Spelling Bee Acceptable

up vote
1
down vote

favorite

This is similar to this question.
However, I have a different goal in mind:
Reduce the wordlist to words that could appear.
The easy part is to delete hyphens and similar.

The hard part, however: Remove words containing more than seven letters.

So, the problem:

Given a set of words containing only lowercase (or uppercase) ASCII letters, remove the words containing eight or more different letters.

Any input and output format is acceptable.

Standard loophole rules apply.

Shortest byte count wins.

Clever solutions are better, and faster is better as well.

My solution (implementing a quite clever O(n) algorithm, noncompetitive ref implementation):

#include <stdio.h>                                                              

#include <stdint.h>                                                             



int main(void) {                                                                

  char str[1024];                                                               

  while(scanf("%sn", str) == 1) {                                              

    uint32_t st = 0;                                                            

    int s = 0;                                                                  

    for(char *p = str; *p && s <= 7; p++) {                                     

      s += ~st >> (*p - 'a') & 1;                                               

      st |= 1 << (*p - 'a');                                                    

    }                                                                           

    if(s <= 7)                                                                  

      printf("%sn", str);                                                      

  }                                                                             

}

edited Nov 9 at 19:52

asked Nov 9 at 1:57

NoLongerBreathedIn

643

5

Could you include a test case?
– Dennis♦
Nov 9 at 2:05

12

It's generally bad practice to have an upper limit on the score. The reason is that some languages are especially verbose or difficult to program in, and the 307 byte requirement would arbitrarily disallow them from competing.
– Nathan Merrill
Nov 9 at 2:05

Honestly, this is the easy part. Finding words that fit a partially revealed word is harder
– Jo King
Nov 9 at 11:19

I meant, this is the hard part of finding words that could ever be solutions.
– NoLongerBreathedIn
Nov 9 at 19:56

"quite clever O(n) algorithm": you'd have to be really clever to come up with an algorithm which does worse than a linear program. There's no need to compare different words, and you'd spend a constant time per word.
– Abigail
Nov 10 at 17:25

|
show 1 more comment

up vote
1
down vote

favorite

This is similar to this question.
However, I have a different goal in mind:
Reduce the wordlist to words that could appear.
The easy part is to delete hyphens and similar.

The hard part, however: Remove words containing more than seven letters.

So, the problem:

Given a set of words containing only lowercase (or uppercase) ASCII letters, remove the words containing eight or more different letters.

Any input and output format is acceptable.

Standard loophole rules apply.

Shortest byte count wins.

Clever solutions are better, and faster is better as well.

My solution (implementing a quite clever O(n) algorithm, noncompetitive ref implementation):

#include <stdio.h>                                                              

#include <stdint.h>                                                             



int main(void) {                                                                

  char str[1024];                                                               

  while(scanf("%sn", str) == 1) {                                              

    uint32_t st = 0;                                                            

    int s = 0;                                                                  

    for(char *p = str; *p && s <= 7; p++) {                                     

      s += ~st >> (*p - 'a') & 1;                                               

      st |= 1 << (*p - 'a');                                                    

    }                                                                           

    if(s <= 7)                                                                  

      printf("%sn", str);                                                      

  }                                                                             

}

edited Nov 9 at 19:52

asked Nov 9 at 1:57

NoLongerBreathedIn

643

5

Could you include a test case?
– Dennis♦
Nov 9 at 2:05

12

It's generally bad practice to have an upper limit on the score. The reason is that some languages are especially verbose or difficult to program in, and the 307 byte requirement would arbitrarily disallow them from competing.
– Nathan Merrill
Nov 9 at 2:05

Honestly, this is the easy part. Finding words that fit a partially revealed word is harder
– Jo King
Nov 9 at 11:19

I meant, this is the hard part of finding words that could ever be solutions.
– NoLongerBreathedIn
Nov 9 at 19:56

"quite clever O(n) algorithm": you'd have to be really clever to come up with an algorithm which does worse than a linear program. There's no need to compare different words, and you'd spend a constant time per word.
– Abigail
Nov 10 at 17:25

|
show 1 more comment

up vote
1
down vote

favorite

This is similar to this question.
However, I have a different goal in mind:
Reduce the wordlist to words that could appear.
The easy part is to delete hyphens and similar.

The hard part, however: Remove words containing more than seven letters.

So, the problem:

Given a set of words containing only lowercase (or uppercase) ASCII letters, remove the words containing eight or more different letters.

Any input and output format is acceptable.

Standard loophole rules apply.

Shortest byte count wins.

Clever solutions are better, and faster is better as well.

My solution (implementing a quite clever O(n) algorithm, noncompetitive ref implementation):

#include <stdio.h>                                                              

#include <stdint.h>                                                             



int main(void) {                                                                

  char str[1024];                                                               

  while(scanf("%sn", str) == 1) {                                              

    uint32_t st = 0;                                                            

    int s = 0;                                                                  

    for(char *p = str; *p && s <= 7; p++) {                                     

      s += ~st >> (*p - 'a') & 1;                                               

      st |= 1 << (*p - 'a');                                                    

    }                                                                           

    if(s <= 7)                                                                  

      printf("%sn", str);                                                      

  }                                                                             

}

edited Nov 9 at 19:52

asked Nov 9 at 1:57

NoLongerBreathedIn

643

This is similar to this question.
However, I have a different goal in mind:
Reduce the wordlist to words that could appear.
The easy part is to delete hyphens and similar.

The hard part, however: Remove words containing more than seven letters.

So, the problem:

Given a set of words containing only lowercase (or uppercase) ASCII letters, remove the words containing eight or more different letters.

Any input and output format is acceptable.

Standard loophole rules apply.

Shortest byte count wins.

Clever solutions are better, and faster is better as well.

My solution (implementing a quite clever O(n) algorithm, noncompetitive ref implementation):

#include <stdio.h>                                                              

#include <stdint.h>                                                             



int main(void) {                                                                

  char str[1024];                                                               

  while(scanf("%sn", str) == 1) {                                              

    uint32_t st = 0;                                                            

    int s = 0;                                                                  

    for(char *p = str; *p && s <= 7; p++) {                                     

      s += ~st >> (*p - 'a') & 1;                                               

      st |= 1 << (*p - 'a');                                                    

    }                                                                           

    if(s <= 7)                                                                  

      printf("%sn", str);                                                      

  }                                                                             

}

code-golf

edited Nov 9 at 19:52

asked Nov 9 at 1:57

NoLongerBreathedIn

643

edited Nov 9 at 19:52

asked Nov 9 at 1:57

NoLongerBreathedIn

643

edited Nov 9 at 19:52

asked Nov 9 at 1:57

NoLongerBreathedIn

643

asked Nov 9 at 1:57

NoLongerBreathedIn

643

asked Nov 9 at 1:57

NoLongerBreathedIn

643

5

Could you include a test case?
– Dennis♦
Nov 9 at 2:05

12

It's generally bad practice to have an upper limit on the score. The reason is that some languages are especially verbose or difficult to program in, and the 307 byte requirement would arbitrarily disallow them from competing.
– Nathan Merrill
Nov 9 at 2:05

Honestly, this is the easy part. Finding words that fit a partially revealed word is harder
– Jo King
Nov 9 at 11:19

I meant, this is the hard part of finding words that could ever be solutions.
– NoLongerBreathedIn
Nov 9 at 19:56

