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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<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Score</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

Answer 1

jq, score 60

(40 characters code + 3 characters command line option)

[./""|group_by(.)[]|range(length+1)]|add

Sample run:

bash-4.4$ jq -R '[./""|group_by(.)[]|range(length+1)]|add' <<< 'Programming Puzzles & Code Golf'
47

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jq, score 63

[./""|group_by(.)[]|range(length+1)]|add+0

In case the result for empty string must be numeric 0. There are a few other solutions that use alternative representation of nothing.

Answer 2

Ruby, score: 38

->z{z.sum{|k|z.count(k)+1}/2}

Try it online!

Answer 3

Stax, 8 6

ç╕┌α▒'

Run and debug online!

Saved 2 points by using runs per comment by @recursive.

Explanation

Uses the unpacked version to explain.

o:GF:T+
o          Sort array
 :G        Run lengths
   F       For each run length `n`
    :T     The nth triangular number
      +    Add to accumulator
           Implicit output

Answer 4

R, 349

Here is my R solution bit of tidy

s=function(x) {c=0
for (i in unique(strsplit(x,"")[[1]])) {
c=c(c,sum(1:sum(charToRaw(x)==charToRaw(i))))}
sum(c)
}

Answer 5

SHELL ( 67 bytes)

Last Solution : ( 68 bytes)

 f(){ bc<<<`echo -n "$1"|od -bAn|sed "s/\([^ ]\+\)/s+=++v\1;/g"`s;}

Old solutions :

Old Solution 1 : by creating function f (83 bytes ) :

f(){ bc<<<`echo "$1"|sed "s/./ss+=++&;/g"|sed "s/[A-Z]/\L&_/g"|sed "s/ /bb/g"`ss;}

Old Solution 2 : by creating alias f ( 81 bytes )

alias f='bc<<<`cat -|sed "s/./ss+=++&;/g"|sed "s/[A-Z]/\L&_/g"|sed "s/ /bb/g"`ss'

Explanations :
I used ( sed "s/./ss+=++&;/g" ) to replace any char a by ( ss+=++a; ); where ss is the sum to calculate, then concatenate with ss, and calculate all expression by the command bc :

echo "a" | sed "s/./ss+=++&;/g"
ss+=++a;

for a string in input :

echo "aab" | sed "s/./ss+=++&;/g"
ss+=++a;ss+=++a;ss+=++b;

step by step :

ss+=++a;    // a=1  -->  ss=1
ss+=++a;    // a=2  -->  ss=3
ss+=++b;    // b=1  -->  ss=4

for lowercase only input, the simpler function need only ( 47 bytes ) to define :

f(){ bc<<<`echo "$1"|sed "s/./ss+=++&;/g"`ss;}

to take uppercase in consideration, Without spaces ( 68 bytes) :

f(){ bc<<<`echo "$1"|sed "s/./ss+=++&;/g"|sed "s/[A-Z]/\L&_/g"`ss;}

Complete Solution ( 83 bytes):

f(){ bc<<<`echo "$1"|sed "s/./ss+=++&;/g"|sed "s/[A-Z]/\L&_/g"|sed "s/ /bb/g"`ss;}

fuction tests :

f "abaacab"
14

f "abcdefg"
7

f "       "
28

f "Aa"
2

using an alias instead the function, we can reduce code size to ( 68 bytes )

alias f='bc<<<`cat -|sed "s/./ss+=++&;/g"|sed "s/[A-Z]/\L&_/g"|sed "s/ /bb/g"`ss'

alias tests :

echo "       " | f
28

echo "Ab" | f
2

Answer 6

Kotlin, score 93

{it.foldIndexed(0){z,q,v->q+it.take(z+1).count{it==v}}}

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Answer 7

Reticular, score 56

iSd:A=>ddV@c~dc$:A=+:A`L0EQ6jj+1*37_$O;

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Answer 8

JavaScript, 150 141 133

f=(g,i=[])=>g?(i[j=g.charCodeAt()]=(i[j]|0)+1)+f(g.substr(1),i):0

Ungolfed:

f=(g,i=[])=>                                                       //function declaration, a is list of appearance of characters so far.
            g?                                                     //check if string is empty
              (i[j=g.charCodeAt()]=                                //update appearance list
                                    (i[j]|0)+1)+                   //handle undefined, then increment
                                                f(g.substr(1),i)   //next character
                                                                :0 //base case for the empty string

150->141 - -6 for renaming variables to unique characters (thanks to Ørjan Johansen), and -3 for removing implicit 0 in charCodeAt.

141->133 - -8 for using actual unique variable names(thanks again to Ørjan Johansen).

