Jewels and stones

From Rosetta Code


Task
Jewels and stones
You are encouraged to solve this task according to the task description, using any language you may know.
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. Only letters in the ISO basic Latin alphabet i.e. 'A to Z' or 'a to z' need be considered.
  2. A lower case letter is considered to be different from its upper case equivalent for this purpose, i.e., 'a' != 'A'.
  3. The parameters do not need to have exactly the same names.
  4. 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.


Other tasks related to string operations:
Metrics
Counting
Remove/replace
Anagrams/Derangements/shuffling
Find/Search/Determine
Formatting
Song lyrics/poems/Mad Libs/phrases
Tokenize
Sequences



11l[edit]

Translation of: Python
F count_jewels(s, j)
   R sum(s.map(x -> Int(x C @j)))

print(count_jewels(‘aAAbbbb’, ‘aA’))
print(count_jewels(‘ZZ’, ‘z’))
Output:
3
0

8080 Assembly[edit]

	org	100h
	jmp	demo
	;;;	Count jewels.
	;;;	Input: BC = jewels, DE = stones.
	;;;	Output: BC = count
	;;;	Destroyed: A, DE, HL
jewel:	lxi	h,jarr	; Zero out the page of memory
	xra	a
jzero:	mov	m,a
	inr	l
	jnz	jzero
jrjwl:	ldax	b	; Get jewel
	inx	b
	mov	l,a	; Mark the corresponding byte in the array
	inr	m
	ana	a	; If 'jewel' is 0, we've reached the end
	jnz	jrjwl	; Otherwise, do next jewel
	lxi	b,0	; BC = count (we no longer need the jewel string)
jrstn:	ldax	d	; Get stone
	inx	d
	ana	a	; If zero, we're done
	rz
	mov	l,a	; Get corresponding byte in array 
	mov	a,m
	ana	a
	jz	jrstn	; If zero, it is not a jewel
	inx	b	; But otherwise, it is a jewel
	jmp	jrstn
	;;;	Demo code
demo:	lxi	b,jewels	; Set up registers
	lxi	d,stones
	call	jewel		; Call the function
	;;;	Print the number
	lxi	h,num	; Pointer to number string
	push	h	; Push to stack
	mov	h,b	; HL = number to print
	mov	l,c
	lxi	b,-10	; Divisor
dgt:	lxi	d,-1	; Quotient
dgtlp:	inx	d	; Divide using trial subtraction
	dad	b
	jc	dgtlp
	mvi	a,'0'+10
	add	l	; HL = remainder-10
	pop	h	; Get pointer
	dcx	h	; Decrement pointer
	mov	m,a	; Store digit
	push	h	; Put pointer back
	xchg		; Go on with new quotient
	mov	a,h	; If 0, we're done
	ana	l
	jnz	dgt	; If not 0, next digit
	pop	d	; Get pointer and put it in DE
	mvi	c,9	; CP/M syscall to print string
	jmp	5
	db	'*****'	; Placeholder for ASCII number output
num:	db	'$'
	;;;	Example from the task
jewels:	db	'aA',0
stones:	db	'aAAbbbb',0
	;;;	Next free page of memory is used for the jewel array
jpage:	equ	$/256+1
jarr:	equ	jpage*256
Output:
3


8086 Assembly[edit]

	cpu	8086
	bits	16
	org	100h
section	.text
	jmp	demo
	;;;	Count jewels.
	;;;	Input: DS:SI = jewels, DS:DX = stones
	;;;	Output: CX = how many stones are jewels
	;;;	Destroyed: AX, BX, SI, DI
jewel:	xor	ax,ax
	mov	cx,128		; Allocate 256 bytes (128 words) on stack
.zloop:	push	ax		; Set them all to zero
	loop	.zloop
	mov	di,sp		; DI = start of array
	xor	bh,bh
.sjwl:	lodsb			; Get jewel
	mov	bl,al
	inc	byte [ss:di+bx]	; Set corresponding byte
	test	al,al		; If not zero, there are more jewels
	jnz	.sjwl
	mov	si,dx		; Read stones
.sstn:	lodsb			; Get stone
	mov	bl,al		; Get corresponding byte
	mov	bl,[ss:di+bx]
	add	cx,bx		; Add to count (as word)
	test	al,al		; If not zero, there are more stones
	jnz	.sstn
	add	sp,256		; Otherwise, we are done - free the array
	dec	cx		; The string terminator is a 'jewel', so remove
	ret
	;;;	Demo
demo:	mov	si,jewels	; Set up registers
	mov	dx,stones
	call	jewel		; Call the function
	;;;	Print number 
	mov	ax,10		; Result is in CX
	xchg	ax,cx		; Set AX to result and CX to divisor (10)
	mov	bx,num		; Number pointer
dgt:	xor	dx,dx
	div	cx		; Divide AX by 10
	add	dl,'0'		; Remainder is in DX - add ASCII 0
	dec	bx		; Store digit in string
	mov	[bx],dl
	test	ax,ax		; Any more digits?
	jnz	dgt		; If so, next digit
	mov	dx,bx		; When done, print string
	mov	ah,9
	int	21h
	ret
section	.data
	db	'*****'		; Placeholder for ASCII number output
num:	db	'$'
stones:	db	'aAAbbbb',0	; Example from the task
jewels:	db	'aA',0
Output:
3

Ada[edit]

with Ada.Text_IO;

procedure Jewels_And_Stones is

   function Count (Jewels, Stones : in String) return Natural is
      Sum : Natural := 0;
   begin
      for J of Jewels loop
         for S of Stones loop
            if J = S then
               Sum := Sum + 1;
               exit;
            end if;
         end loop;
      end loop;
      return Sum;
   end Count;

   procedure Show (Jewels, Stones : in String) is
      use Ada.Text_IO;
   begin
      Put (Jewels);
      Set_Col (12); Put (Stones);
      Set_Col (20); Put (Count (Jewels => Jewels,
                                Stones => Stones)'Image);
      New_Line;
   end Show;

begin
   Show ("aAAbbbb", "aA");
   Show ("ZZ",      "z");
end Jewels_And_Stones;
Output:
aAAbbbb    aA       3
ZZ         z        0