"quite clever O(n) algorithm": you'd have to be really clever to come up with an algorithm which does worse than a linear program. There's no need to compare different words, and you'd spend a constant time per word.
– Abigail
Nov 10 at 17:25

|
show 1 more comment

5

Could you include a test case?
– Dennis♦
Nov 9 at 2:05

12

It's generally bad practice to have an upper limit on the score. The reason is that some languages are especially verbose or difficult to program in, and the 307 byte requirement would arbitrarily disallow them from competing.
– Nathan Merrill
Nov 9 at 2:05

Honestly, this is the easy part. Finding words that fit a partially revealed word is harder
– Jo King
Nov 9 at 11:19

I meant, this is the hard part of finding words that could ever be solutions.
– NoLongerBreathedIn
Nov 9 at 19:56

"quite clever O(n) algorithm": you'd have to be really clever to come up with an algorithm which does worse than a linear program. There's no need to compare different words, and you'd spend a constant time per word.
– Abigail
Nov 10 at 17:25

Could you include a test case?
– Dennis♦
Nov 9 at 2:05

It's generally bad practice to have an upper limit on the score. The reason is that some languages are especially verbose or difficult to program in, and the 307 byte requirement would arbitrarily disallow them from competing.
– Nathan Merrill
Nov 9 at 2:05

Honestly, this is the easy part. Finding words that fit a partially revealed word is harder
– Jo King
Nov 9 at 11:19

I meant, this is the hard part of finding words that could ever be solutions.
– NoLongerBreathedIn
Nov 9 at 19:56

"quite clever O(n) algorithm": you'd have to be really clever to come up with an algorithm which does worse than a linear program. There's no need to compare different words, and you'd spend a constant time per word.
– Abigail
Nov 10 at 17:25

|
show 1 more comment

19 Answers
19

active

oldest

votes

up vote
4
down vote

APL (Dyalog Unicode), 11 bytes

{⍵/⍨8>≢∪⍵}¨

Takes input as a list of strings.

Explanation:

{⍵/⍨8>≢∪⍵}¨

{       ⍵}¨ for each word

       ∪    take unique letters

    8>≢     length less than 8 as a boolean (0 or 1)

 ⍵/⍨        repeat word that many times

Try it online!

edited Nov 9 at 23:28

answered Nov 9 at 2:27

Quintec

1,195518

the result contains empty strings. you can fix that preserving the byte count: ⊢(/⍨)8>≢∘∪¨
– ngn
Nov 11 at 7:44

add a comment |

up vote
3
down vote

Perl 6, 20 bytes

*.grep(8>*.comb.Set)

Try it online!

Filters by words that have a set of letters with size less than 8.

answered Nov 9 at 2:06

Jo King

19.3k245102

add a comment |

up vote
3
down vote

05AB1E, 5 bytes

ʒÙg8‹

Try it online.

Explanation:

ʒ        # Filter the (implicit) input-list by:

 Ù       #  Only leave distinct letters of the word

  g      #  Take its length

   8‹    #  And only leave those with a length smaller than 8

         # (And output implicitly after we're done filtering)

answered Nov 9 at 7:59

Kevin Cruijssen

34.4k554182

add a comment |

up vote
2
down vote

Jelly, 6 bytes

Qṫ¥Ðḟ8

Try it online!

How it works

Qṫ¥Ðḟ8  Main link. Argument: A (array or words)



   Ðḟ   Filterfalse; only keep the words W of A for which the chain to the left

        returns a falsy value.

  ¥         Combine the two links to the left into a dyadic chain.

Q               Unique; remove duplicate letters from W.

 ṫ   8          Tail 8; remove the first 7 letters of the result.

answered Nov 9 at 2:13

Dennis♦

184k32295731

add a comment |

up vote
2
down vote

MathGolf, 6 bytes

Ç{▀£7>

Try it online!

Explanation

Really similar to the 05AB1E solution, but I lose one byte thanks to explicitly having to define the code block for filtering.

Ç       Implicit faulty filter by block

 {      Start block

  ▀     Get unique characters of string

   £    Get length

    7>  Is greater than 7

answered Nov 9 at 10:51

maxb

2,1081923

add a comment |

up vote
2
down vote

JavaScript, 33 bytes

a=>a.filter(s=>new Set(s).size<8)

Try it online, using Dennis' test cases

answered Nov 9 at 10:54

Shaggy

18.2k21663

add a comment |

up vote
2
down vote

J, 10 bytes

#~8>#@~.@>

explanation

#~ 8 > #@~.@>

#~                 NB. filter the input based on...

   8 >             NB. is 8 greater than...

       #@          NB. the length of...

         ~.@       NB. the unique characters of...

            >      NB. the unboxed input.

Try it online!

edited Nov 9 at 14:18

answered Nov 9 at 3:48

Jonah

1,981816

1

Shouldn't 9 be 8?
– Galen Ivanov
Nov 9 at 8:03

1

Fixed. Weird, I could have sworn the OP said 8 was the max length allowed before. Anyway, ty...
– Jonah
Nov 9 at 14:19

add a comment |

up vote
2
down vote

Java 8, 46 bytes

s->s.filter(w->w.chars().distinct().count()<8)

Try it online.

Explanation:

s->               // Method with String-Stream as both parameter and return-type

  s.filter(w->    //  Filter the words in the input-Stream by:

    w.chars()     //  Convert the String to characters

     .distinct()  //  Only leave distinct characters

     .count()<8)  //  Only leave words with less than 8 distinct characters

edited Nov 10 at 10:16

answered Nov 9 at 9:42

Kevin Cruijssen

34.4k554182

Wow, w.chars.distinct.count. Turn each of those words into a symbol and you have a golfing language! xD
– Quintec
Nov 9 at 12:56

@Quintec Hehe, it's what I do in my 05AB1E answer. :) s->s is implicit input in 05AB1E; filter(w-> is ʒ; .chars() is implicitly again; .distinct() is Ù; .count() (or length) is g; <8 is 8‹; and the closing ) is implicitly since it's at the end of the 05AB1E program (if I wanted to do something after the filter I'd have to close it with }); and finally outputting is implicitly again. So my Java and 05AB1E are similar, one is a 46 bytes function and the other a 5 bytes full program, though. xD
– Kevin Cruijssen
Nov 9 at 13:06

Hehe, also basically same for my APL answer. It meets in the middle: 11 bytes :P
– Quintec
Nov 9 at 15:24

add a comment |

up vote
1
down vote

Python 2, 40 bytes

lambda i:[x for x in i if len(set(x))<8]

Try it online!

Test cases borrowed from @Dennis. Input and output are both lists.

answered Nov 9 at 7:05

ElPedro

3,4131023

add a comment |

up vote
1
down vote

Red, 54 bytes

func[b][foreach a b[if 8 > length? unique a[print a]]]

Try it online!

The first test set was taken from Dennis'

answered Nov 9 at 8:04

Galen Ivanov

5,94711032

add a comment |

up vote
1
down vote

C# (.NET Core), 58 bytes

 a=>a.Where(x=>x.GroupBy(y=>y).Count()<8)

Try It Online!

answered Nov 9 at 13:34

LiefdeWen

2,502936

add a comment |

up vote
1
down vote

C (gcc), 104 95 bytes

Dennis's word set was also used here.

