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Unique characters

From Rosetta Code
Unique characters 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

Given a list of strings,   find characters appearing only in one string and once only.

The result should be given in alphabetical order.


Use the following list for this task:

        ["133252abcdeeffd",  "a6789798st",  "yxcdfgxcyz"]


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



8080 Assembly[edit]

puts:	equ	9		; CP/M print syscall
TERM: equ '$' ; CP/M string terminator
org 100h
jmp demo
;;; Given a list of strings, find characters appearing only
;;; in one string and once only.
;;; Input: DE = list of strings, BC = start of output
unique: xra a ; Zero out the workspace
mvi h,upage
mov l,a
uzbuf: mov m,a
inr l
jnz uzbuf
push d
ustr: pop h
mov e,m ; Load next string pointer
inx h
mov d,m
inx h
mov a,d ; End of list?
ora e
jz uclat ; Then go find the uniques
push h ; Otherwise, keep list pointer
mvi h,upage
uchar: ldax d ; Get character
cpi TERM ; Done?
jz ustr ; Then do next string
mov l,a ; Otherwise, count the character
inr m
inx d ; Next character
jmp uchar
uclat: lxi h,upage*256 ; Start of page
utst: dcr m ; Is this character included?
jnz uskip
mov a,l ; If so add it to the output
stax b
inx b
uskip: inr l ; Try next character
jnz utst
mvi a,TERM ; CP/M string terminator
stax b
ret
;;; Demo code
demo: lxi b,outbuf ; Set BC to location of output buffe
lxi d,list ; Set DE to the list of strings
call unique ; Call the code
mvi c,puts ; Print the result
lxi d,outbuf
jmp 5
;;; List of strings
list: dw str1, str2, str3, 0
str1: db '133252abcdeeffd', TERM
str2: db 'a6789798st', TERM
str3: db 'yxcdfgxcyz', TERM
;;; Memory
upage: equ ($/256)+1 ; Workspace for 'unique'
outbuf: equ (upage+1)*256 ; Output
Output:
156bgstz


8086 Assembly[edit]

	cpu	8086
org 100h
puts: equ 9 ; MS-DOS syscall to print a string
TERM: equ '$' ; String terminator
section .text
jmp demo
;;; Given a list of strings, find characters appearing
;;; only in one string and once only.
;;; Input: BX = list of strings, DX = start of output
;;; Assuming DS = ES
unique: mov cx,128 ; Zero out array
mov di,uniqws
xor ax,ax
rep stosw
.str: mov si,[bx] ; Get next string
inc bx
inc bx
test si,si ; Done?
jz .fltr ; Then start writing to output
.char: lodsb ; Read character
cmp al,TERM ; Done?
je .str ; Then get next string
mov di,ax ; Otherwise, count the character
inc byte [di+uniqws]
jmp .char
.fltr: mov di,dx
xor bx,bx
.fchr: dec byte [bx+uniqws]
jnz .skip ; Character occurs once?
mov al,bl ; If so, write it
stosb
.skip: inc bl ; Any more?
jnz .fchr
mov [di],byte TERM ; Terminate string
ret
;;; Demo code
demo: mov bx,list ; Find unique characters
mov dx,outbuf
call unique
mov ah,puts ; Print result
mov dx,outbuf
int 21h
ret
section .data
list: dw .s1, .s2, .s3, 0
.s1: db '133252abcdeeffd', TERM
.s2: db 'a6789798st', TERM
.s3: db 'yxcdfgxcyz', TERM
section .bss
uniqws: resb 256
outbuf: resb 256
Output:
156bgstz

Ada[edit]

with Ada.Text_Io;
 
procedure Unique_Characters is
 
List : array (Character) of Natural := (others => 0);
 
procedure Count (Item : String) is
begin
for C of Item loop
List (C) := List (C) + 1;
end loop;
end Count;
 
procedure Put_Only_Once is
use Ada.Text_Io;
begin
for C in List'Range loop
if List (C) = 1 then
Put (C);
Put (' ');
end if;
end loop;
New_Line;
end Put_Only_Once;
 
begin
Count ("133252abcdeeffd");
Count ("a6789798st");
Count ("yxcdfgxcyz");
Put_Only_Once;
end Unique_Characters;
Output:
1 5 6 b g s t z

ALGOL 68[edit]

Case sensitive. This assumes a small character set (e.g. ASCII where max abs char is 255). Would probably need some work if CHAR is Unicode.

