicit end of the for loop  • You can take the input with quotes (allowed by default) and remove j. Also, I think you can move vA within the loop and remove ] Apr 12, 2016 at 8:52 • @LuisMendo Thanks! I had messed around with putting Y& inside but that didn't work because fd2< can be empty. Moving vA inside works great though! Also I really wish we had a stable unique that didn't take up tons of bytes. Apr 12, 2016 at 11:24 • Now stable unique takes a little less, using a number instead of the predefined string. I may add a shorter version in the future, though. Or just make u stable by default (you could always include S afterwards, two bytes). What do you think? Apr 12, 2016 at 12:13 # Jelly, 4 bytes i@ÞƑ  Try it online! Semi-based on several of DLosc's 5-byters.  Ƒ The input is not changed by Þ sorting its elements by i@ their first indices in the input.  Kind of like cutting the F out of ¹ƙF⁼, but also kind of like cutting the ṢƑ out of iⱮṢƑ. • @thejonymyster Should be clearer now Aug 21, 2022 at 19:05 # Regex (Perl / PCRE / Java / Boost / Pythonregex / Ruby), 17 bytes ^(?!(.)+\1*+.+\1)  Try it online! - Perl Try it online! - PCRE Try it online! - Java Try it online! - Boost Try it online! - Python import regex Try it online! - Ruby ^ # Anchor to start of string (?! # Assert that the following can't match: (.)+ # Skip any number of characters (minimum zero), then capture # the next character in \1 \1*+ # Skip all subsequent occurrences of \1, and lock in that match # using a possessive quantifier. .+ # Skip at least one character, or as many as necessary to make the # following match: \1 # Match the character captured in \1 )  ## Regex (Perl / PCRE / Java / Boost / Pythonregex / Ruby), 11 bytes (.)\1*+.+\1  Returns a match for false, and a non-match for true. Try it online! - Perl Try it online! - PCRE Try it online! - Java Try it online! - Boost Try it online! - Python import regex Try it online! - Ruby This beats the regex in the previous Java answer, Perl answer, and Ruby answer by 2 bytes. (.) # \1 = a character anywhere in the string \1*+ # Skip all subsequent occurrences of \1, and lock in that match using a # possessive quantifier. .+ # Skip at least one character, or as many as necessary to make the # following match: \1 # Match the character captured in \1  ## Regex (ECMAScript (or better) / Python / .NET), 13 bytes (.)(?!\1).+\1  Returns a match for false, and a non-match for true. Try it online! - ECMAScript Try it online! - Perl Try it online! - PCRE Try it online! - Java Try it online! - Boost Try it online! - Python Try it online! - Python import regex Try it online! - Ruby Try it online! - .NET ## $$\\large\textit{Anonymous functions}\$$ # Ruby, 19 bytes ->s{!/(.)\1*+.+\1/}  Try it online! # JavaScript (ES6), 27 bytes s=>!/(.)(?!\1).+\1/.test(s)  Try it online! Included for completeness; already done in Neil's answer. # PowerShell, 29 bytes !($args-match'(.)(?!\1).+\1')


Try it online!

# Julia, 30 bytes

s->!occursin(r"(.)\1*+.+\1",s)


Attempt This Online!

# R, 32 bytes

\(s)!grepl('(.)\\1*+.+\\1',s,,1)


Attempt This Online!

Beaten by Alex A.'s non-regex answer by 2 bytes, or by 6 bytes with pajonk's improvement.

# Java, 33 bytes

s->!s.matches("(.)+\\1*+.+\\1.*")


Try it online!

# PHP, 39 bytes

fn($s)=>!preg_match('/(.)\1*+.+\1/',$s)


Try it online!

# Perl-p, 17 bytes (18 bytes including flag)

$_=!/(.)\1*+.+\1/  Try it online! # Retina 0.8.2, 17 bytes M(.)(?!\1).+\1 0  Try it online! Included for completeness; already done in Martin Ender's answer. # Retina 1.0, 17 bytes C(.)(?!\1).+\1 0  Try it online! # Pip, 19 18 bytes (.)(?!\1).+\1NIa  Try it online! Thanks to DLosc. • Aug 21, 2022 at 18:36 • @pajonk Thanks, edited. It also works with utf8ToInt() instead of el(strsplit(,"")). Aug 21, 2022 at 18:51 • 13 bytes in Python re flavor with match = falsey / no match = truthy: (.)(?!\1).+\1 (Try it online!) Aug 21, 2022 at 20:47 • (Note: the Pip program can be 18 bytes because the space isn't necessary.) Sep 3, 2022 at 5:16 ## Haskell, 44 bytes import Data.List ((==)<*>nub).map head.group  Usage example: ((==)<*>nub).map head.group$ "44999911" -> True.