Thanks to nwellnhof for the suggestions.

char*s,*t;f(i,j,k){for(;~scanf("%ms",&s);i<8&&puts(s))for(i=j=0,t=s;*t;j=k)k=j|1<<*t++,i+=j<k;}

Try it online!

edited Nov 9 at 17:15

answered Nov 9 at 9:36

ErikF

1,25917

96 bytes, if you're OK with implementation- and undefined behavior.
– nwellnhof
Nov 9 at 10:15

@nwellnhof What's a little UD between friends for code golf? :-)
– ErikF
Nov 9 at 17:16

add a comment |

up vote
1
down vote

perl -nlE, 32 bytes

my%h;@h{/./g}=();say if keys%h<8

This reads words from STDIN, printing out those with less than 7 different characters.

answered Nov 10 at 17:13

Abigail

41717

1

25 bytes: my%h;@h{/./g}=1;%h<8&&say (%h<8 requires Perl 5.26)
– nwellnhof
Nov 12 at 12:25

add a comment |

up vote
0
down vote

Racket, 95 bytes

(require racket/set)(define(f l)(filter(lambda(x)(<(set-count(list->set(string->list x)))8))l))

Try it online!

The test set was taken from Dennis'

answered Nov 9 at 8:45

Galen Ivanov

5,94711032

add a comment |

up vote
0
down vote

Charcoal, 12 bytes

ΦＡ›⁸ＬΦι⁼μ⌕ιλ

Try it online! Link is to verbose version of code. Explanation:

 Ａ              Input array

Φ               Filter strings where

   ⁸            Literal 8

  ›             Is greater than

    Ｌ           Length of

      ι         Current string

     Φ          Filtered on characters where

        μ       Inner index

       ⁼        Equals

           λ    Current character's

         ⌕      First index in

          ι     Current string

                Implicitly print matching strings

answered Nov 9 at 9:33

Neil

78.1k744175

10?
– ASCII-only
Nov 18 at 5:12

add a comment |

up vote
0
down vote

Japt `-f`, 6 bytes

¬â Ê<8

Try it, using Dennis' test cases

answered Nov 9 at 10:57

Shaggy

18.2k21663

add a comment |

up vote
0
down vote

JavaScript (ES6), 53 bytes

Without using a set:

a=>a.filter(w=>[...w].every(o=c=>o[c]=o[c]||--k,k=8))

Try it online! (using Dennis' test set)

edited Nov 9 at 11:23

answered Nov 9 at 10:32

Arnauld

69.5k586294

add a comment |

up vote
0
down vote

Snap! 4, scratchblocks2 syntax

(pretending that Snap! exclusive blocks are valid in scratchblocks2 but functions use the Scratch define)

69 bytes

b takes a list of strings.

define((b)

report((#)keep items such that(<(length of )<[7]>)from(b

answered Nov 9 at 23:49

Silas Reel

316

add a comment |

up vote
0
down vote

C++ (gcc), 105 104 bytes

#import<bits/stdc++.h>

f(){for(char*n;~scanf("%ms",&n);)std::set<int>(n,n+strlen(n)).size()<8&&puts(n);}

Try it online!

-1 byte thanks to @ceilingcat

edited yesterday

answered Nov 9 at 19:19

Logern

70546

add a comment |

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19 Answers
19

active

oldest

votes

19 Answers
19

active

oldest

votes

up vote
4
down vote

APL (Dyalog Unicode), 11 bytes

{⍵/⍨8>≢∪⍵}¨

Takes input as a list of strings.

Explanation:

{⍵/⍨8>≢∪⍵}¨

{       ⍵}¨ for each word

       ∪    take unique letters

    8>≢     length less than 8 as a boolean (0 or 1)

 ⍵/⍨        repeat word that many times

Try it online!

edited Nov 9 at 23:28

answered Nov 9 at 2:27

Quintec

1,195518

the result contains empty strings. you can fix that preserving the byte count: ⊢(/⍨)8>≢∘∪¨
– ngn
Nov 11 at 7:44

add a comment |

up vote
4
down vote

APL (Dyalog Unicode), 11 bytes

{⍵/⍨8>≢∪⍵}¨

Takes input as a list of strings.

Explanation:

{⍵/⍨8>≢∪⍵}¨

{       ⍵}¨ for each word

       ∪    take unique letters

    8>≢     length less than 8 as a boolean (0 or 1)

 ⍵/⍨        repeat word that many times

Try it online!

edited Nov 9 at 23:28

answered Nov 9 at 2:27

Quintec

1,195518

the result contains empty strings. you can fix that preserving the byte count: ⊢(/⍨)8>≢∘∪¨
– ngn
Nov 11 at 7:44

add a comment |

up vote
4
down vote

APL (Dyalog Unicode), 11 bytes

{⍵/⍨8>≢∪⍵}¨

Takes input as a list of strings.

Explanation:

{⍵/⍨8>≢∪⍵}¨

{       ⍵}¨ for each word

       ∪    take unique letters

    8>≢     length less than 8 as a boolean (0 or 1)

 ⍵/⍨        repeat word that many times

Try it online!

edited Nov 9 at 23:28

answered Nov 9 at 2:27

Quintec

1,195518

APL (Dyalog Unicode), 11 bytes

{⍵/⍨8>≢∪⍵}¨

Takes input as a list of strings.

Explanation:

{⍵/⍨8>≢∪⍵}¨

{       ⍵}¨ for each word

       ∪    take unique letters

    8>≢     length less than 8 as a boolean (0 or 1)

 ⍵/⍨        repeat word that many times

Try it online!

edited Nov 9 at 23:28

answered Nov 9 at 2:27

Quintec

1,195518

edited Nov 9 at 23:28

answered Nov 9 at 2:27

Quintec

1,195518

answered Nov 9 at 2:27

Quintec

1,195518

answered Nov 9 at 2:27

Quintec

1,195518

the result contains empty strings. you can fix that preserving the byte count: ⊢(/⍨)8>≢∘∪¨
– ngn
Nov 11 at 7:44

add a comment |

the result contains empty strings. you can fix that preserving the byte count: ⊢(/⍨)8>≢∘∪¨
– ngn
Nov 11 at 7:44

the result contains empty strings. you can fix that preserving the byte count: ⊢(/⍨)8>≢∘∪¨
– ngn
Nov 11 at 7:44

add a comment |

up vote
3
down vote

Perl 6, 20 bytes

*.grep(8>*.comb.Set)

Try it online!

Filters by words that have a set of letters with size less than 8.

answered Nov 9 at 2:06

Jo King

19.3k245102

add a comment |

up vote
3
down vote

Perl 6, 20 bytes

*.grep(8>*.comb.Set)

Try it online!

Filters by words that have a set of letters with size less than 8.

answered Nov 9 at 2:06

Jo King

19.3k245102

add a comment |

up vote
3
down vote

Perl 6, 20 bytes

*.grep(8>*.comb.Set)

Try it online!

Filters by words that have a set of letters with size less than 8.

answered Nov 9 at 2:06

Jo King

19.3k245102

Perl 6, 20 bytes

*.grep(8>*.comb.Set)

Try it online!

Filters by words that have a set of letters with size less than 8.

answered Nov 9 at 2:06

Jo King

19.3k245102

answered Nov 9 at 2:06

Jo King

19.3k245102

answered Nov 9 at 2:06

Jo King

19.3k245102

answered Nov 9 at 2:06

Jo King

19.3k245102

add a comment |

up vote
3
down vote

05AB1E, 5 bytes

ʒÙg8‹

Try it online.

Explanation:

ʒ        # Filter the (implicit) input-list by:

 Ù       #  Only leave distinct letters of the word

  g      #  Take its length

   8‹    #  And only leave those with a length smaller than 8

         # (And output implicitly after we're done filtering)

answered Nov 9 at 7:59

Kevin Cruijssen

34.4k554182

add a comment |

up vote
3
down vote

05AB1E, 5 bytes

ʒÙg8‹

Try it online.