BEGIN # find the characters that occur once only in a list of stings        #
# returns the characters that occur only once in the elements of s #
OP UNIQUE = ( []STRING s )STRING:
BEGIN
[ 0 : max abs char ]INT counts; # counts of used characters #
FOR i FROM LWB counts TO UPB counts DO counts[ i ] := 0 OD;
# count the occurances of the characters in the elements of s #
FOR i FROM LWB s TO UPB s DO
FOR j FROM LWB s[ i ] TO UPB s[ i ] DO
counts[ ABS s[ i ][ j ] ] +:= 1
OD
OD;
# construct a string of the characters that occur only once #
STRING result := "";
FOR i FROM LWB counts TO UPB counts DO
IF counts[ i ] = 1 THEN result +:= REPR i FI
OD;
result
END; # UNIQUE #
# task test case #
print( ( UNIQUE []STRING( "133252abcdeeffd", "a6789798st", "yxcdfgxcyz" ), newline ) )
END
Output:
156bgstz

AppleScript[edit]

AppleScriptObjC[edit]

The filtering here is case sensitive, the sorting dependent on locale.

on uniqueCharacters(listOfStrings)
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to ""
set countedSet to current application's class "NSCountedSet"'s setWithArray:((listOfStrings as text)'s characters)
set AppleScript's text item delimiters to astid
set mutableSet to current application's class "NSMutableSet"'s setWithSet:(countedSet)
tell countedSet to minusSet:(mutableSet)
tell mutableSet to minusSet:(countedSet)
set sortDescriptor to current application's class "NSSortDescriptor"'s sortDescriptorWithKey:("self") ¬
ascending:(true) selector:("localizedStandardCompare:")
 
return (mutableSet's sortedArrayUsingDescriptors:({sortDescriptor})) as list
end uniqueCharacters
Output:
{"1", "5", "6", "b", "g", "s", "t", "z"}

Core language only[edit]

This isn't quite as fast as the ASObjC solution above, but it can be case-insensitive if required. (Simply leave out the 'considering case' statement round the call to the handler). The requirement for AppleScript 2.3.1 is just for the 'use' command which loads the "Heap Sort" script. If "Heap Sort"'s loaded differently or compiled directly into the code, this script will work on systems at least as far back as Mac OS X 10.5 (Leopard) and possibly earlier. Same output as above.

use AppleScript version "2.3.1" -- OS X 10.9 (Mavericks) or later
use sorter : script "Heap Sort" -- <https://www.rosettacode.org/wiki/Sorting_algorithms/Heapsort#AppleScript>
 
on uniqueCharacters(listOfStrings)
script o
property allCharacters : {}
property uniques : {}
end script
 
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to ""
set o's allCharacters to text items of (listOfStrings as text)
set AppleScript's text item delimiters to astid
 
set characterCount to (count o's allCharacters)
tell sorter to sort(o's allCharacters, 1, characterCount)
 
set i to 1
set currentCharacter to beginning of o's allCharacters
repeat with j from 2 to characterCount
set thisCharacter to item j of o's allCharacters
if (thisCharacter is not currentCharacter) then
if (j - i = 1) then set end of o's uniques to currentCharacter
set i to j
set currentCharacter to thisCharacter
end if
end repeat
if (i = j) then set end of o's uniques to currentCharacter
 
return o's uniques
end uniqueCharacters
 
considering case
return uniqueCharacters({"133252abcdeeffd", "a6789798st", "yxcdfgxcyz"})
end considering


Functional[edit]

Composing a solution from existing generic primitives, for speed of drafting and refactoring, and for high levels of code reuse.

use framework "Foundation"
 
