Jewels and stones: Difference between revisions
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=={{header|Tcl}}== |
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<lang Tcl>proc shavej {stones jewels} { |
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set n 0 |
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foreach c [split $stones {}] { |
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incr n [expr { [string first $c $jewels] >= 0 }] |
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} |
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return $n |
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} |
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puts [shavej aAAbbbb aA] |
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puts [shavej ZZ z]</lang> |
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{{out}} |
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3 |
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0 |
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=={{header|VBA}}== |
=={{header|VBA}}== |
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Revision as of 23:54, 24 April 2020
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:
- Only letters in the ISO basic Latin alphabet i.e. 'A to Z' or 'a to z' need be considered.
- A lower case letter is considered to be different to its upper case equivalent for this purpose i.e. 'a' != 'A'.
- The parameters do not need to have exactly the same names.
- 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.
Ada
<lang Ada>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;</lang>
- Output:
aAAbbbb aA 3 ZZ z 0
ALGOL 68
<lang algol68>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</lang>
- Output:
+3
+52
+0
+0
AppleScript
<lang applescript>-- 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</lang>
- Output:
3 0
AutoHotkey
<lang AutoHotkey>JewelsandStones(ss, jj){ for each, jewel in StrSplit(jj) for each, stone in StrSplit(ss) if (stone == jewel) num++ return num }</lang> Example:<lang AutoHotkey>MsgBox % JewelsandStones("aAAbbbbz", "aAZ") return</lang>
Outputs:
3
AWK
<lang AWK># 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
BASIC256
<lang BASIC256> 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", "") </lang>
- Output:
Igual que la entrada de FreeBASIC.
C
<lang c>#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;
}</lang>
- Output:
3 0
C#
<lang csharp>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);
}
}</lang>
- Output:
3 0
C++
<lang cpp>#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;
}</lang>
- Output:
3 0
D
<lang d>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;
}</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
FreeBASIC
<lang freebasic> 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", "") </lang>
- Output:
3 0 53 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: <lang go>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"))
}</lang>
- Output:
3
Outer loop jewels, count stones: <lang go>func js(stones, jewels string) (n int) {
for _, b := range []byte(jewels) {
n += strings.Count(stones, string(b))
}
return
}</lang>
Construct jewel set, then loop over stones: <lang go>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
}</lang>
Construct stone multiset, then loop over jewels: <lang go>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
}</lang>
Haskell
<lang haskell>jewelCount :: Eq a => [a] -> [a] -> 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")
]</lang>
- Output:
3 0
Or in terms of filter rather than foldr
<lang haskell>jewelCount :: Eq a => [a] -> [a] -> 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"</lang>
- Output:
3 0
J
<lang J>
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
</lang>
Java
<lang 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"));
}
}</lang>
- Output:
3 0
JavaScript
<lang 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))
})();</lang>
- Output:
[3, 0]
Julia
Module: <lang julia>module Jewels
count(s, j) = Base.count(x ∈ j for x in s)
end # module Jewels</lang>
Main: <lang julia>@show Jewels.count("aAAbbbb", "aA") @show Jewels.count("ZZ", "z")</lang>
- Output:
Jewels.count("aAAbbbb", "aA") = 3
Jewels.count("ZZ", "z") = 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
Lua
<lang lua>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"))</lang>
- Output:
3 0
Maple
<lang 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")</lang>
- Output:
3
Modula-2
<lang modula2>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.</lang>
- Output:
3 0
Objeck
<lang Objeck>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;
}
}</lang>
- Output:
3 0
Perl
<lang 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';</lang>
- Output:
3 0
Phix
<lang 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 res
end function ?count_jewels("aAAbbbb","aA") ?count_jewels("ZZ","z")</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
Python 3 Alternative
<lang python>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"))</lang>
- Output:
3 0
Racket
<lang 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"))
</lang>
- Output:
3 0
Raku
(formerly 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
REXX
Programming note: a check is made so that only (Latin) letters are counted as a match. <lang rexx>/*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. */</lang>
- output when using the default inputs:
3 0
Ring
<lang ring># 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
</lang> Output:
3 0
Ruby
<lang ruby>stones, jewels = "aAAbbbb", "aA"
stones.count(jewels) # => 3 </lang>
Rust
<lang rust>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"));
} </lang>
Output:
3 0
Scala
<lang 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"))
}</lang>
- Output:
See it in running in your browser by ScalaFiddle (JavaScript) or by Scastie (JVM).
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>
Swift
<lang 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"))</lang>
- Output:
3 0
Tcl
<lang Tcl>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]</lang>
- Output:
3 0
VBA
<lang vb>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</lang>
- Output:
3 0
Visual Basic .NET
<lang vbnet>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</lang>
- Output:
3 0
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