33
\$\begingroup\$

Given a string as input find the longest contiguous substring that does not have any character twice or more. If there are multiple such substrings you may output either. You may assume that the input is on the printable ASCII range if you wish.

Scoring

Answers will first be ranked by the length of their own longest non-repeating substring, and then by their total length. Lower scores will be better for both criteria. Depending on the language this will probably feel like a challenge with a source restriction.

Triviality

In some languages achieving a score of 1, x (lenguage) or 2, x (Brain-flak and other turing tarpits) is pretty easy, however there are other languages in which minimizing the longest non-repeating substring is a challenge. I had a lot of fun getting a score of 2 in Haskell, so I encourage you to seek out languages where this task is fun.

Test cases

"Good morning, Green orb!" -> "ing, Gre"
"fffffffffff" -> "f"
"oiiiiioiiii" -> "io", "oi"
"1234567890"  -> "1234567890"
"11122324455" -> "324"

Scoring submission

You can score your programs using the following snippet:

input.addEventListener("input", change);

// note: intentionally verbose :)
function isUnique(str) {
  var maxCount = 0;
  var counts = {};
  for(var i = 0; i < str.length; i++) {
    var c = str.charAt(i);
    counts[c] |= 0;
    counts[c]++;
    if(maxCount < counts[c]) {
      maxCount = counts[c];
    }
  }
  return maxCount <= 1;
}

function maximizeSubstring(str, pred, cost) {
  var mostExpensive = -1;
  // iterate over substrings
  function iterate(start, end) {
    var slice = str.slice(start, end);
    if(pred(slice)) {
      var profit = cost(slice);
      if(profit > mostExpensive) {
        mostExpensive = profit;
      }
    }
    end++;
    if(start >= str.length) {
      return;
    }
    else if(end > str.length) {
      start++;
      iterate(start, start);
    }
    else {
      iterate(start, end);
    }
  }
  iterate(0, 0);
  return mostExpensive;
}

function size(x) {
  return x.length;
}

function longestNonRepeatingSize(str) {
  return maximizeSubstring(str, isUnique, size);
}

function change() {
  var code = input.value;
  output.value = "Your score is: " + longestNonRepeatingSize(code);
}

change();
* {
  font-family: monospace;
}
Input code here:
<br>
<textarea id="input"></textarea>
<br>
<textarea id="output"></textarea>

\$\endgroup\$
  • \$\begingroup\$ Proposed test case: 11122324455 Jonathan Allan realized that my first revision didn't handle it correctly. \$\endgroup\$ – Dennis Jan 30 '18 at 0:08
  • \$\begingroup\$ @Dennis Test case added. I'm curious as to how that happened. \$\endgroup\$ – Wheat Wizard Jan 30 '18 at 0:09
  • 2
    \$\begingroup\$ I generated all substrings (already sorted by length), then deduplicated the substrings and kept those that remained substrings. Unfortunately, that alterates the order; 11122 ocurrs after 324, but gets deduplicated to 12. \$\endgroup\$ – Dennis Jan 30 '18 at 0:17
  • \$\begingroup\$ I'm wondering where the whitespace answer is. \$\endgroup\$ – Magic Octopus Urn Feb 2 '18 at 20:50

21 Answers 21

13
\$\begingroup\$

C, score 2,  747   720  662 bytes

L  [  1  <<  7  ]  ,  *  q  ,  *  r  ,  l  ,  d  ,  i  ,  c  ,  j  ,  s  ,  t  ,  k  =  1  <<  7  ;  h  (  )  {  q  =  s  +  i  +  j  ++  ;  *  q  %  k  &&  !  L  [  *  q  %  k  ]  ++  &&  h  (  ++  c  )  ;  }  g  (  )  {  q  =  s  +  i  ;  *  q  %  k  ?  z  (  k  )  ,  h  (  j  =  c  =  0  )  ,  c  >  d  &&  (  d  =  c  )  &&  (  l  =  i  )  ,  g  (  ++  i  )  :  0  ;  }  f  (  S  ,  T  )  {  s  =  S  ;  l  =  i  =  d  =  0  ;  g  (  t  =  T  )  ;  p  (  i  =  0  )  ;  }  p  (  )  {  q  =  s  +  l  +  i  ;  r  =  t  +  i  ;  i  ++  <  d  ?  p  (  *  r  =  *  q  )  :  (  *  r  =  0  )  ;  }  z  (  i  )  {  L  [  --  i  ]  =  0  ;  i  &&  z  (  i  )  ;  }

Works at least on 32-bit MinGW (with optimizations disabled). Doesn't use a single keyword.