ALGOL 68[edit]

BEGIN
    # procedure that counts the number of times the letters in jewels occur in stones #
    PROC count jewels = ( STRING stones, jewels )INT:
         BEGIN
             # count the occurences of each letter in stones #
             INT upper a pos = 0;
             INT lower a pos = 1 + ( ABS "Z" - ABS "A" );
             [ upper a pos : lower a pos + 26 ]INT letter counts;
             FOR c FROM LWB letter counts TO UPB letter counts DO letter counts[ c ] := 0 OD;
             FOR s pos FROM LWB stones TO UPB stones DO
                 CHAR s = stones[ s pos ];
                 IF   s >= "A" AND s <= "Z" THEN letter counts[ upper a pos + ( ABS s - ABS "A" ) ] +:= 1
                 ELIF s >= "a" AND s <= "z" THEN letter counts[ lower a pos + ( ABS s - ABS "a" ) ] +:= 1
                 FI
             OD;
             # sum the counts of the letters that appear in jewels #
             INT count := 0;
             FOR j pos FROM LWB jewels TO UPB jewels DO
                 CHAR j = jewels[ j pos ];
                 IF   j >= "A" AND j <= "Z" THEN count +:= letter counts[ upper a pos + ( ABS j - ABS "A" ) ]
                 ELIF j >= "a" AND j <= "z" THEN count +:= letter counts[ lower a pos + ( ABS j - ABS "a" ) ]
                 FI 
             OD;
             count
         END # count jewels # ;

    print( ( count jewels( "aAAbbbb", "aA" ), newline ) );
    print( ( count jewels( "ABCDEFGHIJKLMNOPQRSTUVWXYZ@abcdefghijklmnopqrstuvwxyz"
                         , "ABCDEFGHIJKLMNOPQRSTUVWXYZ@abcdefghijklmnopqrstuvwxyz"
                         )
           , newline
           )
         );
    print( ( count jewels( "AB", "" ), newline ) );
    print( ( count jewels( "ZZ", "z" ), newline ) )

END
Output:
         +3
        +52
         +0
         +0

APL[edit]

Works with: Dyalog APL
jewels  +/
Output:
      'aA' jewels 'aAAbbbb'
3


AppleScript[edit]

Functional[edit]

-- jewelCount :: String -> String -> Int
on jewelCount(jewels, stones)
    set js to chars(jewels)
    script
        on |λ|(a, c)
            if elem(c, jewels) then
                a + 1
            else
                a
            end if
        end |λ|
    end script
    foldl(result, 0, chars(stones))
end jewelCount

-- OR in terms of filter
-- jewelCount :: String -> String -> Int
on jewelCount2(jewels, stones)
    script
        on |λ|(c)
            elem(c, jewels)
        end |λ|
    end script
    length of filter(result, stones)
end jewelCount2

-- TEST --------------------------------------------------
on run
    
    unlines(map(uncurry(jewelCount), ¬
        {Tuple("aA", "aAAbbbb"), Tuple("z", "ZZ")}))
    
end run


-- GENERIC FUNCTIONS -------------------------------------

-- Tuple (,) :: a -> b -> (a, b)
on Tuple(a, b)
    {type:"Tuple", |1|:a, |2|:b}
end Tuple

-- chars :: String -> [Char]
on chars(s)
    characters of s
end chars

-- elem :: Eq a => a -> [a] -> Bool
on elem(x, xs)
    considering case
        xs contains x
    end considering
end elem

-- filter :: (a -> Bool) -> [a] -> [a]
on filter(f, xs)
    tell mReturn(f)
        set lst to {}
        set lng to length of xs
        repeat with i from 1 to lng
            set v to item i of xs
            if |λ|(v, i, xs) then set end of lst to v
        end repeat
        return lst
    end tell
end filter

-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
    tell mReturn(f)
        set v to startValue
        set lng to length of xs
        repeat with i from 1 to lng
            set v to |λ|(v, item i of xs, i, xs)
        end repeat
        return v
    end tell
end foldl

-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
    tell mReturn(f)
        set lng to length of xs
        set lst to {}
        repeat with i from 1 to lng
            set end of lst to |λ|(item i of xs, i, xs)
        end repeat
        return lst
    end tell
end map

-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
    if class of f is script then
        f
    else
        script
            property |λ| : f
        end script
    end if
end mReturn

-- Returns a function on a single tuple (containing 2 arguments)
-- derived from an equivalent function with 2 distinct arguments
-- uncurry :: (a -> b -> c) -> ((a, b) -> c)
on uncurry(f)
    script
        property mf : mReturn(f)'s |λ|
        on |λ|(pair)
            mf(|1| of pair, |2| of pair)
        end |λ|
    end script
end uncurry

-- unlines :: [String] -> String
on unlines(xs)
    set {dlm, my text item delimiters} to ¬
        {my text item delimiters, linefeed}
    set str to xs as text
    set my text item delimiters to dlm
    str
end unlines
Output:
3
0

Idiomatic[edit]

on jewelsAndStones(stones, jewels)
    set counter to 0
    considering case
        repeat with thisCharacter in stones
            if (thisCharacter is in jewels) then set counter to counter + 1
        end repeat
    end considering
    
    return counter
end jewelsAndStones

jewelsAndStones("aAAbBbb", "aAb")
Output:
6

Arturo[edit]

count: function [jewels,stones][
    size select split stones => [in? & jewels]
]

print count "aA" "aAAbbbb"
Output:
3

AutoHotkey[edit]

JewelsandStones(ss, jj){
	for each, jewel in StrSplit(jj)
		for each, stone in StrSplit(ss)
			if (stone == jewel)
				num++
	return num
}
Example:
MsgBox % JewelsandStones("aAAbbbbz", "aAZ")
return
Outputs:
3

AWK[edit]

# 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)
}
Output:
3
0

BASIC[edit]

10 READ N%
20 FOR A%=1 TO N%
30 READ J$,S$
40 GOSUB 100
50 PRINT S$;" in ";J$;":";J%
60 NEXT
70 END

100 REM Count how many stones (S$) are jewels (J$). 
110 DIM J%(127)
120 J%=0
130 FOR I%=1 TO LEN(J$): J%(ASC(MID$(J$,I%,1)))=1: NEXT
140 FOR I%=1 TO LEN(S$): J%=J%+J%(ASC(MID$(S$,I%,1))): NEXT
150 ERASE J%
160 RETURN