A non-pointfree version:

f x = q == nub q                -- True if q equals q with duplicates removed
where
q = map head $group x -- group identical digits and take the first -- e.g. "44999911" -> ["44","9999","11"] -> "491" -- e.g "3443311" -> ["3","44","33","11"] -> "3431"  # Python, 56 55 bytes a=lambda s:~(s[0]in s.lstrip(s[0]))&a(s[1:])if s else 1  • Fails in Python 3.4.1 (int not subscriptable) Apr 13, 2016 at 1:38 • Saved an extra byte with ~(which literally is equivalent to1-): a=lambda s:~(s[0]in s.lstrip(s[0]))&a(s[1:])if s else 1 Apr 13, 2016 at 1:53 # C#, 119 bytes bool m(String s){for(int i=0;i<9;i++){if(new Regex(i.ToString()+"+").Matches(s).Count>1){return false;}}return true;}  ## Ungolfed bool m(String s) { for(int i=0;i<9;i++) { if(new Regex(i.ToString() + "+").Matches(s).Count > 1) { return false; } } return true; }  • Welcome to PPCG! Instead of deleting a post and making a new post with the fixed version, you could also edit your old post and then undelete it. (No need to do that now that there's two posts already anyway, but just so you know in the future.) Apr 14, 2016 at 13:06 • My bad. When I first intended to participate in this Code Golf I misread the objective and didn't had much time to do another solution ( and knowing myself, I wouldn't try to correct the previous posted solution ). But then I was told I had some more time free and attempted to post the "correct solution". Didn't even thought in doing what you said. Next time I'll have that in mind! Apr 14, 2016 at 14:11 • No problem at all, I hope you'll have a good time in the community. :) Apr 14, 2016 at 14:13 # Vyxal, 4 bytes vḟÞṠ  Try it Online! vḟ # Index of B in A, vectorized (B will be each individual character in the # string / item in the list, and A will be the string/list itself) ÞṠ # Is the result sorted?  Answers that use this method, in chronological order: # Vyxal, 5 bytes ₌ÞǓU=  Try it Online! ₌ # Parallel Apply - apply the following two elements to the input and keep # both results on the stack. ÞǓ # Connected Uniquify - Remove occurences of adjacent duplicates. U # Uniquify - Remove all duplicates. = # Are the two results equal?  Turned out to be a port of Adnan's 05AB1E answer. # K (ngn/k), 11 10 bytes -1 byte thanks to @ovs ~/,/'='^:\  Try it online! ^: test for spaces - always returns 0s, as the input consists only of digits. The : here means monadic. ^:\ repeat until convergence. This will return a pair of the original argument and a list of 0s of the same length as the argument. =' group each. To "group" a list means to make a dictionary that maps each distinct element of the list to the positions where it occurs. ,/' raze each, i.e. ignore the keys and concatenate the values. For the list of all 0s this will result in 0,1,...,length-1. For the other list, it will depend on the presence of non-continuous blocks - if there aren't any, the result will be 0,1,...,length-1 too. ~/ do they match? • I think the #:'\ can be ^:\ – ovs Aug 21, 2022 at 12:52 • @ovs well spotted, thanks! – ngn Aug 21, 2022 at 13:57 # Prolog (SWI), 40 bytes \[A|S]:-S=[];(S=[A|_];\+member(A,S)),\S.  Try it online! -3 thanks to Jo King Similar to xnor's second Haskell answer. This won't work on TIO because it is too old of a version, and it's longer, but I think it's more interesting: # Prolog (SWI), 57 bytes \A:-clumped(A,B),maplist([X-_,X]>>!,B,C),list_to_set(A,C).  # Julia, 35 bytes s->issorted(s,by=x->findfirst(s,x))  For whatever reason, sort does not take a string, but issorted does... • ...Are strings not immutable arrays in Julia like Python? That would make me really sad. – cat Apr 12, 2016 at 19:21 • Yes, strings are immutable. That's probably why issorted works, but sort doesn't. Apr 12, 2016 at 19:24 • There isn't a sorting method defined for strings, but it wouldn't work if they were processed in the same way as one-dimensional arrays because those are sorted by performing an in-place sort of a copy, and as you said, strings are immutable. It's not a problem for checking for sorted order though because it's implemented as a simple loop over an iterable, which is fine for strings. Just some trivia. ¯\_(ツ)_/¯ Apr 13, 2016 at 16:48 • @AlexA. So very much like Python in fact; the difference is that Python's builtin sorted turns its iterable argument into a mutable list first -- that's why sorted(string) returns a list of strings – cat Apr 13, 2016 at 20:09 # Factor, 22 bytes [ dup natural-sort = ]  Does what it says on the tin. As an anonymouse function, you should call this, or make it a : word ;. • it scares me when a cat brings a mouse into the game Apr 13, 2016 at 14:35 • @downrep_nation :P – cat Apr 13, 2016 at 15:10 ## Haskell, 37 bytes f l=(==)=<<scanl1 min$(<$>l).(==)<$>l


Uses the same approach as Luis Mendo's MATL answer: creates a vector for each entry which indices equal it, and checks that the result is sorted in decreasing order.