Works apparently on TIO with gcc and clang, too: Try it online! (Thanks @Dennis!)

Call with:

int main()
{
    char str[1024];

    f("Good morning, Green orb!", str);
    puts(str);

    f("fffffffffff", str);
    puts(str);

    f("oiiiiioiiii", str);
    puts(str);

    f("1234567890", str);
    puts(str);

    f("L  [  1  <<  7  ]  ,  *  q  ,  *  r  ,  l  ,  d  ,  i  ,  c  ,  j  ,  s  ,  t  ,  k  =  1  <<  7  ;  h  (  )  {  q  =  s  +  i  +  j  ++  ;  *  q  %  k  &&  !  L  [  *  q  %  k  ]  ++  &&  h  (  ++  c  )  ;  }  g  (  )  {  q  =  s  +  i  ;  *  q  %  k  ?  z  (  k  )  ,  h  (  j  =  c  =  0  )  ,  c  >  d  &&  (  d  =  c  )  &&  (  l  =  i  )  ,  g  (  ++  i  )  :  0  ;  }  f  (  S  ,  T  )  {  s  =  S  ;  l  =  i  =  d  =  0  ;  g  (  t  =  T  )  ;  p  (  i  =  0  )  ;  }  p  (  )  {  q  =  s  +  l  +  i  ;  r  =  t  +  i  ;  i  ++  <  d  ?  p  (  *  r  =  *  q  )  :  (  *  r  =  0  )  ;  }  z  (  i  )  {  L  [  --  i  ]  =  0  ;  i  &&  z  (  i  )  ;  }");
    puts(str);
}

Output:

The code with slightly more readable formatting:

L[1<<7],
*q, *r, l, d, i, c, j, s, t, k=1<<7;

h()
{
    q = s+i+j++;
    *q%k && !L[*q%k]++ && h(++c);
}

g()
{
    q = s+i;
    *q%k ? z(k), h(j=c=0), c>d && (d=c) && (l=i), g(++i) : 0;
}

f(S, T)
{
    s = S;
    l = i = d = 0;
    g(t=T);
    p(i=0);
}

p()
{
    q = s+l+i;
    r = t+i;
    i++<d ? p(*r=*q) : (*r=0);
}

z(i)
{
    L[--i] = 0;
    i && z(i);
}

And this can be used to generate proper spacing to get to the formatting with score 2: Try it online!


C, score 3, 309 bytes

i
,
j
,
l
,
c
,
d
;
f
(
\
c\
\
h\
\
a\
\
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)
{
\
f\
\
o\
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(
i
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l
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0
;
s
[
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]
;
c
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d
&&
(
d
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)
&&
(
l
=
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)
,
++
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)
\
f\
\
o\
\
r
(
\
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\
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\
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r

L
[
\
1\
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2\
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8
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]
=
{
j
=
c
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0
}
;
s
[
i
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&&
!
L
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\
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(
1
,
s
+
l
,
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)
;
}

Try it online!

\$\endgroup\$
10
+500
\$\begingroup\$

Haskell, score 2, 492 ... 307 224 212 209 207 bytes

((yy:yyy))??ss|ss==yy  =  ""  |  yy==yy=yy:yyy??ss
ss??sss=ss
ss""=""

ss((ff:fff))  =  ff  :  ss  fff??ff
ff""=""

ff((xxx:xx))  =  ss((xxx:xx))##ff  xx
xx##xxx  |  ((((xx>>xx))<))  $  xxx>>xx=xxx|xx==xx=xx

Try it online!

Golfed literally hundreds of bytes thanks to W W and Ørjan Johansen!