Explanation:

ʒ        # Filter the (implicit) input-list by:

 Ù       #  Only leave distinct letters of the word

  g      #  Take its length

   8‹    #  And only leave those with a length smaller than 8

         # (And output implicitly after we're done filtering)

answered Nov 9 at 7:59

Kevin Cruijssen

34.4k554182

add a comment |

up vote
3
down vote

05AB1E, 5 bytes

ʒÙg8‹

Try it online.

Explanation:

ʒ        # Filter the (implicit) input-list by:

 Ù       #  Only leave distinct letters of the word

  g      #  Take its length

   8‹    #  And only leave those with a length smaller than 8

         # (And output implicitly after we're done filtering)

answered Nov 9 at 7:59

Kevin Cruijssen

34.4k554182

05AB1E, 5 bytes

ʒÙg8‹

Try it online.

Explanation:

ʒ        # Filter the (implicit) input-list by:

 Ù       #  Only leave distinct letters of the word

  g      #  Take its length

   8‹    #  And only leave those with a length smaller than 8

         # (And output implicitly after we're done filtering)

answered Nov 9 at 7:59

Kevin Cruijssen

34.4k554182

answered Nov 9 at 7:59

Kevin Cruijssen

34.4k554182

answered Nov 9 at 7:59

Kevin Cruijssen

34.4k554182

answered Nov 9 at 7:59

Kevin Cruijssen

34.4k554182

add a comment |

up vote
2
down vote

Jelly, 6 bytes

Qṫ¥Ðḟ8

Try it online!

How it works

Qṫ¥Ðḟ8  Main link. Argument: A (array or words)



   Ðḟ   Filterfalse; only keep the words W of A for which the chain to the left

        returns a falsy value.

  ¥         Combine the two links to the left into a dyadic chain.

Q               Unique; remove duplicate letters from W.

 ṫ   8          Tail 8; remove the first 7 letters of the result.

answered Nov 9 at 2:13

Dennis♦

184k32295731

add a comment |

up vote
2
down vote

Jelly, 6 bytes

Qṫ¥Ðḟ8

Try it online!

How it works

Qṫ¥Ðḟ8  Main link. Argument: A (array or words)



   Ðḟ   Filterfalse; only keep the words W of A for which the chain to the left

        returns a falsy value.

  ¥         Combine the two links to the left into a dyadic chain.

Q               Unique; remove duplicate letters from W.

 ṫ   8          Tail 8; remove the first 7 letters of the result.

answered Nov 9 at 2:13

Dennis♦

184k32295731

add a comment |

up vote
2
down vote

Jelly, 6 bytes

Qṫ¥Ðḟ8

Try it online!

How it works

Qṫ¥Ðḟ8  Main link. Argument: A (array or words)



   Ðḟ   Filterfalse; only keep the words W of A for which the chain to the left

        returns a falsy value.

  ¥         Combine the two links to the left into a dyadic chain.

Q               Unique; remove duplicate letters from W.

 ṫ   8          Tail 8; remove the first 7 letters of the result.

answered Nov 9 at 2:13

Dennis♦

184k32295731

Jelly, 6 bytes

Qṫ¥Ðḟ8

Try it online!

How it works

Qṫ¥Ðḟ8  Main link. Argument: A (array or words)



   Ðḟ   Filterfalse; only keep the words W of A for which the chain to the left

        returns a falsy value.

  ¥         Combine the two links to the left into a dyadic chain.

Q               Unique; remove duplicate letters from W.

 ṫ   8          Tail 8; remove the first 7 letters of the result.

answered Nov 9 at 2:13

Dennis♦

184k32295731

answered Nov 9 at 2:13

Dennis♦

184k32295731

answered Nov 9 at 2:13

Dennis♦

184k32295731

answered Nov 9 at 2:13

Dennis♦

184k32295731

add a comment |

up vote
2
down vote

MathGolf, 6 bytes

Ç{▀£7>

Try it online!

Explanation

Really similar to the 05AB1E solution, but I lose one byte thanks to explicitly having to define the code block for filtering.

Ç       Implicit faulty filter by block

 {      Start block

  ▀     Get unique characters of string

   £    Get length

    7>  Is greater than 7

answered Nov 9 at 10:51

maxb

2,1081923

add a comment |

up vote
2
down vote

MathGolf, 6 bytes

Ç{▀£7>

Try it online!

Explanation

Really similar to the 05AB1E solution, but I lose one byte thanks to explicitly having to define the code block for filtering.

Ç       Implicit faulty filter by block

 {      Start block

  ▀     Get unique characters of string

   £    Get length

    7>  Is greater than 7

answered Nov 9 at 10:51

maxb

2,1081923

add a comment |

up vote
2
down vote

MathGolf, 6 bytes

Ç{▀£7>

Try it online!

Explanation

Really similar to the 05AB1E solution, but I lose one byte thanks to explicitly having to define the code block for filtering.

Ç       Implicit faulty filter by block

 {      Start block

  ▀     Get unique characters of string

   £    Get length

    7>  Is greater than 7

answered Nov 9 at 10:51

maxb

2,1081923

MathGolf, 6 bytes

Ç{▀£7>

Try it online!

Explanation

Really similar to the 05AB1E solution, but I lose one byte thanks to explicitly having to define the code block for filtering.

Ç       Implicit faulty filter by block

 {      Start block

  ▀     Get unique characters of string

   £    Get length

    7>  Is greater than 7

answered Nov 9 at 10:51

maxb

2,1081923

answered Nov 9 at 10:51

maxb

2,1081923

answered Nov 9 at 10:51

maxb

2,1081923

answered Nov 9 at 10:51

maxb

2,1081923

add a comment |

up vote
2
down vote

JavaScript, 33 bytes

a=>a.filter(s=>new Set(s).size<8)

Try it online, using Dennis' test cases

answered Nov 9 at 10:54

Shaggy

18.2k21663

add a comment |

up vote
2
down vote

JavaScript, 33 bytes

a=>a.filter(s=>new Set(s).size<8)

Try it online, using Dennis' test cases

answered Nov 9 at 10:54

Shaggy

18.2k21663

add a comment |

up vote
2
down vote

JavaScript, 33 bytes

a=>a.filter(s=>new Set(s).size<8)

Try it online, using Dennis' test cases

answered Nov 9 at 10:54

Shaggy

18.2k21663

JavaScript, 33 bytes

a=>a.filter(s=>new Set(s).size<8)

Try it online, using Dennis' test cases

answered Nov 9 at 10:54

Shaggy

18.2k21663

answered Nov 9 at 10:54

Shaggy

18.2k21663

answered Nov 9 at 10:54

Shaggy

18.2k21663

answered Nov 9 at 10:54

Shaggy

18.2k21663

add a comment |

up vote
2
down vote

J, 10 bytes

#~8>#@~.@>

explanation

#~ 8 > #@~.@>

#~                 NB. filter the input based on...

   8 >             NB. is 8 greater than...

       #@          NB. the length of...

         ~.@       NB. the unique characters of...

            >      NB. the unboxed input.

Try it online!

edited Nov 9 at 14:18

answered Nov 9 at 3:48

Jonah

1,981816

1

Shouldn't 9 be 8?
– Galen Ivanov
Nov 9 at 8:03

1

Fixed. Weird, I could have sworn the OP said 8 was the max length allowed before. Anyway, ty...
– Jonah
Nov 9 at 14:19

add a comment |

up vote
2
down vote

J, 10 bytes

#~8>#@~.@>

explanation

#~ 8 > #@~.@>

#~                 NB. filter the input based on...