 
-------------------- UNIQUE CHARACTERS -------------------
 
-- uniques :: [String] -> String
on uniques(xs)
script single
on |λ|(x)
if 1 = length of x then
item 1 of x
else
{}
end if
end |λ|
end script
 
concatMap(single, ¬
group(sort(concatMap(my chars, xs))))
end uniques
 
 
 
--------------------------- TEST -------------------------
on run
uniques({"133252abcdeeffd", "a6789798st", "yxcdfgxcyz"})
 
--> {"1", "5", "6", "b", "g", "s", "t", "z"}
end run
 
 
------------------------- GENERIC ------------------------
 
-- chars :: String -> [Char]
on chars(s)
characters of s
end chars
 
 
-- concatMap :: (a -> [b]) -> [a] -> [b]
on concatMap(f, xs)
set lng to length of xs
set acc to {}
tell mReturn(f)
repeat with i from 1 to lng
set acc to acc & (|λ|(item i of xs, i, xs))
end repeat
end tell
if {text, string} contains class of xs then
acc as text
else
acc
end if
end concatMap
 
 
-- eq (==) :: Eq a => a -> a -> Bool
on eq(a, b)
a = b
end eq
 
 
-- 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
 
 
-- group :: Eq a => [a] -> [[a]]
on group(xs)
script eq
on |λ|(a, b)
a = b
end |λ|
end script
 
groupBy(eq, xs)
end group
 
 
-- groupBy :: (a -> a -> Bool) -> [a] -> [[a]]
on groupBy(f, xs)
-- Typical usage: groupBy(on(eq, f), xs)
set mf to mReturn(f)
 
script enGroup
on |λ|(a, x)
if length of (active of a) > 0 then
set h to item 1 of active of a
else
set h to missing value
end if
 
if h is not missing value and mf's |λ|(h, x) then
{active:(active of a) & {x}, sofar:sofar of a}
else
{active:{x}, sofar:(sofar of a) & {active of a}}
end if
end |λ|
end script
 
if length of xs > 0 then
set dct to foldl(enGroup, {active:{item 1 of xs}, sofar:{}}, rest of xs)
if length of (active of dct) > 0 then
sofar of dct & {active of dct}
else
sofar of dct
end if
else
{}
end if
end groupBy
 
 
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
-- The list obtained by applying f
-- to each element of 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
 
 
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
-- 2nd class handler function lifted into 1st class script wrapper.
if script is class of f then
f
else
script
property |λ| : f
end script
end if
end mReturn
 
 
-- sort :: Ord a => [a] -> [a]
on sort(xs)
((current application's NSArray's arrayWithArray:xs)'s ¬
sortedArrayUsingSelector:"compare:") as list
end sort
Output:
{"1", "5", "6", "b", "g", "s", "t", "z"}

APL[edit]

uniques ← (⊂∘⍋⌷⊣)∘(∪(/⍨)(1=(≢⊢))⌸)∘∊
Output:
      uniques '133252abcdeeffd' 'a6789798st' 'yxcdfgxcyz'
156bgstz

Arturo[edit]

arr: ["133252abcdeeffd" "a6789798st" "yxcdfgxcyz"]
str: join arr
 
print sort select split str 'ch -> 1 = size match str ch
Output:
1 5 6 b g s t z

AWK[edit]

 
# syntax: GAWK -f UNIQUE_CHARACTERS.AWK
#
# sorting:
# PROCINFO["sorted_in"] is used by GAWK
# SORTTYPE is used by Thompson Automation's TAWK
#
BEGIN {
PROCINFO["sorted_in"] = "@ind_str_asc" ; SORTTYPE = 1
n = split("133252abcdeeffd,a6789798st,yxcdfgxcyz",arr1,",")
for (i=1; i<=n; i++) {
str = arr1[i]
printf("%s\n",str)
total_c += leng = length(str)
for (j=1; j<=leng; j++) {
arr2[substr(str,j,1)]++
}
}
for (c in arr2) {
if (arr2[c] == 1) {
rec = sprintf("%s%s",rec,c)
}
}
printf("%d strings, %d characters, %d different, %d unique: %s\n",n,total_c,length(arr2),length(rec),rec)
exit(0)
}
 
Output:
133252abcdeeffd
a6789798st
yxcdfgxcyz
3 strings, 35 characters, 20 different, 8 unique: 156bgstz