Explanation

The function (??) takes a character c and a string s and returns the longest prefix of s that does not contain c. Ungolfed and not optimized for score:

c ?? (y:s)  
    | c==y = ""
    | True = y : c ?? s
c ?? s = s

The function ss uses (??) to find the longest prefix of unique chars of a given string:

ss (x:r) = x : (x ?? ss r)
ss "" = ""

(##) is a function which takes two strings and returns the longer one. The length comparison works by repeating the string x as often as x is long (x>>y) and as y is long (y>>x) and checking which of the resulting strings is lexicographically larger.

x ## y
  | (x>>x) < (y>>x) = y
  | True = x

Finally ff recurses over the input string, generates the longest prefix with ss, recursively determines the longest non-repeating substring of the tail of the string and returns the longer of the two with (##):

ff "" = ""
ff (x:r) = ss(x:r) ## ff r
\$\endgroup\$
  • 4
    \$\begingroup\$ 224, mainly by fusing away the intermediate list. \$\endgroup\$ – Ørjan Johansen Jan 30 '18 at 6:57
  • 2
    \$\begingroup\$ I combined this answer with the one I posted in chat earlier to get 216. \$\endgroup\$ – Wheat Wizard Jan 30 '18 at 18:31
  • 3
    \$\begingroup\$ 209 by reordering things. \$\endgroup\$ – Ørjan Johansen Jan 31 '18 at 1:46
  • 3
    \$\begingroup\$ With the bounty announcement I took another look and realized the @ trick actually costs 2 bytes over just making ? two chars: 207 \$\endgroup\$ – Ørjan Johansen Aug 6 '18 at 7:38
5
\$\begingroup\$

Lua, score 3, 274 bytes

g='g'..'s'..'u'..'b'  _G  [  'l'..'o'..'a'..'d'  ](  g[g  ](  "s  =...f  o  r d = # s - 1 , 0 , - 1 d  o f  or r = 1 , # s - d d  o t = s :s  ub  (r  ,r  +d  )i  f n  ot t:  fi  nd  '(  .)  .*  %1  't  he  n p  ri  nt  (t  )r  et  ur  n en  d e  n  d e  nd  ","  ",""))(...)

Note: Lua 5.2 or Lua 5.3 is required

Usage:

$ lua lnrs.lua "Good morning, Green orb!"
ing, Gre
$ lua lnrs.lua "fffffffffff"
f
$ lua lnrs.lua "oiiiiioiiii"
oi
$ lua lnrs.lua "1234567890"
1234567890
$ lua lnrs.lua "11122324455"
324

Main idea: interleave everything with spaces, insert " " (two spaces) to split long identifiers

Ungolfed code:

g = "gsub"
_G["load"](
   g[g](      -- g[g] == string.gsub - a function for substitution of substrings
      "The source of actual program, but two-space sequences were inserted in some places", 
      "  ",   -- we are replacing all two-space substrings
      ""      -- with an empty string
   )
)(...)

Actual program (after removing all pairs of spaces):

s = ...
for d = #s - 1, 0, -1 do
   for r = 1, #s - d do
      t = s:sub(r, r+d)
      if not t:find"(.).*%1" then
         print(t)
         return
      end
   end
end

BTW, the JS snippet for calculating the score fails on my code.

\$\endgroup\$
4
\$\begingroup\$

Retina 0.8.2, 37 bytes, score 9

.
$&$'¶
(.)(?<=\1.+).*

O#$^`
$.&
1G`

Try it online! The direct translation of this answer to Retina 1 saves a byte by using N instead of O#. However, if you naïvely golf the Retina 1 answer down to 28 bytes, its score actually rises to 10! Explanation:

.
$&$'¶

Generate all suffixes of the input.

(.)(?<=\1.+).*

For each suffix, take the prefix up to the first duplicated character.

O#$^`
$.&

Sort the remaining strings in reverse order of length (i.e. longest first).

1G`

Take the longest.

\$\endgroup\$
4
\$\begingroup\$

Jelly, score 2, 14 bytes

Ẇµµff  Q  €  Ṫ

Thanks to @JonathanAllan for score -1, +7 bytes and for noticing a bug.

