Is this n-speak?

Question

Inspired by Is it double speak?, I devised a harder challenge. Given a string, determine if the string is n-speak, for any \$n\geq 2\$.

N-speak is defined by repeating each letter \$n\$ times. With \$n = 4\$, the string Hello is transformed to HHHHeeeelllllllloooo. Your goal is to figure out if the input is a valid output for any n-speak transformation.

It should be noted that any sentence which is valid n-speak, for \$n = 2k\$, is also valid k-speak. Thus, the hard parts to solve will be odd values of \$n\$.

Input

A string consisting of at least 2 characters. Input could also be a list of characters. Input is case sensitive.

Output

Truthy if the string is n-speak, falsey otherwise.

Examples

True cases

HHeelllloo,,  wwoorrlldd!!
TTTrrriiipppllleee   ssspppeeeaaakkk
QQQQuuuuaaaaddddrrrruuuupppplllleeee    ssssppppeeeeaaaakkkk
7777777-------ssssssspppppppeeeeeeeaaaaaaakkkkkkk
999999999
aaaabb
aaaaaaaabbbbcc
aaaaabbbbb
@@@

If you want to generate additional truthy cases, you can use this MathGolf script. Place the string within the quotation marks, and the value of \$n\$ as the input.

False cases

Hello, world!
TTTrrriiipppllleee   speak
aaaaaaaaaaaaaaaab
Ddoouubbllee  ssppeeaakk
aabbab
aaaabbb
a (does not need to be handled)
(empty string, does not need to be handled)

Of course, since this is code golf, get ready to trim some bytes!

Answer 1

APL (Dyalog Unicode), 12 bytes

Runs with ⎕io←0

1≠∨/⍸2≠/∊0⍞0

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Golfed together with Adám.

On the input (example: "aaccccaaaaaabb", using "" to denote a string (an array of chars) and '' to denote a char)

∊0⍞0 surround with 0s and flatten, 0 'a' 'a' 'c' 'c' 'c' 'c' 'a' 'a' 'a' 'a' 'a' 'a' 'b' 'b' 0

2≠/ perform pairwise not-equal, 1 0 1 0 0 0 1 0 0 0 0 0 1 0 1

⍸ get the 0-indexed indices, 0 2 6 12 14

∨/ compute the GCD, 2

1≠ is this not equal to 1?

Answer 2

Java 10, 85 bytes

s->{var r=0>1;for(int i=0;++i<s.length();)r|=s.matches("((.)\\2{"+i+"})*");return r;}

Regex ported from @Arnauld's JavaScript answer.

Try it online.

Explanation:

s->{                          // Method with String parameter and boolean return-type
  var r=0>1;                  //  Result-boolean, starting at false
  for(int i=0;++i<s.length();)//  Loop `i` in the range [1, input-length):
    r|=                       //   Change the result to true if:
      s.matches("((.)\\2{"+i+"})*");
                              //    The input-String matches this regex
                              // NOTE: String#matches implicitly adds a leading ^ and 
                              //       trailing $ to match the full String
  return r;}                  // After the loop, return the result-boolean

Regex explanation:

^((.)\2{i})*$                 // Full regex to match, where `i` is the loop-integer
^           $                 // If the full String matches:
  (.)                         //  A character
     \2{i}                    //  Appended with that same character `i` amount of times
 (        )*                  //  And that repeated zero or more times for the entire string

Answer 3

Jelly, 5 bytes

Œɠg/’

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Answer 4

JavaScript (ES6), 53 bytes

Derived from the regular expression used by @wastl in Is it double speak?.

s=>[...s].some((_,n)=>s.match(`^((.)\\2{${++n}})*$`))

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---

Recursive version, 55 bytes

s=>(g=n=>s[++n]&&!!s.match(`^((.)\\2{${n}})*$`)|g(n))``

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Commented

s => (                    // s = input string
  g = n =>                // g is a recursive function taking a repetition length n
    s[++n] &&             // increment n; abort if s[n] is not defined
    !!s.match(            // otherwise, test whether s consists of groups of:
      `^((.)\\2{${n}})*$` //   some character, followed by n copies of the same character
    )                     //
    | g(n)                // or whether it works for some greater n
)``                       // initial call to g with n = [''] (zero-ish)

Answer 5

05AB1E, 5 bytes

γ€g¿≠

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Answer 6

Python 2, 73 70 69 67 bytes

lambda s:s in[''.join(c*n for c in s[::n])for n in range(2,len(s))]

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_{-4 bytes, thanks to Jitse}

Answer 7

Python 3, 69 bytes

lambda s:any(s=="".join(i*k for i in s[::k])for k in range(2,len(s)))

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Answer 8

QuadS, 16 bytes^SBCS

1≠∨/⍵
(.)\1*
⊃⍵L

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1≠ is 1 different from

∨/ the GCD

⍵ of the result of

(.)\1* PCRE Searching for any character followed by 0 or more repetitions thereof

⊃⍵L and returning the first of the match lengths (i.e. the length of the match)

Answer 9

Stax, 5 bytes

╢b}▄;

Run and debug it

Procedure:

• Calculate run-lengths.
• GCD of array
• Is > 1?