200 DATA 2
210 DATA "aA","aAAbbbb"
220 DATA "z","ZZZZ"
Output:
aAAbbbb in aA: 3
ZZZZ in z: 0

BASIC256[edit]

function contar_joyas(piedras, joyas)
	cont = 0
	for i = 1 to length(piedras)
		bc = instr(joyas, mid(piedras, i, 1), 1)
		if bc <> 0 then cont += 1
	next i
	return cont
end function

print contar_joyas("aAAbbbb", "aA")
print contar_joyas("ZZ", "z")
print contar_joyas("ABCDEFGHIJKLMNOPQRSTUVWXYZ@abcdefghijklmnopqrstuvwxyz", "ABCDEFGHIJKLMNOPQRSTUVWXYZ@abcdefghijklmnopqrstuvwxyz")
print contar_joyas("AB", "")
Output:
Igual que la entrada de FreeBASIC.

BCPL[edit]

get "libhdr"

let jewels(j, s) = valof
$(  let jewel = vec 255
    let count = 0
    for i = 0 to 255 do jewel!i := false
    for i = 1 to j%0 do jewel!(j%i) := true
    for i = 1 to s%0 do 
        if jewel!(s%i) then
            count := count + 1
    resultis count
$)

let show(j, s) be
    writef("*"%S*" in *"%S*": %N*N", j, s, jewels(j, s))

let start() be
$(  show("aA", "aAAbbbb")
    show("z", "ZZ")
$)
Output:
"aA" in "aAAbbbb": 3
"z" in "ZZ": 0

BQN[edit]

Similar in nature to APL, mostly due to how trivial the problem is in an array language.

Jewels←+´∊˜
  "aA" Jewels "aAAbbbb"
3

Bracmat[edit]

  ( f
  =   stones jewels N
    .   !arg:(?stones.?jewels)
      & 0:?N
      & ( @( !stones
           :   ?
               (   %@
                 : [%(   @(!jewels:? !sjt ?)
                       & 1+!N:?N
                     |
                     )
               & ~
               )
               ?
           )
        | !N
        )
  )
& f$(aAAbbbb.aA)

Output

3

C[edit]

Translation of: Kotlin
#include <stdio.h>
#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;
}
Output:
3
0

C#[edit]

using System;
using System.Linq;

public class Program
{
    public static void Main() {
        Console.WriteLine(Count("aAAbbbb", "Aa"));
        Console.WriteLine(Count("ZZ", "z"));
    }

    private static int Count(string stones, string jewels) {
        var bag = jewels.ToHashSet();
        return stones.Count(bag.Contains);
    }
}
Output:
3
0

C++[edit]

Translation of: D
#include <algorithm>
#include <iostream>

int countJewels(const std::string& s, const std::string& j) {
    int count = 0;
    for (char c : s) {
        if (j.find(c) != std::string::npos) {
            count++;
        }
    }
    return count;
}

int main() {
    using namespace std;

    cout << countJewels("aAAbbbb", "aA") << endl;
    cout << countJewels("ZZ", "z") << endl;

    return 0;
}
Output:
3
0

CLU[edit]

count_jewels = proc (jewels, stones: string) returns (int)
    is_jewel: array[bool] := array[bool]$fill(0, 256, false)
    for c: char in string$chars(jewels) do
        is_jewel[char$c2i(c)] := true
    end
    
    n_jewels: int := 0
    for c: char in string$chars(stones) do
        if is_jewel[char$c2i(c)] then n_jewels := n_jewels + 1 end
    end
    return (n_jewels)
end count_jewels

show = proc (jewels, stones: string)
    po: stream := stream$primary_output()
    
    stream$putl(po, "\"" || jewels || "\" in \"" || stones || "\": "
                || int$unparse(count_jewels(jewels, stones)))
end show

start_up = proc ()
    show("aA", "aAAbbbb")
end start_up
Output:
"aA" in "aAAbbbb": 3

Cowgol[edit]

include "cowgol.coh";

sub count_jewels(stones: [uint8], jewels: [uint8]): (count: uint16) is
    var jewel_mark: uint8[256];
    MemZero(&jewel_mark as [uint8], 256);

    while [jewels] != 0 loop
        jewel_mark[[jewels]] := 1;
        jewels := @next jewels;
    end loop;

    count := 0;
    while [stones] != 0 loop
        count := count + jewel_mark[[stones]] as uint16;
        stones := @next stones;
    end loop;
end sub;

sub print_and_count(stones: [uint8], jewels: [uint8]) is
    print(jewels);
    print(" in ");
    print(stones);
    print(": ");
    print_i16(count_jewels(stones, jewels));
    print_nl();
end sub;

print_and_count("aAAbbbb", "aA");
print_and_count("ZZ", "z");
Output:
aA in aAAbbbb: 3
z in ZZ: 0

Crystal[edit]

stones, jewels = "aAAbbbb", "aA"
stones.count(jewels) # => 3

# The above solution works for that case, but fails with certain other "jewels":
stones, jewels = "aA^Bb", "^b"
stones.count(jewels) # => 4
# '^b' in the "jewels" is read as "characters other than 'b'".