Answer 9

Perl 6, Score: 34

{sum [\+](^∞)[.comb.Bag{*}]}

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Takes advantage of the unicode operator ∞ to save on characters.

Explanation

{                          }  # Anonymous code block
              .comb           # Split the input string into characters
                   .Bag{*}    # Get the count of each character
             [            ]   # And inded each into
         (^∞)                 # The infinite list of integers
     [\+]                     # Triangularly summed (0,0+1,0+1+2...)
 sum                          # And return the sum total

Answer 10

05AB1E (legacy), score: 4 (4 bytes)

Ù¢LO

Input as a list of characters.

Try it online or verify all test cases.

Explanation:

Ù     # Get all unique characters of the (implicit) input-list
 ¢    # For each, count the amount of times it occurs in the (implicit) input-list
  L   # Create a list of each of these counts
      # (which implicitly flattens in the legacy version of 05AB1E)
   O  # Take the sum of all those integers
      # (which is output implicitly as result)

Answer 11

Zsh, 36 bytes, score 52

for	p (${(s..)1})$[j+=++a[#p]]
<<<$j

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One tab, one space, one newline. This will fill up the debug log with "command not found". If that must be avoided, 38 bytes, score 55.

Answer 12

Haskell, 150 133 points

(79 75 bytes)

import Data.List
c=foldr(\x->	\y->y+(x*(x+1)`div`2))0.map length.group.sort

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• -17 points by removing some parentheses (thanks to @Laikoni)

Answer 13

JavaScript, 87 score

A=([B,...C],Z=0)=>B?C.map(D=>D!=B||Z++)^A(C)-~Z:0

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C.map(D=>D!=B||Z++) is either empty, 0, true or something with a ,, making it never show value to ^

Answer 14

Vyxal d, 3 bytes

Ċɾ⌊

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Ċ   # Counts of each element within itself
 ɾ  # Convert each value to 1...n
  ⌊ # Parse each to integer, zeroes element values
    # (d flag) take the deep sum

Answer 15

Pyth, 8 bytes, Score: 9

ssSM/LQ{

Test suite

Explanation

ssSM/LQ{  | Full code
ssSM/LQ{Q | with implicit variables filled
----------+-------------------------------
     L {Q | Map over deduplicated input:
    /LQ   |  count in input
  SM      | Map: to [1...N]
 s        | Flatten
s         | Sum

ss can be replaced with s.n for an identical score with no repeat bytes, at the cost of being one byte longer.

Answer 16

Husk, Score: 6

ṁoΣ#¹u

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Explanation

ṁoΣ#¹u   translates to ṁoΣ#⁰u⁰
ṁ    u⁰  map following function on unique input and then sum
 o       composed function
   #⁰    count occurrences in input
  Σ      triangular number

Answer 17

Gaia, score: 11 10

$uC¦ₔ┅¦_Σ

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For some reason, _ is necessary else this spits out garbage answers. But the language is defined by its implementation, so what are ya gonna do. Without the _, one of the sublists generated by ┅¦is [1 2] which is converted to 12 by Σ.

$		| split into characters
 		| 
  u		| get unique
   C¦ₔ		| implicitly copy original input; get counts of uniques into original
      ┅¦	| sequence: 1..count
        _Σ	| flatten and sum

Answer 18

C++ (gcc), Score: 209 196 193

98 96 bytes

-3 thanx to l4m2

#include<map>
long	F(std::string	S,int I=0){for(auto&C:S){++I;for(char&D:S)I+=C==D;}return I/2;}

C++-specific tricks:

• #include<map> instead of #include<string>
• Returning long instead of int to use more different characters
• Using auto instead of char in one of the loops has the same effect
• References (&) inside range-for save whitespace

Common tricks:

• Evenly use both tabs and blanks
• Upper case variables to be distinct from keyword characters

Try it online!

Answer 19

Excel, Score: 206

G4 =MID(B3,SEQUENCE(LEN(B3)),1)
C3 =IF(B3<>"",SUM(EXACT(G4#,TRANSPOSE(G4#))/2)+ROWS(G4#)/2,)

Link to Spreadsheet

Answer 20

Excel VBA Immediate Window, Score 159 141

-18 byte thank to @TaylorRaine

FOR I=1TO[len(b3)]:for J=1to i:Y=Y-(MID([B3],i,1)=mid([B3],j,1)):NEXT:next:?y

The mixed case reduces the score by 30.