   8 >             NB. is 8 greater than...

       #@          NB. the length of...

         ~.@       NB. the unique characters of...

            >      NB. the unboxed input.

Try it online!

edited Nov 9 at 14:18

answered Nov 9 at 3:48

Jonah

1,981816

1

Shouldn't 9 be 8?
– Galen Ivanov
Nov 9 at 8:03

1

Fixed. Weird, I could have sworn the OP said 8 was the max length allowed before. Anyway, ty...
– Jonah
Nov 9 at 14:19

add a comment |

up vote
2
down vote

J, 10 bytes

#~8>#@~.@>

explanation

#~ 8 > #@~.@>

#~                 NB. filter the input based on...

   8 >             NB. is 8 greater than...

       #@          NB. the length of...

         ~.@       NB. the unique characters of...

            >      NB. the unboxed input.

Try it online!

edited Nov 9 at 14:18

answered Nov 9 at 3:48

Jonah

1,981816

J, 10 bytes

#~8>#@~.@>

explanation

#~ 8 > #@~.@>

#~                 NB. filter the input based on...

   8 >             NB. is 8 greater than...

       #@          NB. the length of...

         ~.@       NB. the unique characters of...

            >      NB. the unboxed input.

Try it online!

edited Nov 9 at 14:18

answered Nov 9 at 3:48

Jonah

1,981816

edited Nov 9 at 14:18

answered Nov 9 at 3:48

Jonah

1,981816

answered Nov 9 at 3:48

Jonah

1,981816

answered Nov 9 at 3:48

Jonah

1,981816

1

Shouldn't 9 be 8?
– Galen Ivanov
Nov 9 at 8:03

1

Fixed. Weird, I could have sworn the OP said 8 was the max length allowed before. Anyway, ty...
– Jonah
Nov 9 at 14:19

add a comment |

1

Shouldn't 9 be 8?
– Galen Ivanov
Nov 9 at 8:03

1

Fixed. Weird, I could have sworn the OP said 8 was the max length allowed before. Anyway, ty...
– Jonah
Nov 9 at 14:19

Shouldn't 9 be 8?
– Galen Ivanov
Nov 9 at 8:03

Fixed. Weird, I could have sworn the OP said 8 was the max length allowed before. Anyway, ty...
– Jonah
Nov 9 at 14:19

add a comment |

up vote
2
down vote

Java 8, 46 bytes

s->s.filter(w->w.chars().distinct().count()<8)

Try it online.

Explanation:

s->               // Method with String-Stream as both parameter and return-type

  s.filter(w->    //  Filter the words in the input-Stream by:

    w.chars()     //  Convert the String to characters

     .distinct()  //  Only leave distinct characters

     .count()<8)  //  Only leave words with less than 8 distinct characters

edited Nov 10 at 10:16

answered Nov 9 at 9:42

Kevin Cruijssen

34.4k554182

Wow, w.chars.distinct.count. Turn each of those words into a symbol and you have a golfing language! xD
– Quintec
Nov 9 at 12:56

@Quintec Hehe, it's what I do in my 05AB1E answer. :) s->s is implicit input in 05AB1E; filter(w-> is ʒ; .chars() is implicitly again; .distinct() is Ù; .count() (or length) is g; <8 is 8‹; and the closing ) is implicitly since it's at the end of the 05AB1E program (if I wanted to do something after the filter I'd have to close it with }); and finally outputting is implicitly again. So my Java and 05AB1E are similar, one is a 46 bytes function and the other a 5 bytes full program, though. xD
– Kevin Cruijssen
Nov 9 at 13:06

Hehe, also basically same for my APL answer. It meets in the middle: 11 bytes :P
– Quintec
Nov 9 at 15:24

add a comment |

up vote
2
down vote

Java 8, 46 bytes

s->s.filter(w->w.chars().distinct().count()<8)

Try it online.

Explanation:

s->               // Method with String-Stream as both parameter and return-type

  s.filter(w->    //  Filter the words in the input-Stream by:

    w.chars()     //  Convert the String to characters

     .distinct()  //  Only leave distinct characters

     .count()<8)  //  Only leave words with less than 8 distinct characters

edited Nov 10 at 10:16

answered Nov 9 at 9:42

Kevin Cruijssen

34.4k554182

Wow, w.chars.distinct.count. Turn each of those words into a symbol and you have a golfing language! xD
– Quintec
Nov 9 at 12:56

@Quintec Hehe, it's what I do in my 05AB1E answer. :) s->s is implicit input in 05AB1E; filter(w-> is ʒ; .chars() is implicitly again; .distinct() is Ù; .count() (or length) is g; <8 is 8‹; and the closing ) is implicitly since it's at the end of the 05AB1E program (if I wanted to do something after the filter I'd have to close it with }); and finally outputting is implicitly again. So my Java and 05AB1E are similar, one is a 46 bytes function and the other a 5 bytes full program, though. xD
– Kevin Cruijssen
Nov 9 at 13:06

Hehe, also basically same for my APL answer. It meets in the middle: 11 bytes :P
– Quintec
Nov 9 at 15:24

add a comment |

up vote
2
down vote

Java 8, 46 bytes

s->s.filter(w->w.chars().distinct().count()<8)

Try it online.

Explanation:

s->               // Method with String-Stream as both parameter and return-type

  s.filter(w->    //  Filter the words in the input-Stream by:

    w.chars()     //  Convert the String to characters

     .distinct()  //  Only leave distinct characters

     .count()<8)  //  Only leave words with less than 8 distinct characters

edited Nov 10 at 10:16

answered Nov 9 at 9:42

Kevin Cruijssen

34.4k554182

Java 8, 46 bytes

s->s.filter(w->w.chars().distinct().count()<8)

Try it online.

Explanation:

s->               // Method with String-Stream as both parameter and return-type

  s.filter(w->    //  Filter the words in the input-Stream by:

    w.chars()     //  Convert the String to characters

     .distinct()  //  Only leave distinct characters

     .count()<8)  //  Only leave words with less than 8 distinct characters

edited Nov 10 at 10:16

answered Nov 9 at 9:42

Kevin Cruijssen

34.4k554182

edited Nov 10 at 10:16

answered Nov 9 at 9:42

Kevin Cruijssen

34.4k554182

answered Nov 9 at 9:42

Kevin Cruijssen

34.4k554182

answered Nov 9 at 9:42

Kevin Cruijssen

34.4k554182

Wow, w.chars.distinct.count. Turn each of those words into a symbol and you have a golfing language! xD
– Quintec
Nov 9 at 12:56

@Quintec Hehe, it's what I do in my 05AB1E answer. :) s->s is implicit input in 05AB1E; filter(w-> is ʒ; .chars() is implicitly again; .distinct() is Ù; .count() (or length) is g; <8 is 8‹; and the closing ) is implicitly since it's at the end of the 05AB1E program (if I wanted to do something after the filter I'd have to close it with }); and finally outputting is implicitly again. So my Java and 05AB1E are similar, one is a 46 bytes function and the other a 5 bytes full program, though. xD
– Kevin Cruijssen
Nov 9 at 13:06

Hehe, also basically same for my APL answer. It meets in the middle: 11 bytes :P
– Quintec
Nov 9 at 15:24

add a comment |

Wow, w.chars.distinct.count. Turn each of those words into a symbol and you have a golfing language! xD
– Quintec
Nov 9 at 12:56