BASIC[edit]

10 DEFINT A-Z
20 DIM C(255)
30 READ A$: IF A$="" GOTO 90
40 FOR I=1 TO LEN(A$)
50 A=ASC(MID$(A$,I,1))
60 C(A)=C(A)+1
70 NEXT I
80 GOTO 30
90 FOR I=1 TO 255
100 IF C(I)=1 THEN A$=A$+CHR$(I)
110 NEXT I
120 PRINT A$
130 DATA "133252abcdeeffd"
140 DATA "a6789798st"
150 DATA "yxcdfgxcyz"
160 DATA ""
Output:
156bgstz

BCPL[edit]

get "libhdr"
 
let uniques(strings, out) be
$( let counts = vec 255
for i=0 to 255 do counts!i := 0
until !strings = 0
$( let string = !strings
strings := strings + 1
for i=1 to string%0 do
counts!(string%i) := counts!(string%i) + 1
$)
out%0 := 0
for i=0 to 255 if counts!i = 1
$( out%0 := out%0 + 1
out%(out%0) := i
$)
$)
 
let start() be
$( let strings = vec 3 and out = vec 1+255/BYTESPERWORD
 
strings!0 := "133252abcdeeffd"
strings!1 := "a6789798st"
strings!2 := "yxcdfgxcyz"
strings!3 := 0
 
uniques(strings, out)
writef("%S*N", out)
$)
Output:
156bgstz

BQN[edit]

Uniq ← (⍷/˜1=/⁼∘⊐)∧∘∾
Output:
   Uniq "133252abcdeeffd"‿"a6789798st"‿"yxcdfgxcyz"
"156bgstz"

C[edit]

#include <stdio.h>
#include <string.h>
 
char *uniques(char *str[], char *buf) {
static unsigned counts[256];
unsigned i;
char *s, *o = buf;
memset(counts, 0, 256 * sizeof(unsigned));
 
for (; *str; str++)
for (s = *str; *s; s++)
counts[(unsigned) *s]++;
 
for (i=0; i<256; i++)
if (counts[i] == 1)
*o++ = (char) i;
 
*o = '\0';
return buf;
}
 
int main() {
char buf[256];
char *strings[] = {
"133252abcdeeffd",
"a6789798st",
"yxcdfgxcyz",
NULL
};
 
printf("%s\n", uniques(strings, buf));
return 0;
}
Output:
156bgstz

C++[edit]

#include <iostream>
#include <map>
 
int main() {
const char* strings[] = {"133252abcdeeffd", "a6789798st", "yxcdfgxcyz"};
std::map<char, int> count;
for (const char* str : strings) {
for (; *str; ++str)
++count[*str];
}
for (const auto& p : count) {
if (p.second == 1)
std::cout << p.first;
}
std::cout << '\n';
}
Output:
156bgstz

Factor[edit]

Works with: Factor version 0.99 build 2074
USING: io sequences sets.extras sorting ;
 
{ "133252abcdeeffd" "a6789798st" "yxcdfgxcyz" }
concat non-repeating natural-sort print
Output:
156bgstz

Go[edit]

package main
 
import (
"fmt"
"sort"
)
 
func main() {
strings := []string{"133252abcdeeffd", "a6789798st", "yxcdfgxcyz"}
m := make(map[rune]int)
for _, s := range strings {
for _, c := range s {
m[c]++
}
}
var chars []rune
for k, v := range m {
if v == 1 {
chars = append(chars, k)
}
}
sort.Slice(chars, func(i, j int) bool { return chars[i] < chars[j] })
fmt.Println(string(chars))
}
Output:
156bgstz

Haskell[edit]

import Data.List (group, sort)
 
uniques :: [String] -> String
uniques ks =
[c | (c : cs) <- (group . sort . concat) ks, null cs]
 
main :: IO ()
main =
putStrLn $
uniques
[ "133252abcdeeffd",
"a6789798st",
"yxcdfgxcyz"
]
Output:
156bgstz