Try it online!

How it works

Ẇµµff  Q  €  Ṫ  Main link. Argument: s (string)

Ẇ               Window; yield all substrings of s, sorted by length.
 µ              Begin a new chain. Argument: A (array of substrings)
  µ             Begin a new chain. Argument: A (array of substrings)
   f            Filter A by presence in itself. Does nothing.
       Q  €     Unique each; deduplicate all strings in A.
    f           Filter A by presence in the array of deduplicated substrings,
                keeping only substrings composed of unique characters.
             Ṫ  Tail; take the last (longest) kept substring.
\$\endgroup\$
4
\$\begingroup\$

Clean, score 7 5, 276 bytes

@[ss:s]=rr(pp[][ss:s])((@s))
@s=s
ee x[rr:xx]|e x rr=True=ee x xx
ee x xx=f
f=e'e'' '
e::!  Char  !  Char  ->Bool
e  _ _=  code  {

eqC
}
pp p[r:rr]|ee r p=p=pp(a r p)rr
pp a _=a
a  x[ll:l]=[ll:a x  l]
a l ll=[l]
l[]rr=e'l''l'
l ff[]=f

l[r:rr][ss:ll]=l rr ll
rr x y|l x y=y=x

Try it online! Thanks to @Οurous for showing me that it is possible to call ABC machine code directly from within Clean. This allows to get rid of the previous bottle-neck import which set the minimal score to 7, but needs the keyword code which sets the minimal score to 5 for this approach.

An ungolfed and not score-optimized version of the above code can be found here: Try it online!


Previous version with score 7, 158 154 130 bytes

import  StdEnv  
@[xx:rr]=c(%[][xx:rr])(@rr)
@e=e
c s b|  length  s<  length  b=b=s
%s[xx:r]|  isMember xx s=s= %(s++[xx])r
%r _=r

Try it online!

With the import the score cannot go below 7. Without the import one would need to implement equality on strings or chars without any library functions which is probably not possible, as can be seen in the new version above.

\$\endgroup\$
  • 1
    \$\begingroup\$ You can indeed implement equality using inline ABC, which should reduce the score. I'll come back with a suggested modification later today if you're interested. \$\endgroup\$ – Οurous Feb 2 '18 at 20:15
  • \$\begingroup\$ Eg: char equality: tio.run/##S85JTcz7/… \$\endgroup\$ – Οurous Feb 2 '18 at 23:01
  • \$\begingroup\$ @Ourous A code block with raw ABC instructions, which can be used for primitive functions like integer addition, for linking with C, bypassing the type system... welcome down the rabbit hole! (from cloogle) certainly sounds inviting. I'll look into it tomorrow, thanks for the suggestion! \$\endgroup\$ – Laikoni Feb 3 '18 at 0:29
  • 1
    \$\begingroup\$ @Οurous Thanks again, with your char equality test the score is now at 5. \$\endgroup\$ – Laikoni Feb 7 '18 at 17:14
  • \$\begingroup\$ Incidentally, you don't need either -IL flag, since nothing is being imported. \$\endgroup\$ – Οurous Feb 7 '18 at 19:26
3
\$\begingroup\$

Python 3, score 4, 155 bytes

exec(('l=la''mbd''a f'',e=en''ume''rat''e:m''ax''([f[ j  :k]  for  j,i in e ( f)f''or  k,i in e ( f )if  len  ( { *''f[j'':k]''})==k-''j],''key''=le''n)'))

This defines a function l.

Thanks to @xnor for pointing out that strings of length 3 don't raise the score, saving 32 bytes.

Try it online!

\$\endgroup\$
  • \$\begingroup\$ The string can be in chunks of 3, right? \$\endgroup\$ – xnor Jan 30 '18 at 6:21
  • \$\begingroup\$ @xnor Changing the name of the function, indeed. Thanks! \$\endgroup\$ – Dennis Jan 30 '18 at 13:19
3
\$\begingroup\$

Brachylog, score 2, 19 bytes

s  ᶠ  l  ᵒ  ≠  ˢ  t

Try it online!