@Quintec Hehe, it's what I do in my 05AB1E answer. :) s->s is implicit input in 05AB1E; filter(w-> is ʒ; .chars() is implicitly again; .distinct() is Ù; .count() (or length) is g; <8 is 8‹; and the closing ) is implicitly since it's at the end of the 05AB1E program (if I wanted to do something after the filter I'd have to close it with }); and finally outputting is implicitly again. So my Java and 05AB1E are similar, one is a 46 bytes function and the other a 5 bytes full program, though. xD
– Kevin Cruijssen
Nov 9 at 13:06

Hehe, also basically same for my APL answer. It meets in the middle: 11 bytes :P
– Quintec
Nov 9 at 15:24

Wow, w.chars.distinct.count. Turn each of those words into a symbol and you have a golfing language! xD
– Quintec
Nov 9 at 12:56

@Quintec Hehe, it's what I do in my 05AB1E answer. :) s->s is implicit input in 05AB1E; filter(w-> is ʒ; .chars() is implicitly again; .distinct() is Ù; .count() (or length) is g; <8 is 8‹; and the closing ) is implicitly since it's at the end of the 05AB1E program (if I wanted to do something after the filter I'd have to close it with }); and finally outputting is implicitly again. So my Java and 05AB1E are similar, one is a 46 bytes function and the other a 5 bytes full program, though. xD
– Kevin Cruijssen
Nov 9 at 13:06

Hehe, also basically same for my APL answer. It meets in the middle: 11 bytes :P
– Quintec
Nov 9 at 15:24

add a comment |

up vote
1
down vote

Python 2, 40 bytes

lambda i:[x for x in i if len(set(x))<8]

Try it online!

Test cases borrowed from @Dennis. Input and output are both lists.

answered Nov 9 at 7:05

ElPedro

3,4131023

add a comment |

up vote
1
down vote

Python 2, 40 bytes

lambda i:[x for x in i if len(set(x))<8]

Try it online!

Test cases borrowed from @Dennis. Input and output are both lists.

answered Nov 9 at 7:05

ElPedro

3,4131023

add a comment |

up vote
1
down vote

Python 2, 40 bytes

lambda i:[x for x in i if len(set(x))<8]

Try it online!

Test cases borrowed from @Dennis. Input and output are both lists.

answered Nov 9 at 7:05

ElPedro

3,4131023

Python 2, 40 bytes

lambda i:[x for x in i if len(set(x))<8]

Try it online!

Test cases borrowed from @Dennis. Input and output are both lists.

answered Nov 9 at 7:05

ElPedro

3,4131023

answered Nov 9 at 7:05

ElPedro

3,4131023

answered Nov 9 at 7:05

ElPedro

3,4131023

answered Nov 9 at 7:05

ElPedro

3,4131023

add a comment |

up vote
1
down vote

Red, 54 bytes

func[b][foreach a b[if 8 > length? unique a[print a]]]

Try it online!

The first test set was taken from Dennis'

answered Nov 9 at 8:04

Galen Ivanov

5,94711032

add a comment |

up vote
1
down vote

Red, 54 bytes

func[b][foreach a b[if 8 > length? unique a[print a]]]

Try it online!

The first test set was taken from Dennis'

answered Nov 9 at 8:04

Galen Ivanov

5,94711032

add a comment |

up vote
1
down vote

Red, 54 bytes

func[b][foreach a b[if 8 > length? unique a[print a]]]

Try it online!

The first test set was taken from Dennis'

answered Nov 9 at 8:04

Galen Ivanov

5,94711032

Red, 54 bytes

func[b][foreach a b[if 8 > length? unique a[print a]]]

Try it online!

The first test set was taken from Dennis'

answered Nov 9 at 8:04

Galen Ivanov

5,94711032

answered Nov 9 at 8:04

Galen Ivanov

5,94711032

answered Nov 9 at 8:04

Galen Ivanov

5,94711032

answered Nov 9 at 8:04

Galen Ivanov

5,94711032

add a comment |

up vote
1
down vote

C# (.NET Core), 58 bytes

 a=>a.Where(x=>x.GroupBy(y=>y).Count()<8)

Try It Online!

answered Nov 9 at 13:34

LiefdeWen

2,502936

add a comment |

up vote
1
down vote

C# (.NET Core), 58 bytes

 a=>a.Where(x=>x.GroupBy(y=>y).Count()<8)

Try It Online!

answered Nov 9 at 13:34

LiefdeWen

2,502936

add a comment |

up vote
1
down vote

C# (.NET Core), 58 bytes

 a=>a.Where(x=>x.GroupBy(y=>y).Count()<8)

Try It Online!

answered Nov 9 at 13:34

LiefdeWen

2,502936

C# (.NET Core), 58 bytes

 a=>a.Where(x=>x.GroupBy(y=>y).Count()<8)

Try It Online!

answered Nov 9 at 13:34

LiefdeWen

2,502936

answered Nov 9 at 13:34

LiefdeWen

2,502936

answered Nov 9 at 13:34

LiefdeWen

2,502936

answered Nov 9 at 13:34

LiefdeWen

2,502936

add a comment |

up vote
1
down vote

C (gcc), 104 95 bytes

Dennis's word set was also used here.

Thanks to nwellnhof for the suggestions.

char*s,*t;f(i,j,k){for(;~scanf("%ms",&s);i<8&&puts(s))for(i=j=0,t=s;*t;j=k)k=j|1<<*t++,i+=j<k;}

Try it online!

edited Nov 9 at 17:15

answered Nov 9 at 9:36

ErikF

1,25917

96 bytes, if you're OK with implementation- and undefined behavior.
– nwellnhof
Nov 9 at 10:15

@nwellnhof What's a little UD between friends for code golf? :-)
– ErikF
Nov 9 at 17:16

add a comment |

up vote
1
down vote

C (gcc), 104 95 bytes

Dennis's word set was also used here.

Thanks to nwellnhof for the suggestions.

char*s,*t;f(i,j,k){for(;~scanf("%ms",&s);i<8&&puts(s))for(i=j=0,t=s;*t;j=k)k=j|1<<*t++,i+=j<k;}

Try it online!

edited Nov 9 at 17:15

answered Nov 9 at 9:36

ErikF

1,25917

96 bytes, if you're OK with implementation- and undefined behavior.
– nwellnhof
Nov 9 at 10:15

@nwellnhof What's a little UD between friends for code golf? :-)
– ErikF
Nov 9 at 17:16

add a comment |

up vote
1
down vote

C (gcc), 104 95 bytes

Dennis's word set was also used here.

Thanks to nwellnhof for the suggestions.

char*s,*t;f(i,j,k){for(;~scanf("%ms",&s);i<8&&puts(s))for(i=j=0,t=s;*t;j=k)k=j|1<<*t++,i+=j<k;}

Try it online!

edited Nov 9 at 17:15

answered Nov 9 at 9:36

ErikF

1,25917

C (gcc), 104 95 bytes

Dennis's word set was also used here.