Or folding the strings down to a hash of character frequencies:

import qualified Data.Map.Strict as M
 
--------- UNIQUE CHARACTERS FROM A LIST OF STRINGS -------
 
uniqueChars :: [String] -> String
uniqueChars =
M.keys . M.filter (1 ==)
. foldr (M.unionWith (+) . charCounts) M.empty
 
charCounts :: String -> M.Map Char Int
charCounts =
foldr
(flip (M.insertWith (+)) 1)
M.empty
 
--------------------------- TEST -------------------------
main :: IO ()
main =
putStrLn $
uniqueChars
[ "133252abcdeeffd",
"a6789798st",
"yxcdfgxcyz"
]
Output:
156bgstz

JavaScript[edit]

(() => {
"use strict";
 
// ---------------- UNIQUE CHARACTERS ----------------
 
// uniques :: [String] -> [Char]
const uniques = xs =>
group(
xs.flatMap(x => [...x])
.sort()
)
.flatMap(
x => 1 === x.length ? (
x
) : []
);
 
// ---------------------- TEST -----------------------
// main :: IO ()
const main = () =>
uniques([
"133252abcdeeffd",
"a6789798st",
"yxcdfgxcyz"
]);
 
 
// --------------------- GENERIC ---------------------
 
// group :: Eq a => [a] -> [[a]]
const group = xs => {
// A list of lists, each containing only equal elements,
// such that the concatenation of these lists is xs.
const go = ys =>
0 < ys.length ? (() => {
const
h = ys[0],
i = ys.findIndex(y => h !== y);
 
return i !== -1 ? (
[ys.slice(0, i)].concat(go(ys.slice(i)))
) : [ys];
})() : [];
 
return go(xs);
};
 
 
// MAIN ---
return JSON.stringify(main());
})();
Output:
["1","5","6","b","g","s","t","z"]


Or, folding the strings (with Array.reduce) down to a hash of character frequencies:

(() => {
"use strict";
 
// uniqueChars :: [String] -> [Char]
const uniqueChars = ws =>
Object.entries(
ws.reduce(
(dict, w) => [...w].reduce(
(a, c) => Object.assign({}, a, {
[c]: 1 + (a[c] || 0)
}),
dict
), {}
)
)
.flatMap(
([k, v]) => 1 === v ? (
[k]
) : []
);
 
// ---------------------- TEST -----------------------
const main = () =>
uniqueChars([
"133252abcdeeffd", "a6789798st", "yxcdfgxcyz"
]);
 
 
return JSON.stringify(main());
})();
Output:
["1","5","6","b","s","t","g","z"]

J[edit]

uniques =: [:/:[email protected];#~1=+/"1@[email protected];
Output:
   uniques '133252abcdeeffd';'a6789798st';'yxcdfgxcyz'
156bgstz

jq[edit]

Works with: jq

Works with gojq, the Go implementation of jq

The following "bag-of-words" solution is quite efficient as it takes advantage of the fact that jq implements JSON objects as a hash.
 
# bag of words
def bow(stream):
reduce stream as $word ({}; .[($word|tostring)] += 1);
 
# Input: an array of strings
# Output: an array of the characters that appear just once
def in_one_just_once:
bow( .[] | explode[] | [.] | implode) | with_entries(select(.value==1)) | keys;
 

The task

["133252abcdeeffd", "a6789798st", "yxcdfgxcyz"]
| in_one_just_once
Output:
["1","5","6","b","g","s","t","z"]

Julia[edit]

list = ["133252abcdeeffd", "a6789798st", "yxcdfgxcyz"]
 
function is_once_per_all_strings_in(a::Vector{String})
charlist = collect(prod(a))
counts = Dict(c => count(x -> c == x, charlist) for c in unique(charlist))
return sort([p[1] for p in counts if p[2] == 1])
end
 
println(is_once_per_all_strings_in(list))
 
Output:

['1', '5', '6', 'b', 'g', 's', 't', 'z']

One might think that the method above suffers from too many passes through the text with one pass per count, but with a small text length the dictionary lookup takes more time. Compare times for a single pass version:

function uniquein(a)
counts = Dict{Char, Int}()
for c in prod(list)
counts[c] = get!(counts, c, 0) + 1
end
return sort([c for (c, n) in counts if n == 1])
end
 
println(uniquein(list))
 
using BenchmarkTools
@btime is_once_per_all_strings_in(list)
@btime uniquein(list)
 