Answer 10

T-SQL 2008 query, 193 bytes

DECLARE @ varchar(max)='bbbbbbccc';

WITH C as(SELECT number+2n,@ t
FROM spt_values
WHERE'P'=type
UNION ALL 
SELECT n,stuff(t,1,n,'')FROM C
WHERE left(t,n)collate Thai_Bin=replicate(left(t,1),n))SELECT 1+1/~count(*)FROM C
WHERE''=t

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Answer 11

PHP, 76 75 bytes

while(($x=strspn($argn,$argn[$n+=$x],$n))>1&&($m=max($m,$x))%$x<1);echo!$x;

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First attempt, a somewhat naïve iterative approach.

Ungolfed:

// get the length of the next span of the same char
while( $s = strspn( $argn, $argn[ $n ], $n ) ) {

    // if span is less than 2 chars long, input is not n-speak
    if ( $s < 2 ) {
        break;
    }

    // k is GCD
    $k = max( $k, $s );

    // if span length does not divide evenly into GCD, input is not n-speak
    if( ( $k % $s ) != 0 ) {
        break;
    }

    // increment current input string index
    $n += $s;

}

-1 byte, thx to @Night2!

Answer 12

Perl 6, 30 27 26 bytes

{1-[gcd] m:g/(.)$0*/>>.to}

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Also uses the GCD trick, but uses the index of the end position of each run matched by the regex. Returns a negative number (truthy) if n-speak, zero (falsey) otherwise.

Answer 13

Haskell, 48 bytes

import Data.List
f=(>1).foldr(gcd.length)0.group

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Straightforward; uses the GCD trick.

Answer 14

Red, 80 bytes

func[s][repeat n length? s[if parse/case s[any[copy t skip n t]][return on]]off]

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More idiomatic Red:

Red, 81 bytes

func[s][any collect[repeat n length? s[keep parse/case s[any[copy t skip n t]]]]]

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Answer 15

Japt -¡, 8 bytes

ò¦ mÊrÕÉ

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ò¦ mÊrÕÉ     :Implicit input of string
ò            :Partition by
 ¦           :  Inequality
   m         :Map
    Ê        :  Length
     r       :Reduce by
      Õ      :  GCD
       É     :Subtract 1
             :Implicit output of boolean negation

Answer 16

Brachylog, 5 bytes

ġz₂=Ṁ

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Takes input through the input variable and outputs through success or failure.

At first I thought this would actually be shorter than my solution to Is it double speak?, but then I realized that ġ can and will try a group length of 1.

ġ        It is possible to split the input into chunks of similar length
 z₂      such that they have strictly equal length, and zipped together
    Ṁ    there are multiple results
   =     which are all equal.

Answer 17

Kotlin, 78 bytes

{s->(2..s.length/2).any{i->s.chunked(i).all{z->z.length==i&&z.all{z[0]==it}}}}

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Explanation

{s->                      Take a string as input
  (2..s.length/2)         The each string needs two parts at least, prevents the case "aaa" is 3-speak
    .any{i->              If there is any n (in this case i) that is n-speak return true
      s.chunked(i)        Split into length i substrings
      .all{z->            All substrings z
        z.length==i       Should be completely full, ie. "aaa"->["aa","a"]
        &&                And
        z.all{            All chars (it)
          z[0]==it        Should be the same as the first char
        }
      }
    }
  }

Answer 18

Scala, 80 bytes

s=>"(.)\\1*".r.findAllIn(s).map(_.size).reduce((x,y)=>(BigInt(x) gcd y).toInt)>1

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PS. Original solution was based on split function but it's longer (83 bytes).

s=>(s+s).split("(.)(?!\\1)").map(_.size+1).reduce((x,y)=>(BigInt(x) gcd y).toInt)>1

Answer 19

Wolfram Language (Mathematica), 34 bytes

GCD@@Length/@Split@Characters@#>1&

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Answer 20

Perl 5 -p, 83 79 76 74 bytes

$_=s,(.)\1+,$t=length$&;$t/=2while$t%2-1;$r+=$t==($g||=$t);'',ge==$r&&/^$/

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Answer 21

Brain-Flak, 96 bytes

{<>({}())<>({}[({})]){{}<>({}<>){{(({})){({}[()])<>}{}}<>([{}()]({}<>)<>)}(<>)<>}{}}<>{}({}[()])

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Uses the same GCD trick that many other submissions use. Output is 0 if the input is not n-speak, and a positive integer otherwise.