Thanks to nwellnhof for the suggestions.

char*s,*t;f(i,j,k){for(;~scanf("%ms",&s);i<8&&puts(s))for(i=j=0,t=s;*t;j=k)k=j|1<<*t++,i+=j<k;}

Try it online!

edited Nov 9 at 17:15

answered Nov 9 at 9:36

ErikF

1,25917

edited Nov 9 at 17:15

answered Nov 9 at 9:36

ErikF

1,25917

answered Nov 9 at 9:36

ErikF

1,25917

answered Nov 9 at 9:36

ErikF

1,25917

96 bytes, if you're OK with implementation- and undefined behavior.
– nwellnhof
Nov 9 at 10:15

@nwellnhof What's a little UD between friends for code golf? :-)
– ErikF
Nov 9 at 17:16

add a comment |

96 bytes, if you're OK with implementation- and undefined behavior.
– nwellnhof
Nov 9 at 10:15

@nwellnhof What's a little UD between friends for code golf? :-)
– ErikF
Nov 9 at 17:16

96 bytes, if you're OK with implementation- and undefined behavior.
– nwellnhof
Nov 9 at 10:15

@nwellnhof What's a little UD between friends for code golf? :-)
– ErikF
Nov 9 at 17:16

add a comment |

up vote
1
down vote

perl -nlE, 32 bytes

my%h;@h{/./g}=();say if keys%h<8

This reads words from STDIN, printing out those with less than 7 different characters.

answered Nov 10 at 17:13

Abigail

41717

1

25 bytes: my%h;@h{/./g}=1;%h<8&&say (%h<8 requires Perl 5.26)
– nwellnhof
Nov 12 at 12:25

add a comment |

up vote
1
down vote

perl -nlE, 32 bytes

my%h;@h{/./g}=();say if keys%h<8

This reads words from STDIN, printing out those with less than 7 different characters.

answered Nov 10 at 17:13

Abigail

41717

1

25 bytes: my%h;@h{/./g}=1;%h<8&&say (%h<8 requires Perl 5.26)
– nwellnhof
Nov 12 at 12:25

add a comment |

up vote
1
down vote

perl -nlE, 32 bytes

my%h;@h{/./g}=();say if keys%h<8

This reads words from STDIN, printing out those with less than 7 different characters.

answered Nov 10 at 17:13

Abigail

41717

perl -nlE, 32 bytes

my%h;@h{/./g}=();say if keys%h<8

This reads words from STDIN, printing out those with less than 7 different characters.

answered Nov 10 at 17:13

Abigail

41717

answered Nov 10 at 17:13

Abigail

41717

answered Nov 10 at 17:13

Abigail

41717

answered Nov 10 at 17:13

Abigail

41717

1

25 bytes: my%h;@h{/./g}=1;%h<8&&say (%h<8 requires Perl 5.26)
– nwellnhof
Nov 12 at 12:25

add a comment |

1

25 bytes: my%h;@h{/./g}=1;%h<8&&say (%h<8 requires Perl 5.26)
– nwellnhof
Nov 12 at 12:25

25 bytes: my%h;@h{/./g}=1;%h<8&&say (%h<8 requires Perl 5.26)
– nwellnhof
Nov 12 at 12:25

add a comment |

up vote
0
down vote

Racket, 95 bytes

(require racket/set)(define(f l)(filter(lambda(x)(<(set-count(list->set(string->list x)))8))l))

Try it online!

The test set was taken from Dennis'

answered Nov 9 at 8:45

Galen Ivanov

5,94711032

add a comment |

up vote
0
down vote

Racket, 95 bytes

(require racket/set)(define(f l)(filter(lambda(x)(<(set-count(list->set(string->list x)))8))l))

Try it online!

The test set was taken from Dennis'

answered Nov 9 at 8:45

Galen Ivanov

5,94711032

add a comment |

up vote
0
down vote

Racket, 95 bytes

(require racket/set)(define(f l)(filter(lambda(x)(<(set-count(list->set(string->list x)))8))l))

Try it online!

The test set was taken from Dennis'

answered Nov 9 at 8:45

Galen Ivanov

5,94711032

Racket, 95 bytes

(require racket/set)(define(f l)(filter(lambda(x)(<(set-count(list->set(string->list x)))8))l))

Try it online!

The test set was taken from Dennis'

answered Nov 9 at 8:45

Galen Ivanov

5,94711032

answered Nov 9 at 8:45

Galen Ivanov

5,94711032

answered Nov 9 at 8:45

Galen Ivanov

5,94711032

answered Nov 9 at 8:45

Galen Ivanov

5,94711032

add a comment |

up vote
0
down vote

Charcoal, 12 bytes

ΦＡ›⁸ＬΦι⁼μ⌕ιλ

Try it online! Link is to verbose version of code. Explanation:

 Ａ              Input array

Φ               Filter strings where

   ⁸            Literal 8

  ›             Is greater than

    Ｌ           Length of

      ι         Current string

     Φ          Filtered on characters where

        μ       Inner index

       ⁼        Equals

           λ    Current character's

         ⌕      First index in

          ι     Current string

                Implicitly print matching strings

answered Nov 9 at 9:33

Neil

78.1k744175

10?
– ASCII-only
Nov 18 at 5:12

add a comment |

up vote
0
down vote

Charcoal, 12 bytes

ΦＡ›⁸ＬΦι⁼μ⌕ιλ

Try it online! Link is to verbose version of code. Explanation:

 Ａ              Input array

Φ               Filter strings where

   ⁸            Literal 8

  ›             Is greater than

    Ｌ           Length of

      ι         Current string

     Φ          Filtered on characters where

        μ       Inner index

       ⁼        Equals

           λ    Current character's

         ⌕      First index in

          ι     Current string

                Implicitly print matching strings

answered Nov 9 at 9:33

Neil

78.1k744175

10?
– ASCII-only
Nov 18 at 5:12

add a comment |

up vote
0
down vote

Charcoal, 12 bytes

ΦＡ›⁸ＬΦι⁼μ⌕ιλ

Try it online! Link is to verbose version of code. Explanation:

 Ａ              Input array

Φ               Filter strings where

   ⁸            Literal 8

  ›             Is greater than

    Ｌ           Length of

      ι         Current string

     Φ          Filtered on characters where

        μ       Inner index

       ⁼        Equals

           λ    Current character's

         ⌕      First index in

          ι     Current string

                Implicitly print matching strings

answered Nov 9 at 9:33

Neil

78.1k744175

Charcoal, 12 bytes

ΦＡ›⁸ＬΦι⁼μ⌕ιλ

Try it online! Link is to verbose version of code. Explanation:

 Ａ              Input array

Φ               Filter strings where

   ⁸            Literal 8

  ›             Is greater than

    Ｌ           Length of

      ι         Current string

     Φ          Filtered on characters where

        μ       Inner index

       ⁼        Equals

           λ    Current character's

         ⌕      First index in

          ι     Current string

                Implicitly print matching strings

answered Nov 9 at 9:33

Neil

78.1k744175

answered Nov 9 at 9:33

Neil

78.1k744175

answered Nov 9 at 9:33

Neil

78.1k744175

answered Nov 9 at 9:33

Neil

78.1k744175

10?
– ASCII-only
Nov 18 at 5:12

add a comment |

10?
– ASCII-only
Nov 18 at 5:12

10?
– ASCII-only
Nov 18 at 5:12

add a comment |

up vote
0
down vote

Japt `-f`, 6 bytes

¬â Ê<8

Try it, using Dennis' test cases

answered Nov 9 at 10:57

Shaggy

18.2k21663

add a comment |

up vote
0
down vote

Japt `-f`, 6 bytes

¬â Ê<8

Try it, using Dennis' test cases

answered Nov 9 at 10:57

Shaggy

18.2k21663

add a comment |

up vote
0
down vote

Japt `-f`, 6 bytes

¬â Ê<8

Try it, using Dennis' test cases

answered Nov 9 at 10:57

Shaggy

18.2k21663

Japt `-f`, 6 bytes

¬â Ê<8

Try it, using Dennis' test cases

answered Nov 9 at 10:57

Shaggy

18.2k21663

answered Nov 9 at 10:57

Shaggy

18.2k21663

answered Nov 9 at 10:57

Shaggy

18.2k21663

answered Nov 9 at 10:57

Shaggy

18.2k21663

add a comment |

up vote
0
down vote

JavaScript (ES6), 53 bytes

Without using a set:

a=>a.filter(w=>[...w].every(o=c=>o[c]=o[c]||--k,k=8))