# This works as intended though:
stones.count { |c| jewels.chars.includes?(c) } # => 2

D[edit]

Translation of: Kotlin
import std.algorithm;
import std.stdio;

int countJewels(string s, string j) {
    int count;
    foreach (c; s) {
        if (j.canFind(c)) {
            count++;
        }
    }
    return count;
}

void main() {
    countJewels("aAAbbbb", "aA").writeln;
    countJewels("ZZ", "z").writeln;
}
Output:
3
0

Draco[edit]

proc nonrec count_jewels(*char jewels, stones) word:
    [256] bool jewel;
    word count;
    byte i;
    char c;
    
    for i from 0 upto 255 do jewel[i] := false od;
    while c := jewels*; c ~= '\e' do
        jewel[c] := true;
        jewels := jewels + 1;
    od;
    
    count := 0;
    while c := stones*; c ~= '\e' do
        if jewel[c] then count := count + 1 fi;
        stones := stones + 1
    od;
    count
corp

proc nonrec show(*char jewels, stones) void:
    writeln("'", jewels, "' in '", stones, "': ", count_jewels(jewels, stones))
corp

proc nonrec main() void:
    show("aA", "aAAbbbb");
    show("z", "ZZ")
corp
Output:
'aA' in 'aAAbbbb': 3
'z' in 'ZZ': 0

Dyalect[edit]

Translation of: Swift
func countJewels(stones, jewels) {
    stones.Iterate().Map(x => jewels.Contains(x) ? 1 : 0).Reduce((x,y) => x + y, 0)
}
 
print(countJewels("aAAbbbb", "aA"))
print(countJewels("ZZ", "z"))
Output:
3
0

F#[edit]

let fN jewels stones=stones|>Seq.filter(fun n->Seq.contains n jewels)|>Seq.length
printfn "%d" (fN "aA" "aAAbbbb")
Output:
3

Factor[edit]

USING: kernel prettyprint sequences ;

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

"aAAbbbb" "aA"
"ZZ" "z" [ count-jewels . ] 2bi@
Output:
3
0

Euphoria[edit]

function number_of(object jewels, object stones) -- why limit ourselves to strings?
integer ct = 0
    for i = 1 to length(stones) do
        ct += find(stones[i],jewels) != 0
    end for
    return ct
end function

? number_of("aA","aAAbbbb")
? number_of("z","ZZ")
? number_of({1,"Boo",3},{1,2,3,'A',"Boo",3}) -- might as well send a list of things to find, not just one!
Output:
3
0
4 -- 1 is found once, "Boo" is found once, and 3 is found twice = 4 things in the search list were found in the target list

FreeBASIC[edit]

function contar_joyas(piedras as string, joyas as string) as integer
    dim as integer bc, cont: cont = 0
    for i as integer = 1 to len(piedras)
        bc = instr(1, joyas, mid(piedras, i, 1))
        if bc <> 0 then cont += 1
    next i
    contar_joyas = cont
end function

print contar_joyas("aAAbbbb", "aA")
print contar_joyas("ZZ", "z")
print contar_joyas("ABCDEFGHIJKLMNOPQRSTUVWXYZ@abcdefghijklmnopqrstuvwxyz", _
                   "ABCDEFGHIJKLMNOPQRSTUVWXYZ@abcdefghijklmnopqrstuvwxyz")
print contar_joyas("AB", "")
Output:
3
0
53
0

FutureBasic[edit]

window 1, @"Jewels and Stones"

local fn JewelsAndStones( jewels as CFStringRef, stones as CFStringRef ) as long
  long index, count = 0
  
  for index = 0 to len(stones) - 1
    if ( fn StringContainsString( jewels, mid(stones,index,1) ) ) then count++
  next
end fn = count

print fn JewelsAndStones( @"aA", @"aAAbbbb" )
print fn JewelsAndStones( @"z", @"ZZ" )

HandleEvents
Output:
3
0

Go[edit]

Four solutions are shown here. The first of two simpler solutions iterates over the stone string in an outer loop and makes repeated searches into the jewel string, incrementing a count each time it finds a stone in the jewels. The second of the simpler solutions reverses that, iterating over the jewel string in the outer loop and accumulating counts of matching stones. This solution works because we are told that all letters of the jewel string must be unique. These two solutions are simple but are both O(|j|*|s|).

The two more complex solutions are analogous to the two simpler ones but build a set or multiset as preprocessing step, replacing the inner O(n) operation with an O(1) operation. The resulting complexity in each case is O(|j|+|s|).

Outer loop stones, index into jewels:

package main
  
import (
    "fmt"
    "strings"
)

func js(stones, jewels string) (n int) {
    for _, b := range []byte(stones) {
        if strings.IndexByte(jewels, b) >= 0 {
            n++
        }
    }
    return
}

func main() {
    fmt.Println(js("aAAbbbb", "aA"))
}
Output:
3

Outer loop jewels, count stones:

func js(stones, jewels string) (n int) {
    for _, b := range []byte(jewels) {
        n += strings.Count(stones, string(b))
    }
    return
}

Construct jewel set, then loop over stones:

func js(stones, jewels string) (n int) {
    var jSet ['z' + 1]int
    for _, b := range []byte(jewels) {
        jSet[b] = 1
    }
    for _, b := range []byte(stones) {
        n += jSet[b]
    }
    return
}

Construct stone multiset, then loop over jewels:

func js(stones, jewels string) (n int) {
    var sset ['z' + 1]int
    for _, b := range []byte(stones) {
        sset[b]++
    }
    for _, b := range []byte(jewels) {
        n += sset[b]
    }
    return
}

Haskell[edit]

jewelCount
  :: Eq a
  => [a] -> [a] -> Int
jewelCount jewels = foldr go 0
  where
    go c
      | c `elem` jewels = succ
      | otherwise = id

--------------------------- TEST -------------------------
main :: IO ()
main = mapM_ print $ uncurry jewelCount <$> [("aA", "aAAbbbb"), ("z", "ZZ")]
Output:
3
0

Or in terms of filter rather than foldr

jewelCount
  :: Eq a
  => [a] -> [a] -> Int
jewelCount jewels = length . filter (`elem` jewels)

--------------------------- TEST -------------------------
main :: IO ()
main = do
  print $ jewelCount "aA" "aAAbbbb"
  print $ jewelCount "z" "ZZ"
Output:
3
0

J[edit]

   NB. jewels sums a raveled equality table
   NB. use: x jewels y  x are the stones, y are the jewels.
   intersect =: -.^:2
   jewels =: ([: +/ [: , =/~) ~.@:intersect&Alpha_j_

   'aAAbbbb ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz' jewels&>&;: 'aA ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz'
3 52