Output:

['1', '5', '6', 'b', 'g', 's', 't', 'z']

 1.740 μs (28 allocations: 3.08 KiB)
 3.763 μs (50 allocations: 3.25 KiB)

This can be rectified (see Phix entry) if we don't save the counts as we go but just exclude entries with duplicates:

function uniquein2(a)
s = sort(collect(prod(list)))
l = length(s)
return [p[2] for p in enumerate(s) if (p[1] == 1 || p[2] != s[p[1] - 1]) && (p[1] == l || p[2] != s[p[1] + 1])]
end
 
println(uniquein2(list))
 
@btime uniquein2(list)
 
Output:

['1', '5', '6', 'b', 'g', 's', 't', 'z']

 1.010 μs (14 allocations: 1.05 KiB)

Mathematica/Wolfram Language[edit]

Select[Tally[Sort[Characters[StringJoin[{"133252abcdeeffd", "a6789798st", "yxcdfgxcyz"}]]]], Last /* EqualTo[1]][[All, 1]]
Output:
{"1", "5", "6", "b", "g", "s", "t", "z"}

Nim[edit]

One solution, but others are possible, for instance concatenating the strings and building the count table from it rather than merging several count tables. And to build the last sequence, we could have used something like sorted(toSeq(charCount.pairs).filterIt(it[1] == 1).mapIt(it[0])), which is a one liner but less readable and less efficient than our solution using “collect”.

import algorithm, sugar, tables
 
var charCount: CountTable[char]
 
for str in ["133252abcdeeffd", "a6789798st", "yxcdfgxcyz"]:
charCount.merge str.toCountTable
 
let uniqueChars = collect(newSeq):
for ch, count in charCount.pairs:
if count == 1: ch
 
echo sorted(uniqueChars)
Output:
@['1', '5', '6', 'b', 'g', 's', 't', 'z']

Perl[edit]

Translation of: Raku
# 20210506 Perl programming solution
 
use strict;
use warnings;
use utf8;
use Unicode::Collate 'sort';
 
my %seen;
binmode(STDOUT, ':encoding(utf8)');
map { s/(\X)/$seen{$1}++/egr }
"133252abcdeeffd", "a6789798st", "yxcdfgxcyz", "AАΑSäaoö٥🤔👨‍👩‍👧‍👧";
my $uca = Unicode::Collate->new();
print $uca->sort ( grep { $seen{$_} == 1 } keys %seen )
Output:
👨‍👩‍👧‍👧🤔15٥6AäbgoösStzΑА

Phix[edit]

function once(integer ch, i, string s)
    integer l = length(s)
    return (i=1 or ch!=s[i-1])
       and (i=l or ch!=s[i+1])
end function

sequence set = {"133252abcdeeffd","a6789798st","yxcdfgxcyz"},
         res = filter(sort(join(set,"")),once)
printf(1,"found %d unique characters: %s\n",{length(res),res})
Output:
found 8 unique characters: 156bgstz

PicoLisp[edit]

(de uni (Lst
(let R NIL
(mapc
'((L)
(mapc
'((L) (accu 'R L 1)) L ) )
(mapcar chop Lst) )
(mapcar
car
(by
car
sort
(filter '((L) (=1 (cdr L))) R) ) ) ) )
(println
(uni
(quote
"133252abcdeeffd"
"a6789798st"
"yxcdfgxcyz" ) ) )
Output:
("1" "5" "6" "b" "g" "s" "t" "z")

PL/M[edit]

100H:
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;
PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;
 
/* FIND SORTED UNIQUE CHARACTERS IN ARRAY OF STRINGS */
UNIQUES: PROCEDURE (STRINGS, OUT);
DECLARE (STRINGS, OUT) ADDRESS;
DECLARE STRING BASED STRINGS ADDRESS;
DECLARE IPTR ADDRESS;
DECLARE ICHAR BASED IPTR BYTE;
DECLARE OCHAR BASED OUT BYTE;
DECLARE COUNT (256) BYTE;
DECLARE I ADDRESS;
 
