Jewels and stones

Jewels and stones is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Task

Create a function which takes two string parameters: 'stones' and 'jewels' and returns an integer.

Both strings can contain any number of upper or lower case letters. However, in the case of 'jewels', all letters must be distinct.

The function should count (and return) how many 'stones' are 'jewels' or, in other words, how many letters in 'stones' are also letters in 'jewels'.

Note that:

1. A lower case letter is considered to be different to its upper case equivalent for this purpose i.e. 'a' != 'A'.

2. The parameters do not need to have exactly the same names.

3. Validating the arguments is unnecessary.

So, for example, if passed "aAAbbbb" for 'stones' and "aA" for 'jewels', the function should return 3.

This task was inspired by this problem.

AWK

<lang AWK>

1. syntax: GAWK -f JEWELS_AND_STONES.AWK

BEGIN {

   printf("%d\n",count("aAAbbbb","aA"))
   printf("%d\n",count("ZZ","z"))
   exit(0)

} function count(stone,jewel, i,total) {

   for (i=1; i<length(stone); i++) {
     if (jewel ~ substr(stone,i,1)) {
       total++
     }
   }
   return(total)

} </lang>

Output:

3
0

C

Translation of: Kotlin

<lang c>#include <stdio.h>

1. include <string.h>

int count_jewels(const char *s, const char *j) {

   int count = 0;
   for ( ; *s; ++s) if (strchr(j, *s)) ++count;
   return count;

}

int main() {

   printf("%d\n", count_jewels("aAAbbbb", "aA"));
   printf("%d\n", count_jewels("ZZ", "z"));
   return 0;

}</lang>

Output:

3
0

Factor

<lang factor>USING: kernel prettyprint sequences ;

count-jewels ( stones jewels -- n ) [ member? ] curry count ;

"aAAbbbb" "aA" "ZZ" "z" [ count-jewels . ] 2bi@</lang>

Output:

3
0

Kotlin

<lang scala>// Version 1.2.40

fun countJewels(s: String, j: String) = s.count { it in j }

fun main(args: Array<String>) {

   println(countJewels("aAAbbbb", "aA"))
   println(countJewels("ZZ", "z"))

}</lang>

Output:

3
0

Perl 6

<lang perl6>sub count-jewels ( Str $j, Str $s --> Int ) {

   my %counts_of_all = $s.comb.Bag;
   my @jewel_list    = $j.comb.unique;

   return %counts_of_all ∩ @jewel_list.Bag ?? %counts_of_all{ @jewel_list }.sum !! 0;

}

say count-jewels 'aA' , 'aAAbbbb'; say count-jewels 'z' , 'ZZ';</lang>

Output:

3
0

Python

<lang python>def countJewels(s, j):

   return sum(x in j for x in s)

print countJewels("aAAbbbb", "aA") print countJewels("ZZ", "z")</lang>

Output:

3
0

Racket

<lang racket>

1. lang racket

(define (jewels-and-stones stones jewels)

 (length (filter (curryr member (string->list jewels)) (string->list stones))))

(module+ main

 (jewels-and-stones "aAAbbbb" "aA")
 (jewels-and-stones "ZZ" "z"))

</lang>

Output:

3
0

REXX

This REXX program accepts two characters strings from the command line instead of hardcoding the strings. <lang rexx>/*REXX pgm counts how many letters (in the 1st string) are in common with the 2nd string*/ parse arg stones jewels . /*obtain optional arguments from the CL*/ if stones== then stones= "aAAbbbb" /*Not defined? Then assume the default*/ if jewels== then jewels= "aA" /* " " " " " " */ say 'stones:' stones /*display value of the variable: STONES*/ say 'jewels:' jewels /* " " " " " JEWELS*/

1. =0 /*initalize the variable # to zero. */

    do j=1  for length(stones)                  /*scan STONES for matching JEWELS chars*/
    x=substr(stones, j, 1)                      /*obtain a character of the STONES var.*/
    if datatype(x, 'M') & pos(x, jewels)\==0  then #=# + 1
    end   /*j*/                                 /*if a letter & a match, bump # counter*/

say /*display a blank line, and show number*/ say # /*stick a fork in it, we're all done. */</lang>

output   when using the default input:

stones: aAAbbbb
jewels: aA

3

Ring

<lang ring>

1. Project Jewels and Stones
2. Date 2018/04/25
3. Author Gal Zsolt (~ CalmoSoft ~)
4. Email <calmosoft@gmail.com>

jewels = "aA" stones = "aAAbbbb" see jewelsandstones(jewels,stones) + nl jewels = "z" stones = "ZZ" see jewelsandstones(jewels,stones) + nl

func jewelsandstones(jewels,stones)

       num = 0
       for n = 1 to len(stones)
            pos = substr(jewels,stones[n])
            if pos > 0
               num = num + 1
            ok
       next
       return num

</lang> Output:

3
0

Sidef

<lang ruby>func countJewels(s, j) {

   s.chars.count { |c|
       j.contains(c)
   }

}

say countJewels("aAAbbbb", "aA") #=> 3 say countJewels("ZZ", "z") #=> 0</lang>

zkl

<lang zkl>fcn countJewels(a,b){ a.inCommon(b).len() }</lang> <lang zkl>println(countJewels("aAAbbbb", "aA")); println(countJewels("ZZ", "z"));</lang>

Output:

3
0