Just a boring old "space everything out" answer. At least I learnt that metapredicates can be spaced away from the predicates and still work (and the (parametric) subscripts and superscripts can't).

s ᶠ - find all substrings of the given string

l ᵒ - order them by their length (ascending by default)

≠ ˢ - select those that have all distinct elements

t - get the tail (last element) of that - the one with the biggest length

\$\endgroup\$
2
\$\begingroup\$

Pyth, 11 bytes, score 4

-4 score thanks to Dennis

e lD {I# .:

elD{I#.:Q      Full program, inputs "string" from stdin and outputs to stdout
e              The last element of the list generated by taking
      .:Q      All substrings of the input
     #         Filtered for
   {I          Being invariant over deduplicate i.e. being "non-repeating"
 lD            and sorted by length

Try it online!

\$\endgroup\$
2
\$\begingroup\$

Husk, score 2, 10 bytes

►IIËII≠IIQ

Try it online!

Explanation

The program is equivalent to this:

►Ë≠Q  Implicit input.
   Q  List of substrings.
►     Find one that maximizes:
 Ë    all ordered pairs
  ≠   are inequal.

The built-in Ë evaluates on all ordered pairs of its argument x, and returns length(x)+1 if every result is truthy, otherwise 0. When we maximize this, we find the longest string that has no repeated characters.

In the submission, I just insert the identity function I between each function, twice. Since is the same as Ë, I≠ is the same as and so on, this does not change the semantics. The only danger is that a higher order function could decide to use one of the Is as its argument, but luckily that leads to a type error in our program, so it doesn't happen.

\$\endgroup\$
2
\$\begingroup\$

Clojure, score 4

#(  let  [N  (fn  [[_ & r]] r) R  (fn  R [f v c]  (if  c (R f (f v (  nth  c 0))  ( N  c)) v)) C  (fn  C  (  [i]  (C (  seq  i) 0)) ( [i  n]  (if i (C ( N  i )  (  inc n)) n)))  J  (fn  [c  i]  (assoc c (C  c) i)) I  (fn  F [f i n R]  (if ( =  (C  R) n) R (F f (f  i) n ( J  R (f  i)))))] ( apply  str  (R ( fn  [a  b] ( if  (< (C  a)  (C  b)) b a )) "" (  for  [k  (I N % (C  % ) [])]  (R  ( fn [ t  c ] ( if ( or ( = t (  str t) ) ((  set t)c))(apply  str t) ( J  t c)))[]k)))))

Oh man this was painful! N implements next, R is reduce, C is count, J is conj (works only for vectors) and I is iterate. apply str is there twice because otherwise "aaaa" input wouldn't return a string but a vector [\a]. Luckily I got to use apply and assoc, I didn't know you could assoc one index beyond a vector's last element :o

\$\endgroup\$
1
\$\begingroup\$

Jelly, score 5, 10 bytes

ẆµQQ⁼µÐfµṪ

Try it online!

\$\endgroup\$
  • 2
    \$\begingroup\$ You can lower the score to 3 by adding spaces in your code. For example: ẆµQQ ⁼ µ Ðf µ Ṫ (probably added too many spaces now, but it's just an example. I'll leave optimizing byte-count versus spaces up to you). \$\endgroup\$ – Kevin Cruijssen Jan 30 '18 at 8:43
1
\$\begingroup\$

Python 3, score 4, 317 bytes

exec(('%s'  *58  %(  's=','in','pu','t(',');','pr','in','t(','so','rt','ed','((s','[i',':j',']f','or',' j',' i','n ','ra','ng','e(','1,','le','n(','s)','+1',')f','or',' i',' i','n ','ra','ng','e(','j)','if',' l','en','(s','et','(s','[i',':j',']))','==l','en','(s','[i',':j',']))',',k','ey','=l','en',')[','-1','])')))

Try it online!