DO I=0 TO 255; COUNT(I)=0; END;
I = 0;
DO WHILE STRING(I) <> 0;
IPTR = STRING(I);
DO WHILE ICHAR <> '$';
COUNT(ICHAR) = COUNT(ICHAR) + 1;
IPTR = IPTR + 1;
END;
I = I + 1;
END;
 
DO I=0 TO 255;
IF COUNT(I) = 1 THEN DO;
OCHAR = I;
OUT = OUT + 1;
END;
END;
 
OCHAR = '$';
END UNIQUES;
 
/* INPUT ARRAY */
/* USING UPPERCASE LETTERS BECAUSE PLM-80 DOES NOT SUPPORT LOWERCASE */
DECLARE STRINGS (4) ADDRESS;
STRINGS(0) = .'133252ABCDEEFFD$';
STRINGS(1) = .'A6789798ST$';
STRINGS(2) = .'YXCDFGXCYZ$';
STRINGS(3) = 0;
 
DECLARE BUFFER (255) BYTE;
CALL UNIQUES(.STRINGS, .BUFFER);
CALL PRINT(.BUFFER);
CALL EXIT;
EOF
Output:
156BGSTZ

Python[edit]

'''Unique characters'''
 
from itertools import chain, groupby
 
 
# uniques :: [String] -> [Char]
def uniques(xs):
'''Characters which occur only once
across the given list of strings.
'''

return [
h for h, (_, *tail) in
groupby(sorted(chain(*xs)))
if not tail
]
 
 
# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''Characters occurring only once
across a list of 3 given strings.
'''

print(
uniques([
"133252abcdeeffd",
"a6789798st",
"yxcdfgxcyz"
])
)
 
 
# MAIN ---
if __name__ == '__main__':
main()
Output:
['1', '5', '6', 'b', 'g', 's', 't', 'z']

Or reducing the given strings down to a hash of character frequencies:

'''Unique characters'''
 
from functools import reduce
 
 
# uniqueChars :: [String] -> [Char]
def uniqueChars(ws):
'''Characters which occur only once
across the given list of strings.
'''

def addedWord(dct, w):
return reduce(updatedCharCount, w, dct)
 
def updatedCharCount(a, c):
return dict(
a, **{
c: 1 + a[c] if c in a else 1
}
)
 
return sorted([
k for k, v in reduce(addedWord, ws, {}).items()
if 1 == v
])
 
 
# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''Test'''
print(
uniqueChars([
"133252abcdeeffd",
"a6789798st",
"yxcdfgxcyz"
])
)
 
 
# MAIN ---
if __name__ == '__main__':
main()
Output:
['1', '5', '6', 'b', 'g', 's', 't', 'z']

Raku[edit]

One has to wonder where the digits 0 through 9 come in the alphabet... 🤔 For that matter, What alphabet should they be in order of? Most of these entries seem to presuppose ASCII order but that isn't specified anywhere. What to do with characters outside of ASCII (or Latin-1)? Unicode ordinal order? Or maybe DUCET Unicode collation order? It's all very vague.

my @list = <133252abcdeeffd a6789798st yxcdfgxcyz>;
 
for @list, (@list, 'AАΑSäaoö٥🤔👨‍👩‍👧‍👧') {
say "$_\nSemi-bogus \"Unicode natural sort\" order: ",
.map( *.comb ).Bag.grep( *.value == 1 )».key.sort( { .unival, .NFKD[0], .fc } ).join,
"\n (DUCET) Unicode collation order: ",
.map( *.comb ).Bag.grep( *.value == 1 )».key.collate.join, "\n";
}
Output:
133252abcdeeffd a6789798st yxcdfgxcyz
Semi-bogus "Unicode natural sort" order: 156bgstz
        (DUCET) Unicode collation order: 156bgstz

133252abcdeeffd a6789798st yxcdfgxcyz AАΑSäaoö٥🤔👨‍👩‍👧‍👧
Semi-bogus "Unicode natural sort" order: 15٥6ASäbgoöstzΑА👨‍👩‍👧‍👧🤔
        (DUCET) Unicode collation order: 👨‍👩‍👧‍👧🤔ä15٥6AbögosStzΑА

REXX[edit]

This REXX program doesn't assume ASCII (or any other) order.   This example was run on an ASCII machine.

