All together now

Question

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;

body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

Answer 1

Python 3, 38 34 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!

Answer 2

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.

Answer 3

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!

Answer 4

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.

Answer 5

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.

Answer 6

R, 66 48 46 43 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!

Answer 7

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 (with rows considered as atomic) are in decreasing lexicographical 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 in order.

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

Answer 8

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.

Answer 9

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.

Answer 10

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());
    }
}

Answer 11

Regex (Perl / PCRE / Java / Boost / Python^regex / 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 / Python^regex / 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, 21 bytes

->s{/(.)\1*+.+\1/!~s}

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 v0.7+, 30 bytes

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

Attempt This Online! / Try it online!

Julia v1.3+, 28 bytes

s->count(r"(.)\1*+.+\1",s)<1

Attempt This Online!

-2 bytes compared to v0.7+, thanks to MarcMush

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!

\$\large\textit{Full programs}\$

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).+\1`NIa

Try it online!

Thanks to DLosc.

Answer 12

Pyth, 7 bytes

{IeMrz8

Test Suite.

Answer 13

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.

Answer 14

Mathematica, 26 bytes

0<##&@@Sort[#&@@@Split@#]&

Answer 15

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

Answer 16

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 ¹ƙ`F⁼, but also kind of like cutting the `ṢƑ out of iⱮ`ṢƑ.

Answer 17

Haskell, 37 bytes

f l=(==)=<<scanl1 min$(<$>l).(==)<$>l

Try it online!

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 recursive strategy also gave 37 bytes:

f(h:t)=t==[]||f t>(elem h t&&h/=t!!0)

Try it online!

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.

Other approaches:

42 bytes

f l=and[all(/=a)t|a:b:t<-scanr(:)[]l,a/=b]

Try it online!

43 bytes

import Data.List
((==)<*>nub).map nub.group

Try it online!

Answer 18

Vyxal, 4 bytes

)ġf⁼

Try it Online!

Takes input as a list of singleton strings. Happens to be a different algorithm to the existing vyxal answer.

Explained

)ġf⁼­⁡​‎‎⁡⁠⁡‏⁠‎⁡⁠⁢‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁣‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁤‏‏​⁡⁠⁡‌­
)ġ    # ‎⁡Group by the identity function, by order of first appearance. 
  f   # ‎⁢Flatten the list
   ⁼  # ‎⁣Does that equal the input?
💎

Answer 19

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"

Answer 20

Python, 56 55 bytes

a=lambda s:~(s[0]in s.lstrip(s[0]))&a(s[1:])if s else 1

Answer 21

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;
}

Answer 22

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:

• Dennis – Python, 33 bytes
• isaacg – Pyth, 5 bytes
• Dennis - Julia, 30 bytes
• Dennis - J, 8 bytes
• DLosc – BQN, 5 bytes
• DLosc – Jelly, 5 bytes
• Unrelated String – Jelly, 4 bytes

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.

Answer 23

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?

Answer 24

Brachylog, 5 4 bytes

≡ᵍc?

Try it online!

-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.

Answer 25

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).

Answer 26

Julia, 35 bytes

s->issorted(s,by=x->findfirst(s,x))

For whatever reason, sort does not take a string, but issorted does...

Answer 27

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 ;.

Answer 28

Lua, 107 94 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

Answer 29

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.

Answer 30

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