Unexeced code:

s=input();print(sorted((s[i:j]for j in range(1,len(s)+1)for i in range(j)if len(set(s[i:j]))==len(s[i:j])),key=len)[-1])

lambda a contains mbda which has score 5, and a function needs return which apparently can't be execed (so takes a score of at least 5 for eturn), so a full program was necessary. It's probably possible to golf down the unexeced code size quite a bit, but I can't see a quick clear improvement.

\$\endgroup\$
1
\$\begingroup\$

Alice, 40 bytes

/ii..nn$$@@BBww..DD~~FF..!!nn$$KK??oo@@

(Trailing newline)

Try it online!

The instruction pointer moves diagonally in ordinal mode, so only every other character is executed.

i.n$@Bw.D~F.!n$K?o@

i     take input
.n$@  terminate if empty
B     push all nonempty substrings, with the longest on the top of the stack
w     push return address (start main loop)
.     make copy of current substring
D     deduplicate characters
~     swap: this places the original above the deduplicated copy
F     Push the original string if it is a substring of the deduplicated copy
      (which can only happen if they're equal); otherwise push empty string
.!    place a copy on the tape
n$K   if the empty string was pushed, return to start of loop
o     output
@     terminate
\$\endgroup\$
1
\$\begingroup\$

Perl 6, score: 15 10 8, length: 46 55 62 bytes

{~m:ov/(.+)<!{$0.comb.repeated}>/.max(&chars)}

Test it

{~m:ov/(..*)<!{(($0)).comb.repeated}>{{}}/.max(&chars)}

Test it

{m:ov:i/(..*)<!{(($0)).comb.repeated}>{{}}/.max((&chars)).Str}

Test it

Expanded:

{    # bare block lambda with implicit parameter 「$_」

    m                          # match (implicitly against 「$_」)
    :overlap                   # in every single way possible
    :ignorecase                # add a 「:」 to break up substring
    /

      (..*)                    # match at least one character

      <!{
        (($0)).comb.repeated  # backtrack if there were repeats
      }>

      {{}}                    # anon hash in code block (no-op)
    /

    .max((&chars))            # get the longest

    .Str                      # coerce to a Str (from a Match object)
}
\$\endgroup\$
  • \$\begingroup\$ Score of 5 for 88 bytes. There might be a few places to golf bytes though \$\endgroup\$ – Jo King Aug 6 '18 at 7:59
1
\$\begingroup\$

Java 8, score 9 (384 B) 7 (401 B)

S -> { int s = 0 , e = 0 , l = 0 , x = 0 , y = 0 , b [ ] = new int [ 256 ] ; for ( ; x <S.  length  & y <S.  length  & l <S.  length  - x ; x ++ ) { b [S[x]] = 1 ; for ( y ++ ; y <S.  length  && b [S[y]] < 1 ; b [S[y ++]] = 1 ) ; if ( l < y - x ) { s = x ; e = y ; l = y - x ; } for ( ; y <S.  length  && x < y & S[x] != S[y  ];)b [S[x ++]] = 0 ; }  String g=""; for( ; s<e ; g+= S[s++]);  return  g;}
  • Initial version. Will go down from here. Score is 9 due to "ubstring ", so substring will be the first part to replace.
  • Score is now 7 due to " length", which I probably won't be able to reduce further.. I doubt it is possible to drop the four uses of length. If it is possible, " eturn" (6) might lower the score by 1 as final improvement, but I guess this is it (except maybe a small reduce in byte-count..)

Try it online.

\$\endgroup\$
0
\$\begingroup\$

Haskell, score 7

-4 thanks to Laikoni.

import  Data.List  
f s=snd $ maximum [ (0<$i ,i)|i<-  tails  =<<inits s, nub i==i]

Try it online!