# For each character in the input
{

  # Add 1 to current run length
  <>({}())<>

  # If current and next characters differ:
  ({}[({})]){

    # Clean up unneeded difference
    {}<>

    # Move current run length to left stack, exposing current GCD on right stack
    ({}<>)

    # GCD routine: repeat until L=0
    {

      # Compute L mod R
      {(({})){({}[()])<>}{}}<>

      # Move R to left stack; finish computing L mod R and push to right stack
      ([{}()]({}<>)<>)

    }

    # Push 0 for new run length
    (<>)<>

  }{}

}

# Output GCD-1
<>{}({}[()])

Answer 22

Oracle SQL, 182 bytes

select+1-sign(min(length(x)-(select sum(length(regexp_substr(x,'(.)\1{'||i||'}',1,level)))from t connect by level<length(x))))from(select x,level i from t connect by level<length(x))

It works with an assumption that input data is stored in a table t(x), e.g.

with t(x) as (select 'HHeelllloo,,  wwoorrlldd!!' from dual)

Answer 23

K (ngn/k), 29 23 bytes

{~|/(&/s@&1<s)!s:#'=:x}

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edit: removed some unnecessary colons (i know when a monadic is required but it's not always clear to me if there's ambiguity so i default to including the colon) and changed the mod x-y*x%y to ngn/k's y!x, which meant i could remove a variable assignment

Answer 24

APL (Dyalog Unicode), 24 22 bytes^SBCS

Anonymous tacit prefix function.

⊂∊1↓⍳∘≢{⍵/⍨(≢⍵)⍴⍺↑⍺}¨⊂

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⊂ enclose the string to treat map using the entire string
 e.g. "aaabbb"

⍳∘≢{…}¨ for each of the ⍳ ɩndices 1 through the tally of characters in the string:
 e.g. 3

 ⍺↑⍺ take the current number of elements from the current number, padding with 0s
 e.g. [3,0,0]

 (≢⍵)⍴ cyclically reshape into the shape of the tally of characters in the string
  e.g. [3,0,0,3,0,0]

 ⍵/⍨ use that to replicate the string's characters
  "aaabbb"

1↓ drop the first one (n = 1)

⊂∊ is the the entire string a member of that list?

Answer 25

Japt -mR, 12 bytes

ÊÆóXäd_äe e

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Answer 26

Retina 0.8.2, 28 bytes

M!`(.)\1*
.
.
^(..+)(\1|¶)*$

Try it online! Link includes test cases. Explanation:

M!`(.)\1*

Split the text into runs of identical characters.

.
.

Replace them all with the same character.

^(..+)(\1|¶)*$

Check whether the GCD of the lengths of the runs is greater than 1.

Answer 27

MathGolf, 14 bytes

£─╞möl╠mÅ▀£╙╓┴

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Explanation

Checks all possible divisions of the input string into equal length chunks, and checks if there is a partition in which all chunks have just one unique character.

£                length of string with pop
 ─               get divisors
  ╞              discard from left of string/array (removes 1)
   mö            explicit map using 7 operators
     l           push input
      ╠          divide input into chunks of size k
       mÅ        explicit map using 2 operators
         ߜ      number of unique elements of list
           ╙     get maximum number of unique characters per chunk
                 loop ends here
            ╓    get the minimum of all maximums
             ┴   check if equal to 1

Answer 28

Gaia, 10 bytes

ẋl¦d¦&⊢⌉1>

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Same "GCD of run-lengths > 1" as many other submissions use.

There is a bug in ė (run-length encoding) that drops the last unique element of the list, otherwise we could have the following 9 byte solution: ė(¦d¦&⊢1>.

Answer 29

Pyth, 7 bytes

Outputs 0 for falsy inputs or a positive integer otherwise.

tiFhCr8

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Answer 30

Pyth, 8 bytes

<1iFhMr8

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<1iFhMr8Q   Implicit: Q=eval(input())
            Trailing Q inferred
      r8Q   Run length encode Q into [count, character]
    hM      Take first element of each
  iF        Reduce by GCD
<1          Is 1 less than the above? Implicit print

Answer 31

Perl 5 -n, 38 bytes

for$i(1..y///c){print/^((.)\2{$i})*$/}

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The print"\n" in the footer is needed to separate the outputs.

Straightforward loop through all possible ns. Outputs nothing for "1-speak", anything else for n-speak where n > 1.

Answer 32

C# (Visual C# Interactive Compiler), 76 bytes

s=>s.Where((_,n)=>s.Count%(n+=2)<1&!s.Where((c,i)=>c!=s[i/n*n]).Any()).Any()

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Generate all n from 2, 3, 4, ... and do the following:

• Check if the length of input is divisible by n
• Compare each character of input to the corresponding character of an n-speak string

If both checks pass, the input string is n-speak.

Answer 33

D , 126 bytes

bool f(string s){import std.algorithm;auto a=s.group.minElement!(a=>a[1])[1];foreach(g;s.group)if(g[1]%a)return 0;return a>1;}

First code golf I've done in D.

Does not handle empty input strings (causes an assertion failure in the standard library).

Ungolfed version:

bool f(string s) {
    import std.algorithm; // for group and minElement

    // splits the string into a groups of the same character
    // eg. "HHHiii".group returns
    // [Tuple!(char, uint)('H', 3), Tuple!(char, uint)('i', 3)]
    auto groups = s.group;

    // gets the smallest group length
    auto min_length = groups.minElement!(a => a[1])[1];

    foreach(g; groups) // for each group
        if(g[1] % min_length) // if it's length is not divisible by smallest length
            return 0; // return false

    return min_length > 1; // return true if smallest length was above 1
}