Unique is Cheap

Question

Write a function or program that determines the cost of a given string, where

• the cost of each character equals the number of how many times the character has occurred up to this point in the string, and
• the cost of the string is the sum of its characters' costs.

Example

For an input of abaacab, the cost is computed as follows:

a b a a c a b
1   2 3   4    occurrence of a
  1         2  occurrence of b
        1      occurrence of c
1+1+2+3+1+4+2 = 14

Thus the cost for the string abaacab is 14.

Rules

• The score of your submission is the cost of your code as defined above, that is your submission run on its own source code, with a lower score being better.
• Your submission should work on strings containing printable ASCII-characters, plus all characters used in your submission.
• Characters are case-sensitive, that is a and A are different characters.

Testcases

input -> output
"abaacab" -> 14
"Programming Puzzles & Code Golf" -> 47
"" -> 0
"       " -> 28
"abcdefg" -> 7
"aA" -> 2

Leaderboard

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Answer 1

MATL, score 4

&=Rz

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Explanation

Consider input 'ABBA' as an example.

&=   % Implicit input. Matrix of all equality comparisons
     % STACK: [1 0 0 1;
               0 1 1 0;
               0 1 1 0;
               1 0 0 1]
R    % Upper triangular part
     % STACK: [1 0 0 1;
               0 1 1 0;
               0 0 1 0;
               0 0 0 1]
z    % Number of nonzeros. Implicitly display
     % STACK: 6

Answer 2

Python, score 49

lambda S:sum(1+S.count(C)for[C]in	S)/2

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There's a tab after in.

Score breakdown:

• +27 for 27 unique chars
• +16 for 8 double chars: ()Camnou
• +6 for 1 tripled char: S

Answer 3

T-SQL, score 775 579! 580

declaRe @ char(876),@x int,@v int=0Select @=q+CHAR(9)from z X:seleCT @x=len(@),@=REPLACE(@,LEFT(@,1),''),@v+=(@x-LEN(@))*(@x-LEN(@)+1)/2IF LEN(@)>0GOTO X prINT @v-1

EDIT: Dropped a couple of variables, compacted a bit. Down to 16 @ symbols instead of 22, that by itself reduces my score by a whopping 117 points!

Nice contest, I like the requirement to optimize for something besides total character count.

Input is via varchar field q in pre-existing table z, per our IO rules. The database containing this input table must be set to a case-sensitive collation.

Formatted:

declaRe @ char(876), @x int, @v int=0
Select @=q+CHAR(9)from z
X:
    seleCT @x=len(@)
          ,@=REPLACE(@,LEFT(@,1),'')
          ,@v+=(@x-LEN(@))*(@x-LEN(@)+1)/2
IF LEN(@)>0 GOTO X
prINT @v-1

SQL keywords aren't case sensitive, so I used mixed case to minimize the count of duplicate letters (aaAA generates a better/lower score than aaaa).

The main loop compares the length before and after stripping all instances of the first character out. That difference n*(n+1)/2 is added to a running total.

The SQL LEN() function annoyingly ignores trailing spaces, so I had to append a control character and subtract 1 at the end.

EDIT: Fixed a miscalculation of my own score by 2 points (issue with quoting quotes), reduced by 1 by changing casing of one R. Also working on a completely different strategy, I'll be posting that as its own answer.

Answer 4

C (gcc), score:  113  103 100   96  91

Thanks to @ugoren, @CalculatorFeline, @gastropner, @l4m2, and @JS1 for their tips.

g(char*s){int y[238]={};while(*s)*y-=--y[*s++];*y/=1;}

Initializes an array of zeros, then uses the ASCII values of the characters in the string as indices to that array to keep track of the number of instances of each character in the string.

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Answer 5

JavaScript (ES6), score 81 78

Saved 3 points thanks to @Arnauld

s=>s.replace(d=/./g,z=>q+=d[z]=-~d[z],q=0)&&q

My original score-81 recursive solution:

f=([c,...s],d={})=>c?(d[c]=-~d[c])+f(s,d):0

Answer 6

Haskell, score 42

f l=sum[1|c<-l,d<-c:l,d==c]/2

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Anonymizing \l-> gives the same score.

Answer 7

Jelly, score 6

;\ċ"⁸S

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Answer 8

Retina, score 34

s(O`.
M&!`^|(?<=(.))\1*
.

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Explanation

s(O`.

We start by sorting all the characters in the input so that identical characters are grouped together into a single run. The s( activates singleline mode for all stages (i.e. makes . match linefeeds).