\$\endgroup\$
  • 3
    \$\begingroup\$ f s=snd$maximum[(0<$i,i)|i<-tails=<<inits s,nub i==i] saves a byte and two on the score. \$\endgroup\$ – Laikoni Jan 29 '18 at 23:00
  • 3
    \$\begingroup\$ Adding some spaces brings the score down to 7: Try it online! \$\endgroup\$ – Laikoni Jan 29 '18 at 23:29
0
\$\begingroup\$

Mathematica, score 11 9

Length@Last@Select[Subsequences[Characters@#],#==DeleteDuplicates  @#&]&

Shaving a couple of bytes off of the longest nonrepeating string by obscuring the function's name:

Length@Last@Select[Subsequences[Characters  @#],#==(  ToExpression@ 
StringJoin@@FromCharacterCode@{{68},{101},{108},{101},{116},{101},{68},{117},
{112},{108},{105},{99},{97},{116},{101},{115}}))@#&]&
\$\endgroup\$
0
\$\begingroup\$

Kotlin, score: 11 10 9 bytes, length: 227 246 245 bytes

indices
  .flatMap { p -> indices . map { p to p + it } }
  .  filter { (r,i) -> i < length  }
  .map { ( s , a )->substring  (  s,  a  ) }
  .  filter { it  .  groupBy  { it } .  none { ( k , v )->v . size>1 } }
  .maxBy { it.length }

The longest is ubstring, which is 9 chars

It is called like this:

val c = "Good morning, Green orb!"

fun String.c(): String? = indices
    .flatMap { p -> indices . map { p to p + it } }
    .  filter { (r,i) -> i < length  }
    .map { ( s , a )->substring  (  s,  a  ) }
    .  filter { it  .  groupBy  { it } .  none { ( k , v )->v . size>1 } }
    .maxBy { it.length }

fun main(args: Array<String>) {
    val text = """indices
    .flatMap { p -> indices . map { p to p + it } }
    .  filter { (r,i) -> i < length  }
    .map { ( s , a )->substring  (  s,  a  ) }
    .  filter { it  .  groupBy  { it } .  none { ( k , v )->v . size>1 } }
    .maxBy { it.length }"""
    val message = text.c()!!
    println(message)
    println(text.length)
    println(message.length)
    println(c.c())
}
\$\endgroup\$
  • \$\begingroup\$ Aren't you able to reduce it to 10 by adding an additional space between roupingBy and {? \$\endgroup\$ – Kevin Cruijssen Jan 30 '18 at 10:32
  • 1
    \$\begingroup\$ Nice find, I changed the other 11s and and got down to 10 \$\endgroup\$ – jrtapsell Jan 30 '18 at 10:38
  • \$\begingroup\$ It is 10 chars, but the longest substring isn't roupingBy (which is 9 chars) but eachCount (with trailing space). \$\endgroup\$ – Erik the Outgolfer Jan 30 '18 at 11:46
  • \$\begingroup\$ roupingBy has a trailing space (visible in the markdown, but the renderer seems to strip it) \$\endgroup\$ – jrtapsell Jan 30 '18 at 11:48
  • \$\begingroup\$ Managed to reduce it to 9, fixing the trimming issue \$\endgroup\$ – jrtapsell Jan 30 '18 at 11:56
0
\$\begingroup\$

Pyth, score 3 ( 18  14 bytes)

e  fq  T{T  .:

Try it online!

The longest non-repeating substrings is  .:.

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  • \$\begingroup\$ A cute-ish alternative: e f {I T .:. \$\endgroup\$ – Mr. Xcoder Jan 30 '18 at 14:59
0
\$\begingroup\$

05AB1E, 22 bytes | Score: 2

Œ  ʒ  D  Ù  Q  }  é  ¤

-1 score + 7 bytes thanks to HeebyJeeby

Try it online!


05AB1E, 15 bytes | Score: 3

Œ ʒ D Ù Q } é ¤

Try it online!


05AB1E, 8 bytes | Score: 8

ŒʒDÙQ}é¤

Try it online!


05AB1E can actually do something rather cheap... adding whitespace into 05AB1E does nothing.

If there is a rule against this, I can also use ´ and like 7 other chars.

\$\endgroup\$
  • 1
    \$\begingroup\$ @HeebyJeebyMan because I'm a moron, got a problem with that? \$\endgroup\$ – Magic Octopus Urn Feb 2 '18 at 20:47
  • \$\begingroup\$ @HeebyJeebyMan kidding haha, thanks for the idea. \$\endgroup\$ – Magic Octopus Urn Feb 2 '18 at 20:47

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