(<$>l).(==)<$>l is shorter version of [map(==a)l|a<-l]. The function (<\$>l).(==) that takes a to map(==a)l is mapped onto l.

scanl1 min takes the cumulative smallest elements of l, which equals the original only if l is reverse-sorted. (==)=<< checks if the list is indeed invariant under this operation.

A different recursive strategy gave 40 bytes:

f(a:b:t)=f(b:t)>(elem a t&&a/=b)
f _=1>0


This checks each suffix to see if its first element doesn't appear in the remainder, excusing cases where the first two elements are equal as part of a contiguous block.

## Lua, 10794 85 Bytes

13 bytes saved thanks to @LeakyNun

At least, it beats Java :D. Lua sucks at manipulating strings, but I think it is good enough :).

It takes its input as a command-line argument, and outputs 1 for truthy cases and false for falsy ones. Now outputs using its exit code. Exit code 0 for truthy, and 1 for falsy

o=os.exit;(...):gsub("(.)(.+)%1",function(a,b)if 0<#b:gsub(a,"")then o(1)end end)o(0)


### Ungolfed

Be care, there's two magic-variables called ..., the first one contains the argument of the program, the second one is local to the anonymous function and contains its parameters

o=os.exit;               -- ; mandatory, else it would exit instantly
(...):gsub("(.)(.+)%1",  -- iterate over each group of the form x.*x and apply an anonymous
function(a,b)          -- function that takes the captured group as parameters
if 0<#b:gsub(a,"")     -- if the captured group (.+) contain other character than
then                   -- the one captured by (.)
o(1)                 -- exit with falsy
end
end)
o(0)                     -- exit with truthy, reached only when the string is okay

• If it is permitted, you can replace os.exit() with i=#0... Jun 10, 2016 at 9:38

# Wolfram Language (Mathematica), 18 bytes

Gather@#==Split@#&


Try it online!

Gather gathers a list into sublists of identical elements, and Split splits a list into sublists of consecutive identical elements. They give the same result if and only if each value appears in only one contiguous block.

# BQN, 9 5 bytes

≡⟜∧⊐˜


Anonymous tacit function. Takes a string or a list of integers. Try it at BQN online!

### Explanation

Inspired by Dennis's J answer, though the details are a bit different.

Given two lists, ⊐ works as follows: for each element of its right argument, it finds the first index of that value in its left argument.

We're using ⊐˜, which passes the same list as both left and right arguments. The result is a list, for each digit, of the index of that digit's first occurrence.

If the digits are grouped together, these indices will occur in ascending order. However, if some copies of digit A are separated by digit B, B's (larger) index will come between the copies of A's (smaller) index.

≡⟜∧⊐˜
⊐˜  First index of each digit in the argument list
≡⟜      The result is identical to
∧     The result, sorted ascending


# Jelly, 5 bytes

iⱮṢƑ


Takes a list of integers. Try it online!

### Explanation

Port of my BQN solution.

iⱮṢƑ
i      First index of right argument in left argument
Ɱ     mapped over right argument
using the single argument as both left and right arguments
Ƒ  The result is equal to
Ṣ   itself sorted


Here's some other 5-byte solutions (see also Dennis's 5-byte solution):

ĠṢFṢƑ
Ġ      Group indices by their value in the list
Ṣ     Sort list of lists of indices
F    Flatten
Ƒ  The result is equal to
Ṣ   itself sorted


Try it online!

ĠṢF⁼J
Ġ      Group indices by their value in the list
Ṣ     Sort list of lists of indices
F    Flatten
⁼   The result is equal to
J  Range from 1 up to len of original argument


Try it online!

¹ƙF⁼
¹ƙ    Sort into buckets of identical items, ordered by first occurrence
F   Flatten
⁼  The result is equal to the original list


Try it online!

# Brachylog, 5 4 bytes

≡ᵍc?


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-1 thanks to DLosc

Takes input as a list of digits through the input variable, and succeeds or fails as output.

≡ᵍ      The input with identical elements grouped in order of first appearance
c     with the groups concatenated
?    is the input.