M&!s`^|(?<=(.))\1*

The goal is to turn a run of n characters into T_n characters (the nth triangular number) because that's the score of the occurrences of this character. To do so, we find overlapping matches. In particular, for each i in [1,n], we're going to include i-1 characters in the match. We get all those matches due to the overlapping flag &. That gives us n*(n-1)/2 = T_n-1 = T_n - n characters just from the matches. But the match stage will join these with linefeeds, which are n linefeeds for n matches. There's only one problem. There won't be a linefeed after the last match, so the overall number of characters in the output is one less than we need. We fix this by also matching the beginning of the input, which gives us a single leading linefeed if there is at least one other match.

.

Finally, we just count how many characters there are in the string.

Answer 9

Haskell, score 52 51

f(a:b)=1+sum[1|c<-b,c==a]+f b;f _=0

There's a tab between f and _.

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The value of the empty string is 0. The value of the string s, where a is the first char and b the rest of the string is 1 plus the occurrences of a in b plus a recursive call with b.

Answer 10

J, score 16

1#.,@(*+/\"1)&=

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Explanation

1#.,@(*+/\"1)&=
              =  Self-classify: bit matrix of equality between input
                 and its unique elements.
     (      )&   Apply verb in parentheses to it:
       +/\         running sums
          "1       of each row
      *            multiplied with original matrix.
                 This causes the i'th 1 on each row to be replaced by i.
   ,@            Flatten the resulting matrix
1#.              and interpret as a base-1 number, computing its sum.

Using 1#. instead of +/@ for the sum saved a few points, and & could be used instead of @ in a monadic context to save one more. The repeated 1 costs me one extra point, but I haven't been able to get rid of it.

Answer 11

R, score: 67 83 95 128

-61 thanks to top tips from Giuseppe

function(x,y=table(utf8ToInt(x)))y%*%{y+1}/2

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The string is split using utf8ToInt and each ascii value is counted table. The the result is calculated using doing a matrix multiplication %*% over that at itself + 1 and finally halfed.

Answer 12

05AB1E, score 6

{γ€gLO

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Answer 13

Jelly, score 5

ĠJ€ẎS

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Thanks to Leaky Nun for -2 (previously on his answer)

Answer 14

Pyth, score 6

1 byte thanks to isaacg.

+F/V._

Test suite.

How it works

+F/V._
+F/V._QQ  implicit input
  /V      vectorize count: for each element in the first argument,
                           count the number of occurrences of the
                           second argument:
    ._Q       all prefixes of input
       Q      input
+F        fold (reduce) on +, base case 0.

Answer 15

J, score: 14 12 11

$+/@,2!1#.=

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Answer 16

Jelly, score of 7

ċЀQRFS

Explanation:

   Q    get unique letters
ċЀ     count the occurences of each letter in the original string
    R   [1..n] for n in list of frequencies
     F  flatten list
      S sum
        (implicit output)

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Answer 17

C, 60 bytes, score 108 95

g(char*s){int y[256]={},z=0;while(*s)z-=--y[*s++];return z;}

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Usually pre- and post-increment operators are great for code golf, but they really hurt on this challenge!

EDIT: By subtracting negative counts instead of adding positive ones, I saved a whole bunch of score. Replacing for() with while() eliminated a semicolon as well.

Answer 18

Perl 6, score  61 56 53 46  44

(1 X..*.comb.Bag.values).flat.sum

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{sum flat 1 X.. .comb.Bag.values}

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{sum flat 1 X.. values(.comb.Bag)}

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{[+] flat	1	X.. values(.comb.Bag)}

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{[+] flat	1	X.. values(bag
.comb)}

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Answer 19

C# (.NET Core), score ∞ (I mean, 209)

b=>b.Distinct().Select(z=>{var w=b.Count(p=>p==z);return w*(w+1)/2;}).Sum()

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The score includes the following:

using System.Linq;

Answer 20

PowerShell, score 64

$z=@{}
$ARGS|% getE*|%{$u+=($Z.$_+=1)};$U

(Score is based on a single linefeed newline, which isn't Windows standard but does work in PS).

PS C:\> D:\unique-is-cheap.ps1 (gc D:\unique-is-cheap.ps1 -raw)
64

• Hashtable counter @{}
• Iterate through the letters; $args is an array of parameters - in this case the input string makes it a single item array; |% does a foreach loop over the items, and uses the getE* shortcut to match the GetEnumerator() string method and call it to turn the string into a character stream.
• |% loop over the characters and increment their hashtable entry, add it to a running total. The ($x+=1) form with parens both modifies the variable and outputs the new value for use.
• Output the running total.