Answer 21

JavaScript, 112 95 58 points

A=>A.map(C=>N+=A[C]=-~A[C],N=0)|N

• 7 points saved thanks to ETHProductions

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Answer 22

JavaScript (Firefox 30+), score 68

S=>[for(C of(T=0,S))T+=S.split(C).length]&&T/2

This uses array comprehensions which were originally planned for ES7, but were dropped late in the process. Firefox had implemented them anyway.

Answer 23

Python 2, score 108

lambda S:sum(S.count(Q)*(S.count(Q)+1)/2for Q in set(S))

Still golfing :)

Answer 24

QBIC, score: 128

dim x(126)[_l;||i=asc(_sA,a,1|)┘x(i)=x(i)+1][126|[x(b)|p=p+c}?p

Nice round score of 128.

Explanation

dim x(126)      Create an array of 126 elements (one for each ASCII element)
[_l;||          Read cmd line input, loop over its length
i=asc(_sA,a,1|) Read the next char's ascii value
┘               (Syntactic linebreak)
x(i)=x(i)+1     Raise the count of this ASCII char by 1
]               NEXT
[126|           Loop over the ASCII array again
[x(b)|          Read the count and start a loop that runs from 1 to that count
p=p+c           Add to running total p the loop counter
}               Close both loops
?p              PRINT p

Answer 25

Ruby, score 63 60

Uses the -n flag which ends up adding 2 points due to the presence of another n within the code.

-3 points from Alexis Anderson.

i=x=0
gsub(/./){x+=$_[0,i+=1].count$&};p x

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Answer 26

Swift 3, 153 bytes, score = 637

var u:(String)->Int={var d = Dictionary<Character,Int>()
for c in $0.characters{d[c] = (d[c] ?? 0) + 1}
return d.values.reduce(0,{$0 + $1 * ($1 + 1) / 2})}

Legible version plus explanation

func unique(_ string: String) -> Int {
    var dict = Dictionary<Character, Int>()
    for char in string.characters { dict[char] = (dict[char] ?? 0) + 1 }
    return dict.values.reduce(0, { $0 + $1 * ($1 + 1) / 2 })
}

My solution takes the string, iterates over the characters and tracks each unique character in a dictionary, counting the occurrences per character. Once that's done, it reduces the dictionary's values using the triangle number formula ( n(n+1) / 2 ) and returns the result.

Comments

Swift obviously isn't the best language for golfing, but it was fun to see where I could save bytes despite its rigid syntax. This is my first Swift submission - be gentle!

Answer 27

CJam, score 18

q_,{)W$<}%.e=0+:+

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Explanation

q                  e# Read the input.
 _,                e# Copy it, get its length.
   {               e# Map over the range from 0 .. length-1:
    )              e#  Increment the current number.
     W$            e#  Push the input.
       <           e#  Get the first <current number> characters of it.
        }%         e# (end map)
          .e=      e# Count the occurrences of each character in the corresponding slice.
             0+    e# Append a 0 to the result (because you can't reduce and empty list).
               :+  e# Reduce by addition (sum).

Answer 28

><> (Fish), score 133

!<01-0&i:2(:?&:?n?;&1+&:}}:3(00@@?.~~:{=&+&{$}8e+0.>.!06}~{ 

This code reads the input per character, adds a point and loops through all the read characters, adding an extra point for every equal character.

You can try it out here

Explanation

!<01-0&         // Initialize (put -1 on the stack and 0 in the register)
i:2(:?$:?n?;    // Read the next input. If it's -1, print the register and end the program.
&1+&            // Add one to the register
:}}:            // Duplicate both the input and the check value
3(00@@?.~~      // If the check value is -1, jump the instruction pointer to 0,0 (line 0)
:{=&+&          // Compare the input to the check value and add the result to the register
{$}8e+0.        // Move the checked value to the bottom of the stack and jump to 22,0 (line 4)
>.!06}~{        // Clean up the stack and jump to 0,6 (line 2)

Answer 29

Factor, 2435

(o god my score)

"" over [ 2dup swap in? [ drop ] [ suffix ] if ] each { } swap swapd [ dupd 0 swap swapd swap [ over = swap [ [ 1 + ] [ ] if ] dip ] each drop swap [ suffix ] dip ] each drop [ dup 1 + * 2 / ] map sum (wrong formatting on purpose for multi-line span)

Wow, Factor is hard. Or I'm just bad. Probably a little bit of both.

Answer 30

Pyt, 6 bytes

ĐỤ⇹ɔ△Ʃ

Explanation:

              Implicit input
Đ             Duplicate top of stack
 Ụ            Get unique characters
  ⇹           Swap top two items on stack
   ɔ          Count number of times each unique character occurs in the input
    △         Calculates the total score for each character
     Ʃ        Sum count
              Implicit output

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