   'none' jewels ''
0
   'ZZ' jewels 'z'
0

Java[edit]

import java.util.HashSet;
import java.util.Set;

public class App {
    private static int countJewels(String stones, String jewels) {
        Set<Character> bag = new HashSet<>();
        for (char c : jewels.toCharArray()) {
            bag.add(c);
        }

        int count = 0;
        for (char c : stones.toCharArray()) {
            if (bag.contains(c)) {
                count++;
            }
        }

        return count;
    }

    public static void main(String[] args) {
        System.out.println(countJewels("aAAbbbb", "aA"));
        System.out.println(countJewels("ZZ", "z"));
    }
}
Output:
3
0

JavaScript[edit]

(() => {

    // jewelCount :: String -> String -> Int
    const jewelCount = (j, s) => {
        const js = j.split('');
        return s.split('')
            .reduce((a, c) => js.includes(c) ? a + 1 : a, 0)
    };

    // TEST -----------------------------------------------
    return [
            ['aA', 'aAAbbbb'],
            ['z', 'ZZ']
        ]
        .map(x => jewelCount(...x))
})();
Output:
[3, 0]

jq[edit]

$ jq -n --arg stones aAAbbbb --arg jewels aA '
  [$stones|split("") as $s|$jewels|split("") as $j|$s[]|
  select(. as $c|$j|contains([$c]))]|length'
Output:
3

Julia[edit]

Module:

module Jewels

count(s, j) = Base.count(x  j for x in s)

end  # module Jewels

Main:

@show Jewels.count("aAAbbbb", "aA")
@show Jewels.count("ZZ", "z")
Output:
Jewels.count("aAAbbbb", "aA") = 3
Jewels.count("ZZ", "z") = 0

Kotlin[edit]

// 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"))
}
Output:
3
0

Lambdatalk[edit]

{def countjewels
 {def countjewels.r
  {lambda {:a :b :c}
   {if {A.empty? :a}
    then :c
    else {countjewels.r {A.rest :a}
                        :b
                        {if {= {A.in? {A.first :a} :b} -1}
                         then :c
                         else {+ :c 1}}}}}}
 {lambda {:a :b}
  {countjewels.r {A.split :a} {A.split :b} 0}}} 
-> countjewels

{countjewels aAAbbbb aA} -> 3
{countjewels ZZ z}       -> 0

Lua[edit]

Translation of: C
function count_jewels(s, j)
    local count = 0
    for i=1,#s do
        local c = s:sub(i,i)
        if string.match(j, c) then
            count = count + 1
        end
    end
    return count
end

print(count_jewels("aAAbbbb", "aA"))
print(count_jewels("ZZ", "z"))
Output:
3
0

Maple[edit]

count_jewel := proc(stones, jewels)
	local count, j, letter:
	j := convert(jewels,set):
	count := 0:
	for letter in stones do
		if (member(letter, j)) then
			count++:
		end if:
	end do:
	return count:
end proc:
count_jewel("aAAbbbb", "aA")
Output:
3

Mathematica / Wolfram Language[edit]

JewelsStones[j_String, s_String] := Count[MemberQ[Characters[j], #] & /@ Characters[s], True]
JewelsStones["aA", "aAAbbbb"]
JewelsStones["ZZ", "z"]
Output:
3
0

MATLAB / Octave[edit]

    function s = count_jewels(stones,jewels)
    s=0;
    for c=jewels
        s=s+sum(c==stones);
    end
    %!test
    %! assert(count_jewels('aAAbbbb','aA'),3)
    %!test
    %! assert(count_jewels('ZZ','z'),0)

min[edit]

Works with: min version 0.19.6
(('' '' '') => spread if) :if?

((1 0 if?) concat map sum) :count

(swap indexof -1 !=) :member?

(("" split) dip 'member? cons count) :count-jewels

"aAAbbbb" "aA" count-jewels puts!
"ZZ" "z" count-jewels puts!
Output:
3
0

Modula-2[edit]

MODULE Jewels;
FROM FormatString IMPORT FormatString;
FROM Terminal IMPORT WriteString,WriteLn,ReadChar;

PROCEDURE WriteInt(n : INTEGER);
VAR buf : ARRAY[0..15] OF CHAR;
BEGIN
    FormatString("%i", buf, n);
    WriteString(buf)
END WriteInt;

PROCEDURE CountJewels(s,j : ARRAY OF CHAR) : INTEGER;
VAR c,i,k : CARDINAL;
BEGIN
    c :=0;

    FOR i:=0 TO HIGH(s) DO
        FOR k:=0 TO HIGH(j) DO
            IF (j[k]#0C) AND (s[i]#0C) AND (j[k]=s[i]) THEN
                INC(c);
                BREAK
            END
        END
    END;

    RETURN c
END CountJewels;

BEGIN
    WriteInt(CountJewels("aAAbbbb", "aA"));
    WriteLn;
    WriteInt(CountJewels("ZZ", "z"));
    WriteLn;

    ReadChar
END Jewels.
Output:
3
0

Nim[edit]

import sequtils

func countJewels(stones, jewels: string): Natural =
  stones.countIt(it in jewels)

echo countJewels("aAAbbbb", "aA")
echo countJewels("ZZ", "z")
Output:
3
0

Objeck[edit]

Translation of: Java
use Collection.Generic;

class JewelsStones {
  function : Main(args : String[]) ~ Nil {
    Count("aAAbbbb", "aA")->PrintLine();
    Count("ZZ", "z")->PrintLine();
  }

  function : Count(stones : String, jewels : String) ~ Int {
    bag := Set->New()<CharHolder>;

    each(i : jewels) {
      bag->Insert(jewels->Get(i));
    };

    count := 0;
    each(i : stones) {
      if(bag->Has(stones->Get(i))) {
        count++;
      };
    };
 
    return count;
  }
}
Output:
3
0

Perl[edit]

sub count_jewels {
    my( $j, $s ) = @_;
    my($c,%S);

    $S{$_}++     for split //, $s;
    $c += $S{$_} for split //, $j;
    return "$c\n";
}

print count_jewels 'aA' , 'aAAbbbb';
print count_jewels 'z'  , 'ZZ';
Output:
3
0

Alternate using regex[edit]

#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Jewels_and_Stones#Perl
use warnings;

sub count_jewels { scalar( () =  $_[0] =~ /[ $_[1] ]/gx ) } # stones, jewels

print "$_ = ", count_jewels( split ), "\n" for split /\n/, <<END;
aAAbbbb aA
aAAbbbb abc
ZZ z
END
Output:
aAAbbbb aA = 3
aAAbbbb abc = 5
ZZ z = 0

Phix[edit]

function count_jewels(string stones, jewels)
    integer res = 0
    for i=1 to length(stones) do
        res += find(stones[i],jewels)!=0
    end for
    return res
end function
?count_jewels("aAAbbbb","aA")
?count_jewels("ZZ","z")
Output:
3
0

Picat[edit]

List comprehension[edit]

jewels_and_stones1(Jewels,Stones) = sum([1 : S in Stones, J in Jewels, S == J]).