(When I first wrote it, it was $c=@{};$t=0;[char[]]"$args"|%{$c[$_]++;$t+=$c[$_]};$t with a score of 128, and felt like it wouldn't go much lower. Halving it to 64 is quite pleasing).

Answer 21

Julia 0.6, 45 bytes, Score: 77

Inspired by the MATL solution:

f(w)=sum(UpperTriangular([z==j for z=w,j=w]))

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A less pretty solution, using counts:

Julia 0.6, score: 82

F(w)=sum(l->[l+1]l/2,count(x->x==i,w)for i=Set(w))

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Thanks to Guiseppe for pointing out the scoring and for tips. These comments helped me loads.

Answer 22

Java 10, score: 149 138 137 134 133 130 103 102 101 100

(Bytes: 72 73 74 75 64 62 61) Bytes go up, but score goes down. :D

x->{int j=0,q[]=new int[256];for(var    C:x)j+=++q[C];return
j;}

-28 score (and -11 bytes) thanks to @Nevay.
-1 score (and -2 bytes) thanks to @OlivierGrégoire.
-1 score (and -1 byte) by converting Java 8 to Java 10.

Explanation:

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x->{                     // Method with character-array parameter and integer return-type
  int j=0,               //  Result-integer, starting at 0
      q[]=new int[256];  //  Integer-array with 256 times 0
  for(var   C:x)         //  Loop over the characters of the input array
    j+=++q[C];           //   Raise the value in the array by 1,
                         //   and then add it to the result-integer
  return                 //  Return 
  j;}                    //         the result

Answer 23

F#, score 120 118

let j z=Seq.countBy id z|>Seq.sumBy(fun x->List.sum[0..snd x])

-2 thanks to Kevin Cruijssen!

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Takes a string as an input. Seq.countBy pairs each distinct character with its count (id is the identity function) so you end up with a collection like 'a' = 4, 'b' = 2 etc.

The Seq.sumBy takes the count for every letter and sums all the numbers from 0 to the count for that letter. So if 'a' = 4 the collection would be 0, 1, 2, 3, 4 which summed together is 10. Then Seq.sumBy sums all those totals.

Answer 24

BQN, Score: 4

≠+´⊒

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The occurrence count builtin ⊒ calculates the cost of each character decremented by 1. To adjust for that we can either add 1 to each cost or add the length ≠ to the sum. The latter turns out to be a byte shorter.

Answer 25

APL (Dyalog), score 15

+/1 1⍉+\∘.=⍨⍞

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⍞ get text input

∘.=⍨ equality table with self

+\ cumulative sum across

1 1⍉ diagonal (lit. collapse both dimensions into dimension one)

+/ sum

Answer 26

Retina, score 68 45 43

s`(.)(?<=((\1)|.)+)
$#3$*
1

Try it online! Link shows score. Edit: Thanks to @MartinEnder who saved 20 bytes by using overlapping matches instead of lookaheads and a further three bytes by grouping the stages so that the s flag only needs to be applied once. Saved a further two bytes by calculating the triangular number differently, avoiding the need for a sort.

Answer 27

PHP, 45 bytes, Score 78

WHILE($x=ORD($argn[$v++]))$y+=$$x-=-1;echo$y;

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PHP, 46 bytes, Score 79 Bytes

WHILE(~$xy=$argn[$vz++])$su-=-$$xy+=1;echo$su;

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PHP, 56 bytes, Score 92

FOREAch(COUNT_CHARS($argn)as$z)WHILE($z)$y+=$z--;echo$y;

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Answer 28

Perl 5 score 91 83

Uses the -p flag which adds 2 because of the p in split.

$x=$_;$b+=++$a{$_}for(split//,$x);$_=$b

Answer 29

Octave, 39 bytes, Score 69

@(a)sum((b=hist(a,unique(1*a))).^2+b)/2

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While there is another Octave answer, this one is entirely my own and a different approach, plus it scores less :).

The approach boils down to first finding the count (b) of each unique character, which is achieved using the histogram function. Then for each element we calculate the sum of 1 to b which is done using the formula (b*(b+1))/2. Then the individual sums are all summed into the final score.

In testing it seems brackets are really costly in the scoring because many are needed. I've optimised down from an initial score of about 88 by rearranging the questions to minimise the number of open/close brackets - hence we now do the /2 on the final total rather than individually, and also I've modified the formula to (b^2+b)/2 as that requires fewer brackets.

Answer 30

Octave , score 77 31 bytes

*same as @LuisMendo 's MATL answer.

@(a)nnz(triu(a==a'))

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Previous answer:

@(d,g=accumarray(+d',1)-(0:254))sum(g(g>0));

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