Try it online! (using Dennis' test set)

edited Nov 9 at 11:23

answered Nov 9 at 10:32

Arnauld

69.5k586294

add a comment |

up vote
0
down vote

JavaScript (ES6), 53 bytes

Without using a set:

a=>a.filter(w=>[...w].every(o=c=>o[c]=o[c]||--k,k=8))

Try it online! (using Dennis' test set)

edited Nov 9 at 11:23

answered Nov 9 at 10:32

Arnauld

69.5k586294

add a comment |

up vote
0
down vote

JavaScript (ES6), 53 bytes

Without using a set:

a=>a.filter(w=>[...w].every(o=c=>o[c]=o[c]||--k,k=8))

Try it online! (using Dennis' test set)

edited Nov 9 at 11:23

answered Nov 9 at 10:32

Arnauld

69.5k586294

JavaScript (ES6), 53 bytes

Without using a set:

a=>a.filter(w=>[...w].every(o=c=>o[c]=o[c]||--k,k=8))

Try it online! (using Dennis' test set)

edited Nov 9 at 11:23

answered Nov 9 at 10:32

Arnauld

69.5k586294

edited Nov 9 at 11:23

answered Nov 9 at 10:32

Arnauld

69.5k586294

answered Nov 9 at 10:32

Arnauld

69.5k586294

answered Nov 9 at 10:32

Arnauld

69.5k586294

add a comment |

up vote
0
down vote

Snap! 4, scratchblocks2 syntax

(pretending that Snap! exclusive blocks are valid in scratchblocks2 but functions use the Scratch define)

69 bytes

b takes a list of strings.

define((b)

report((#)keep items such that(<(length of )<[7]>)from(b

answered Nov 9 at 23:49

Silas Reel

316

add a comment |

up vote
0
down vote

Snap! 4, scratchblocks2 syntax

(pretending that Snap! exclusive blocks are valid in scratchblocks2 but functions use the Scratch define)

69 bytes

b takes a list of strings.

define((b)

report((#)keep items such that(<(length of )<[7]>)from(b

answered Nov 9 at 23:49

Silas Reel

316

add a comment |

up vote
0
down vote

Snap! 4, scratchblocks2 syntax

(pretending that Snap! exclusive blocks are valid in scratchblocks2 but functions use the Scratch define)

69 bytes

b takes a list of strings.

define((b)

report((#)keep items such that(<(length of )<[7]>)from(b

answered Nov 9 at 23:49

Silas Reel

316

Snap! 4, scratchblocks2 syntax

(pretending that Snap! exclusive blocks are valid in scratchblocks2 but functions use the Scratch define)

69 bytes

b takes a list of strings.

define((b)

report((#)keep items such that(<(length of )<[7]>)from(b

answered Nov 9 at 23:49

Silas Reel

316

answered Nov 9 at 23:49

Silas Reel

316

answered Nov 9 at 23:49

Silas Reel

316

answered Nov 9 at 23:49

Silas Reel

316

add a comment |

up vote
0
down vote

C++ (gcc), 105 104 bytes

#import<bits/stdc++.h>

f(){for(char*n;~scanf("%ms",&n);)std::set<int>(n,n+strlen(n)).size()<8&&puts(n);}

Try it online!

-1 byte thanks to @ceilingcat

edited yesterday

answered Nov 9 at 19:19

Logern

70546

add a comment |

up vote
0
down vote

C++ (gcc), 105 104 bytes

#import<bits/stdc++.h>

f(){for(char*n;~scanf("%ms",&n);)std::set<int>(n,n+strlen(n)).size()<8&&puts(n);}

Try it online!

-1 byte thanks to @ceilingcat

edited yesterday

answered Nov 9 at 19:19

Logern

70546

add a comment |

up vote
0
down vote

C++ (gcc), 105 104 bytes

#import<bits/stdc++.h>

f(){for(char*n;~scanf("%ms",&n);)std::set<int>(n,n+strlen(n)).size()<8&&puts(n);}

Try it online!

-1 byte thanks to @ceilingcat

edited yesterday

answered Nov 9 at 19:19

Logern

70546

C++ (gcc), 105 104 bytes

#import<bits/stdc++.h>

f(){for(char*n;~scanf("%ms",&n);)std::set<int>(n,n+strlen(n)).size()<8&&puts(n);}

Try it online!

-1 byte thanks to @ceilingcat

edited yesterday

answered Nov 9 at 19:19

Logern

70546

edited yesterday

answered Nov 9 at 19:19

Logern

70546

answered Nov 9 at 19:19

Logern

70546

answered Nov 9 at 19:19

Logern

70546

add a comment |

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Spelling Bee Acceptable

19 Answers 19

APL (Dyalog Unicode), 11 bytes

Perl 6, 20 bytes

05AB1E, 5 bytes

Jelly, 6 bytes

How it works

MathGolf, 6 bytes

Explanation

JavaScript, 33 bytes

J, 10 bytes

explanation

Java 8, 46 bytes

Python 2, 40 bytes

Red, 54 bytes

C# (.NET Core), 58 bytes

C (gcc), 104 95 bytes

perl -nlE, 32 bytes

Racket, 95 bytes

Charcoal, 12 bytes

Japt -f, 6 bytes

JavaScript (ES6), 53 bytes

Snap! 4, scratchblocks2 syntax

69 bytes

C++ (gcc), 105 104 bytes

Your Answer

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19 Answers 19

19 Answers 19

APL (Dyalog Unicode), 11 bytes

APL (Dyalog Unicode), 11 bytes

APL (Dyalog Unicode), 11 bytes

APL (Dyalog Unicode), 11 bytes

Perl 6, 20 bytes

Perl 6, 20 bytes

Perl 6, 20 bytes

Perl 6, 20 bytes

05AB1E, 5 bytes

05AB1E, 5 bytes

05AB1E, 5 bytes

05AB1E, 5 bytes

Jelly, 6 bytes

How it works

Jelly, 6 bytes

How it works

Jelly, 6 bytes

How it works

Jelly, 6 bytes

How it works

MathGolf, 6 bytes

Explanation

MathGolf, 6 bytes

Explanation

MathGolf, 6 bytes

Explanation

MathGolf, 6 bytes

Explanation

JavaScript, 33 bytes

JavaScript, 33 bytes

JavaScript, 33 bytes

JavaScript, 33 bytes

J, 10 bytes

explanation

J, 10 bytes

explanation

J, 10 bytes

explanation

J, 10 bytes

explanation

Java 8, 46 bytes

Java 8, 46 bytes

Java 8, 46 bytes

Java 8, 46 bytes

Python 2, 40 bytes

Python 2, 40 bytes

Python 2, 40 bytes

Python 2, 40 bytes

Red, 54 bytes

19 Answers
19

Japt `-f`, 6 bytes

19 Answers
19

19 Answers
19

Japt `-f`, 6 bytes

Japt `-f`, 6 bytes

Japt `-f`, 6 bytes

Japt `-f`, 6 bytes