Recursion[edit]

jewels_and_stones2(Jewels,Stones) = N =>
  jewels_and_stones2(Jewels,Stones,0,N).

jewels_and_stones2([],_Stones,N,N).
jewels_and_stones2([J|Jewels],[S|Stones],N0,N) :-
  jewels_and_stones2_(J,[S|Stones],0,JN),
  jewels_and_stones2(Jewels,Stones,N0+JN,N).

% Check just this jewel on all the stones
jewels_and_stones2_(_J,[],N,N).
jewels_and_stones2_(J,[S|Stones],N0,N) :-
  ( J == S ->
     N1 = N0+1
    ;
    N1 = N0
  ),
  jewels_and_stones2_(J,Stones,N1,N).

Foreach loop[edit]

jewels_and_stones3(Jewels,Stones) = N =>
  N = 0,
  foreach(J in Jewels, S in Stones)
     if J == S then
       N := N + 1
     end
  end.

Test[edit]

go =>
  Tests = [["aA","aAAbbbb"],
           ["z","ZZ"]
           ],
  println(tests=Tests),
  foreach([Jewels,Stones] in Tests)
    println([jewels=Jewels,stone=Stones]),
    println(js1=jewels_and_stones1(Jewels,Stones)),
    println(js2=jewels_and_stones2(Jewels,Stones)),
    println(js3=jewels_and_stones3(Jewels,Stones)),
    nl
  end,
  nl.
Output:
tests = [[aA,aAAbbbb],[z,ZZ]]
[jewels = aA,stone = aAAbbbb]
js1 = 3
js2 = 3
js3 = 3

[jewels = z,stone = ZZ]
js1 = 0
js2 = 0
js3 = 0

Benchmark[edit]

For a larger test we can see the differences between the three approaches. Here are 100 000 000 random stones and (atmost) 15 jewels (we remove any duplicate jewel).

go2 =>
  Alpha = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ",
  Len = Alpha.len,
  _ = random2(),
  NumStones = 100_000_000,
  NumJewels = 15, % Atmost number of jewels (duplicates are removed)
  
  Stones = [Alpha[random(1,Len)] : _ in 1..NumStones],
  Jewels = [Alpha[random(1,Len)] : _ in 1..NumJewels].sort_remove_dups,
  println(jewels=Jewels),
  nl,  
  time(println(js1=jewels_and_stones1(Jewels,Stones))),
  time(println(js2=jewels_and_stones2(Jewels,Stones))),
  time(println(js3=jewels_and_stones3(Jewels,Stones))),
  nl.
Output:
NumStones: 100_000_000  NumJewels = 15
jewels = TwurtRabSXx

js1 = 21154798

CPU time 15.087 seconds.

js2 = 21154796

CPU time 11.024 seconds.

js3 = 21154798

CPU time 11.94 seconds.

The recursion approach (jewels_and_stones2/2) is a little faster than the loop approach (jewels_and_stones/3).


PL/M[edit]

100H:

/* FIND JEWELS AMONG STONES */
COUNT$JEWELS: PROCEDURE (JEWELS, STONES) BYTE;
    DECLARE (JEWELS, STONES) ADDRESS;
    DECLARE (J BASED JEWELS, S BASED STONES) BYTE;
    DECLARE JFLAG (256) BYTE, I BYTE;
   
    /* ZERO JEWEL FLAGS */
    DO I=0 TO 255;
        JFLAG(I) = 0;
    END;
    
    /* LOOP THROUGH JEWELS AND MARK THEM */
    DO WHILE J <> '$';
        JFLAG(J) = 1;
        JEWELS = JEWELS + 1;
    END;
    
    /* COUNT THE JEWELS AMONG THE STONES */
    I = 0;
    DO WHILE S <> '$';
        I = I + JFLAG(S);
        STONES = STONES + 1;
    END;
    RETURN I;
END COUNT$JEWELS;

/* CP/M BDOS CALL */
BDOS: PROCEDURE (FN, ARG);
    DECLARE FN BYTE, ARG ADDRESS;
    GO TO 5;
END BDOS;

PRINT: PROCEDURE (STR);
    DECLARE STR ADDRESS;
    CALL BDOS(9, STR);
END PRINT;

/* NUMERIC OUTPUT */
PRINT$NUMBER: PROCEDURE (N);
    DECLARE S (4) BYTE INITIAL ('...$');
    DECLARE P ADDRESS, (N, C BASED P) BYTE;
    P = .S(3);
DIGIT:
    P = P-1;
    C = N MOD 10 + '0';
    N = N/10;
    IF N > 0 THEN GO TO DIGIT;
    CALL PRINT(P);
END PRINT$NUMBER;

/* PRINT JEWELS, STONES, AND AMOUNT OF JEWELS IN STONES */
TEST: PROCEDURE (J, S);
    DECLARE (J, S) ADDRESS;
    CALL PRINT(.'''$');
    CALL PRINT(J);
    CALL PRINT(.''' IN ''$');
    CALL PRINT(S);
    CALL PRINT(.''': $');
    CALL PRINT$NUMBER(COUNT$JEWELS(J, S));
    CALL PRINT(.(13,10,'$'));
END TEST;

/* UNFORTUNATELY, THE PL/M COMPILER ACCEPTS A VERY RESTRICTED CHARACTER SET
   IN THE SOURCE CODE. (IT IS THE INTERSECTION OF VARIOUS POPULAR CHARSETS
   FROM THE 1960S.) THIS CHARACTER SET DOES NOT INCLUDE LOWERCASE LETTERS.
   HOWEVER, THIS CODE ASSUMES THE ASCII CHARACTER SET ANYWAY.
   WE CAN INCLUDE LOWERCASE LETTERS ... BY WRITING THEIR ASCII CODES. 
   THE OUTPUT WILL BE CORRECT OF COURSE. */

