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!

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.

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.

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Jelly, 5 bytes

ĠIFPỊ 

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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. 

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. 

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 

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!

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.

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. 

Vyxal, 4 bytes

)ġf⁼ 

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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? 💎 

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.

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()); } } 

Regex (Perl / PCRE / Java / Boost / Python^regex / Ruby), 17 bytes

^(?!(.)+\1*+.+\1) 

Try it online! - Perl
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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
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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
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Try it online! - Python _{^{import regex}}
Try it online! - Ruby
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\$\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') 

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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.

Haskell, 37 bytes

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

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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) 

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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!

Jelly, \$_4\$ 3 bytes

nÞƑ 

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Semi-based on several of DLosc's 5-byters.

 Ƒ The input is not changed by Þ sorting its elements by n the list of, for each element, if it is not equal to that element. 

Kind of like cutting the F out of ¹ƙ`F⁼, but also kind of like cutting the `ṢƑ out of iⱮ`ṢƑ. Compared to the previous i@ÞƑ, it simply saves an @ by using lexicographic comparisons with a dyad that doesn't discriminate between its left and right argument instead of numeric comparisons with one that does--the first index of an element is the first position at which its inequality-vector has a zero, so one element has a greater first index than another iff its inequality-vector is all ones up to the position where the other's has its first zero.

It's been almost 3 years, and ðƙF⁼ still makes me so sad...!

Pyth, 7 bytes

{IeMrz8 

Test Suite.

Mathematica, 26 bytes

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

MATL, 13 11 bytes

u"G@=fd2<vA 

Thanks to Luis Mendo for saving two bytes!

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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 

Vyxal, 4 bytes

vḟÞṠ 

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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= 

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₌ # 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.

Uiua, 6 bytes

≍⊝.⊜⊢. 

Try it!

≍⊝.⊜⊢. ⊜⊢. # get the first digit in each like contiguous group . # duplicate ⊝ # deduplicate ≍ # do they match? 

Jelly, 4 bytes

ĠIỊȦ 

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Explanation

An independently derived variation on Dennis's 5-byte answer:

ĠIỊȦ Ġ Group indices 1..N based on their values in the input list I Get the difference of each pair of adjacent values in those lists Ị For each difference, 1 if it is <= 1, 0 if it is greater than 1 Ȧ 0 if the nested list contains any 0s, 1 otherwise 

Regenerate -c, 103 94 87 bytes

((${$2/10}!$~1)($4!)(${$2%10})($0{$3-$4}!(${$6*$7}!1)(${$4*6+5})0{$6-$6/$7*$7-1})){#~1} 

Outputs 0 for falsey, 1 for truthy. As of this writing, ATO doesn't have the -c flag yet, so the program there outputs nothing for falsey and a string of digits for truthy. Attempt This Online!

Explanation

Inspired by Unrelated String's answer, I thought of a different way to track whether a digit has been seen already: mapping each digit to a different prime number and storing their product. Then I realized that the numbers only needed to be coprime, which saved even more bytes.

(...){#~1} 

Repeat the following a number of times equal to the length of the input:

(${$2/10}!$~1) 

Group 2 stores the input number and removes one digit at a time from the right end. If group 2 has been matched before, ${$2/10} matches the previous value int-divided by 10; if not (which is only the case on the first iteration), $~1 matches the input.

($4!) 

Group 3 stores the previous digit. It matches the previous value of group 4, or empty string if group 4 has not matched before.

(${$2%10}) 

Group 4 stores the current digit, i.e. the rightmost digit of the current number in group 2.

($0{$3-$4}!...) 

Group 5 starts by testing if the difference between group 3 and group 4 is zero, i.e. if we have two consecutive identical digits. If so, we repeat the nonexistent group 0 zero times; this succeeds, matching empty string, and we proceed to the next iteration. If not, trying to match group 0 (or repeat it a negative number of times) fails, and we proceed to the alternate branch, where we process a new digit:

(${$6*$7}!1) 

Group 6 stores the product of coprimes. If groups 6 and 7 have been matched before, ${$6*$7} multiplies them and sets that as the new value of group 6. If not, we use 1 as the initial value.

(${$4*6+5}) 

Group 7 stores the coprime number representing the current digit. We map the value of group 4, a digit between 1 and 9, to the corresponding number from this arithmetic progression: 11, 17, 23, 29, 35, 41, 47, 53, 59. All of them are prime except 35, which doesn't have any of the others as factors.

0{$6-$6/$7*$7-1} 

Now we check whether the current digit has been seen before. If it has, we know that it wasn't just now because the previous digit is different, which means we want the match to fail.

$6/$7*$7 int-divides the product of coprimes in group 6 by the most recent coprime in group 7 and then multiplies it by this coprime again. If the current digit has been seen before, the division is exact, and multiplying gives the original value of group 6 back; otherwise, the division truncates the result, and multiplying gives a number smaller than the original value.

0{$6-...-1} subtracts the above quantity from group 6 and then subtracts 1. This results in -1 if the current digit has been seen before and a nonnegative number if it has not. We then use that number as a repetition count for an arbitrary character. If the number is negative, the regex fails, outputting 0 successful matches. If there are no failures in any iteration, the regex outputs 1 successful match.

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 

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; } 

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 

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?

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. 

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). 

Arturo, 28 22 bytes

$=>[=unique<=squeeze&] 

Try it!

$=>[ ; a function where input is assigned to & squeeze& ; squeeze the input; e.g. [1 1 2 1 1 1 3 3] -> [1 2 1 3] <= ; duplicate unique ; remove duplicates = ; are they equal? ] ; end function