# Jewels and Stones

Jewels and Stones
You are encouraged to solve this task according to the task description, using any language you may know.

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 to 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.

## ALGOL 68

`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( "[email protected]"                         , "[email protected]"                         )           , newline           )         );    print( ( count jewels( "AB", "" ), newline ) );    print( ( count jewels( "ZZ", "z" ), newline ) ) END`
Output:
```         +3
+52
+0
+0```

## AppleScript

`-- jewelCount :: String -> String -> Inton 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 -> Inton 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 send chars -- elem :: Eq a => a -> [a] -> Boolon elem(x, xs)    considering case        xs contains x    end consideringend 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 tellend filter -- foldl :: (a -> b -> a) -> a -> [b] -> aon 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 tellend 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 tellend 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 ifend 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 scriptend uncurry -- unlines :: [String] -> Stringon 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    strend unlines`
Output:
```3
0```

## AutoHotkey

`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

`# syntax: GAWK -f JEWELS_AND_STONES.AWKBEGIN {    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```

## C

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++

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```

## C#

`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
```

## D

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```

## Factor

`USING: kernel prettyprint sequences ;: count-jewels ( stones jewels -- n ) [ member? ] curry count ; "aAAbbbb" "aA""ZZ" "z" [ count-jewels . ] [email protected]`
Output:
```3
0
```

## Go

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}`

`jewelCount :: String -> String -> Int jewelCount jewels =   foldr (\c -> if elem c jewels then succ else id) 0  -- TEST ---------------------------------------------- main :: IO ()main = mapM_ print \$   (uncurry jewelCount) <\$> [       ("aA", "aAAbbbb")      ,("z", "ZZ")    ]`
Output:
```3
0```

Or in terms of filter rather than foldr

`jewelCount :: String -> String -> Int jewelCount jewels =   length . filter (flip elem jewels) -- Which could be further reduced to-- jewelCount = (length .) . filter . flip elem -- TEST ---------------------------------------------- main :: IO ()main = do  print \$ jewelCount "aA" "aAAbbbb"  print \$ jewelCount "z" "ZZ"`
Output:
```3
0```

## J

`    NB. jewels sums a raveled equality table   NB. use: x jewels y  x are the stones, y are the jewels.   intersect =: -.^:2   jewels =: ([: +/ [: , =/~) [email protected]:intersect&Alpha_j_    'aAAbbbb ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz' jewels&>&;: 'aA ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz'3 52    'none' jewels ''0   'ZZ' jewels 'z'0  `

## Java

`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

`(() => {     // 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]`

## Julia

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

`// 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
```

## Lua

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

## Maple

`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`

## Modula-2

`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 cEND CountJewels; BEGIN    WriteInt(CountJewels("aAAbbbb", "aA"));    WriteLn;    WriteInt(CountJewels("ZZ", "z"));    WriteLn;     ReadCharEND Jewels.`
Output:
```3
0```

## Perl

`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```

## 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```

## Phix

`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 resend function?count_jewels("aAAbbbb","aA")?count_jewels("ZZ","z")`
Output:
```3
0
```

## Python

`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

`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```

## Racket

`#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```

## REXX

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

`# Project  Jewels and Stones jewels = "aA"stones = "aAAbbbb"see jewelsandstones(jewels,stones) + nljewels = "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

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

## Scala

`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

`func countJewels(s, j) {    s.chars.count { |c|        j.contains(c)    }} say countJewels("aAAbbbb", "aA")    #=> 3say countJewels("ZZ", "z")          #=> 0`

## Swift

`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```

## VBA

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 = resEnd FunctionPublic Sub main()    Debug.Print count_jewels("aAAbbbb", "aA")    Debug.Print count_jewels("ZZ", "z")End Sub`
Output:
``` 3
0 ```

## zkl

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