CALL TEST(.( 97,'A$' ), .( 97,'AA',98,98,98,98,'$' )); 
CALL TEST(.( 122,'$' ), .( 'ZZZZ$' ));

CALL BDOS(0,0);
EOF
Output:
'aA' IN 'aAAbbbb': 3
'z' IN 'ZZZZ': 0

Prolog[edit]

:- system:set_prolog_flag(double_quotes,codes) .

count_jewels(STONEs0,JEWELs0,COUNT)
:-
findall(X,(member(X,JEWELs0),member(X,STONEs0)),ALLs) ,
length(ALLs,COUNT)
.
Output:
/*
?- count_jewels("aAAbbbb","aA",N).
N = 3.

?- count_jewels("ZZ","z",N).
N = 0.

?-  count_jewels("aAAbbbb","bcd",N) .
N = 4.

?-
*/

alternative version[edit]

Works with: SWI Prolog
count_jewels(Stones, Jewels, N):-
    string_codes(Stones, Scodes),
    string_codes(Jewels, Jcodes),
    msort(Scodes, SScodes),
    sort(Jcodes, SJcodes),
    count_jewels(SScodes, SJcodes, N, 0).

count_jewels([], _, N, N):-!.
count_jewels(_, [], N, N):-!.
count_jewels([C|Stones], [C|Jewels], N, R):-
    !,
    R1 is R + 1,
    count_jewels(Stones, [C|Jewels], N, R1).
count_jewels([S|Stones], [J|Jewels], N, R):-
    J < S,
    !,
    count_jewels([S|Stones], Jewels, N, R).
count_jewels([_|Stones], Jewels, N, R):-
    count_jewels(Stones, Jewels, N, R).
Output:
Welcome to SWI-Prolog (threaded, 64 bits, version 8.0.2)
SWI-Prolog comes with ABSOLUTELY NO WARRANTY. This is free software.
Please run ?- license. for legal details.

For online help and background, visit http://www.swi-prolog.org
For built-in help, use ?- help(Topic). or ?- apropos(Word).

?- count_jewels("aAAbbbb", "aA", N).
N = 3.

Python[edit]

def countJewels(s, j):
    return sum(x in j for x in s)

print countJewels("aAAbbbb", "aA")
print countJewels("ZZ", "z")
Output:
3
0

Python 3 Alternative[edit]

def countJewels(stones, jewels):
    jewelset = set(jewels)
    return sum(1 for stone in stones if stone in jewelset)

print(countJewels("aAAbbbb", "aA"))
print(countJewels("ZZ", "z"))
Output:
3
0

R[edit]

J_n_S <- function(stones  ="aAAbbbb", jewels = "aA") {
  stones <- unlist(strsplit(stones, split = "")) # obtain a character vector
  jewels <- unlist(strsplit(jewels, split = ""))
  count <- sum(stones %in% jewels)
}

print(J_n_S("aAAbbbb", "aA"))
print(J_n_S("ZZ", "z"))
print(J_n_S("lgGKJGljglghGLGHlhglghoIPOgfdtrdDCHnvbnmBVC", "fFgGhH"))
Output:
> print(J_n_S("aAAbbbb", "aA"))
[1] 3
> print(J_n_S("ZZ", "z"))
[1] 0
> print(J_n_S("lgGKJGljglghGLGHlhglghoIPOgfdtrdDCHnvbnmBVC", "fFgGhH"))
[1] 16

Quackery[edit]

  [ 0 0 rot 
    witheach [ bit | ] 
    rot witheach
      [ bit over & if 
          [ dip 1+ ] ] 
    drop ]             is j&s ( $ $ --> n )

  $ "aAAbbbb" $  "aA" j&s echo
Output:
3

Racket[edit]

#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"))
Output:
3
0

Raku[edit]

(formerly Perl 6)

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';
Output:
3
0

Red[edit]

Red [
    title: "Jewels and stones"
    red-version: 0.6.4
]

count: function [
    "Returns the number of values in a block for which a function returns true"
    values [any-list! string!] "The values from which to count"
    fn [function!] "A function that returns true or false"
][
    count: 0
    foreach value values [if fn value [count: count + 1]]
    count
]

count-jewels: function [
    "Returns the count of 'jewels' in the 'stones'"
    stones "The values to search for jewels"
    jewels "The values to count in the stones"
][
    result: 0
    foreach jewel jewels [
        result: result + count stones function [stone][stone = jewel]
    ]
]

print count-jewels "aAAbbbb" "aA"
print count-jewels "ZZ" "z"
Output:
3
0

REXX[edit]

Programming note:   a check is made so that only (Latin) letters are counted as a match.

/*REXX pgm counts how many letters (in the 1st string) are in common with the 2nd string*/
say  count('aAAbbbb', "aA")
say  count('ZZ'     , "z" )
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
count: procedure;  parse arg stones,jewels       /*obtain the two strings specified.    */
       #= 0                                      /*initialize 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')  then if pos(x, jewels)\==0  then #= # + 1
                   end   /*j*/                   /* [↑]  if a letter and a match, bump #*/
       return #                                  /*return the number of common letters. */
output   when using the default inputs:
3
0

Ring[edit]

# Project  Jewels and Stones
 
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

Output:

3
0

Ruby[edit]

stones, jewels = "aAAbbbb", "aA"

stones.count(jewels)  # => 3

Rust[edit]

fn count_jewels(stones: &str, jewels: &str) -> u8 {
    let mut count: u8 = 0;
    for cur_char in stones.chars() {
        if jewels.contains(cur_char) {
            count += 1;
        }
    }
    count
}
fn main() {
    println!("{}", count_jewels("aAAbbbb", "aA"));
    println!("{}", count_jewels("ZZ", "z"));
}
Output:
3
0

Scala[edit]

object JewelsStones extends App {
  def countJewels(s: String, j: String): Int = s.count(i => j.contains(i))

  println(countJewels("aAAbbbb", "aA"))
  println(countJewels("ZZ", "z"))
}
Output:
See it in running in your browser by ScalaFiddle (JavaScript) or by Scastie (JVM).

Sidef[edit]

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

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

Snobol[edit]

*       See how many jewels are among the stones
        DEFINE('JEWELS(JWL,STN)')               :(JEWELS_END)
JEWELS  JEWELS = 0
        JWL = ANY(JWL)
JMATCH  STN JWL = ''                            :F(RETURN)
        JEWELS = JEWELS + 1                     :(JMATCH)  
JEWELS_END
                                                
*       Example from the task (prints 3)
        OUTPUT = JEWELS('aA','aAAbbbb')
*       Example with no jewels (prints 0)
        OUTPUT = JEWELS('z','ZZ')
END
Output:
3
0

SQL[edit]

-- See how many jewels are among the stones
declare @S varchar(1024) 	=	'AaBbCcAa'
,	@J varchar(1024)	=	'aA';

declare @SLEN	int = len(@S);
declare @JLEN	int = len(@J);
declare @TCNT	int = 0;

declare @SPOS	int = 1;	-- curr position in @S
declare @JPOS	int = 1;	-- curr position in @J
declare @FCHR	char(1);	-- char to find

while @JPOS <= @JLEN
begin

	set @FCHR = substring(@J, @JPOS, 1);

	set @SPOS = 1;

	while @SPOS > 0 and @SPOS <= @SLEN
	begin
		
		set @SPOS = charindex(@FCHR, @S COLLATE Latin1_General_CS_AS, @SPOS);
		
		if @SPOS > 0 begin
			set @TCNT = @TCNT + 1;
			set @SPOS = @SPOS + 1;
		end
	end

	set @JPOS = @JPOS + 1;
end
print 'J='+@J+' S='+@S+' TOTAL = '+cast(@TCNT as varchar(8));
Output:
J=aA S=AaBbCcAa TOTAL = 4

Swift[edit]

func countJewels(_ stones: String, _ jewels: String) -> Int {
  return stones.map({ jewels.contains($0) ? 1 : 0 }).reduce(0, +)
}

print(countJewels("aAAbbbb", "aA"))
print(countJewels("ZZ", "z"))
Output:
3
0

Tcl[edit]

proc shavej {stones jewels} {
    set n 0
    foreach c [split $stones {}] {
        incr n [expr { [string first $c $jewels] >= 0 }]
    }
    return $n
}
puts [shavej aAAbbbb aA]
puts [shavej ZZ z]
Output:
3                                                                                       
0

Terraform[edit]

variable "jewels" {
  default = "aA"
}

variable "stones" {
  default = "aAAbbbb"
}

locals {
   jewel_list = split("", var.jewels)
   stone_list = split("", var.stones)
   found_jewels = [for s in local.stone_list: s if contains(local.jewel_list, s)]
}

output "jewel_count" {
  value = length(local.found_jewels)
}
Output:
$ terraform apply --auto-approve

Apply complete! Resources: 0 added, 0 changed, 0 destroyed.

Outputs:

jewel_count = 3

$ TF_VAR_jewels=z TF_VAR_stones=ZZ terraform apply --auto-approve

Apply complete! Resources: 0 added, 0 changed, 0 destroyed.

Outputs:

jewel_count = 0

Transd[edit]

#lang transd

MainModule: {
    countJewels: (λ j String() st String() locals: n 0
        (for s in st do 
            (if (contains j s) (+= n 1)))
        (ret n)
    ),
    _start: (λ (lout (countJewels "aA" "aAAbbbb"))
               (lout (countJewels "b" "BB")))
}
Output:
3
0

VBA[edit]

Translation of: Phix
Function count_jewels(stones As String, jewels As String) As Integer
    Dim res As Integer: res = 0
    For i = 1 To Len(stones)
        res = res - (InStr(1, jewels, Mid(stones, i, 1), vbBinaryCompare) <> 0)
    Next i
    count_jewels = res
End Function
Public Sub main()
    Debug.Print count_jewels("aAAbbbb", "aA")
    Debug.Print count_jewels("ZZ", "z")
End Sub
Output:
 3 
 0 

Visual Basic .NET[edit]

Translation of: C#
Module Module1

    Function Count(stones As String, jewels As String) As Integer
        Dim bag = jewels.ToHashSet
        Return stones.Count(AddressOf bag.Contains)
    End Function

    Sub Main()
        Console.WriteLine(Count("aAAbbbb", "Aa"))
        Console.WriteLine(Count("ZZ", "z"))
    End Sub

End Module
Output:
3
0

V (Vlang)[edit]

Translation of: Go
fn js(stones string, jewels string) int {
	mut n := 0
    for b in stones.bytes() {
        if jewels.index_u8(b) >= 0 {
            n++
        }
    }
    return n
}
 
fn main() {
    println(js("aAAbbbb", "aA"))
}
Output:
3

Wren[edit]

Translation of: Kotlin
var countJewels = Fn.new { |s, j| s.count { |c| j.contains(c) } }

System.print(countJewels.call("aAAbbbb", "aA"))
System.print(countJewels.call("ZZ", "z"))
Output:
3
0

XPL0[edit]

string 0;       \Use zero-terminated strings

func Count(Stones, Jewels);
\Return number of letters in Stones that match letters in Jewels
char Stones, Jewels;
int  Cnt, I, J;
[Cnt:= 0;
I:= 0;
while Jewels(I) do
    [J:= 0;
    while Stones(J) do
        [if Stones(J) = Jewels(I) then Cnt:= Cnt+1;
        J:= J+1;
        ];
    I:= I+1;
    ];
return Cnt;
];

[IntOut(0, Count("aAAbbbb", "aA"));  CrLf(0);
 IntOut(0, Count("ZZ", "z"));  CrLf(0);
 IntOut(0, Count("pack my box with five dozen liquor jugs", "aeiou"));  CrLf(0);
]
Output:
3
0
11

zkl[edit]

fcn countJewels(a,b){ a.inCommon(b).len() }
println(countJewels("aAAbbbb", "aA"));
println(countJewels("ZZ", "z"));
Output:
3
0