If this REXX program is run on an  ASCII  machine,   it will use the   ASCII   order of characters,   in this case,
decimal digits,   uppercase Latin letters,   and then lowercase Latin letters,   with other characters interspersed.

On an  EBCDIC  machine,   the order would be lowercase Latin letters,   uppercase Latin letters,   and then the
decimal digits,   with other characters interspersed.

On an  EBCDIC  machine,   the lowercase letters and the uppercase letters   aren't   contiguous.

/*REXX pgm finds and shows characters that are unique to only one string  and once only.*/
parse arg $ /*obtain optional arguments from the CL*/
if $='' | $="," then $= '133252abcdeeffd' "a6789798st" 'yxcdfgxcyz' /*use defaults.*/
if $='' then do; say "***error*** no lists were specified."; exit 13; end
@= /*will be a list of all unique chars. */
 
do j=0 for 256; x= d2c(j) /*process all the possible characters. */
if x==' ' then iterate /*ignore blanks which are a delimiter. */
_= pos(x, $); if _==0 then iterate /*character not found, then skip it. */
_= pos(x, $, _+1); if _ >0 then iterate /*Character is a duplicate? Skip it. */
@= @ x
end /*j*/ /*stick a fork in it, we're all done. */
 
@@= space(@, 0); L= length(@@) /*elided superfluous blanks; get length*/
if @@=='' then @= " (none)" /*if none were found, pretty up message*/
if L==0 then L= "no" /*do the same thing for the # of chars.*/
say 'unique characters are: ' @ /*display the unique characters found. */
say
say 'Found ' L " unique characters." /*display the # of unique chars found. */
output   when using the default inputs:
unique characters are:   1 5 6 b g s t z

Found  8  unique characters.

Ring[edit]

 
see "working..." + nl
see "Unique characters are:" + nl
row = 0
str = ""
cList = []
uniqueChars = ["133252abcdeeffd", "a6789798st","yxcdfgxcyz"]
for n = 1 to len(uniqueChars)
str = str + uniqueChars[n]
next
for n = 1 to len(str)
ind = count(str,str[n])
if ind = 1
row = row + 1
add(cList,str[n])
ok
next
cList = sort(cList)
for n = 1 to len(cList)
see "" + cList[n] + " "
next
see nl
 
see "Found " + row + " unique characters" + nl
see "done..." + nl
 
func count(cString,dString)
sum = 0
while substr(cString,dString) > 0
sum++
cString = substr(cString,substr(cString,dString)+len(string(sum)))
end
return sum
 
Output:
working...
Unique characters are:
1 5 6 b g s t z 
Found 8 unique characters
done...

Wren[edit]

Library: Wren-seq
Library: Wren-sort
import "/seq" for Lst
import "/sort" for Sort
 
var strings = ["133252abcdeeffd", "a6789798st","yxcdfgxcyz"]
var totalChars = strings.reduce { |acc, str| acc + str }.toList
var uniqueChars = Lst.individuals(totalChars).where { |l| l[1] == 1 }.map { |l| l[0] }.toList
Sort.insertion(uniqueChars)
System.print("Found %(uniqueChars.count) unique character(s), namely:")
System.print(uniqueChars.join(" "))
Output:
Found 8 unique character(s), namely:
1 5 6 b g s t z

XPL0[edit]

int     List, I, N, C;
char Tbl(128), Str;
string 0;
[List:= ["133252abcdeeffd", "a6789798st","yxcdfgxcyz"];
for I:= 0 to 127 do Tbl(I):= 0;
for N:= 0 to 2 do
[Str:= List(N);
I:= 0;
loop [C:= Str(I);
if C = 0 then quit;
I:= I+1;
Tbl(C):= Tbl(C)+1;
];
];
for I:= 0 to 127 do
if Tbl(I) = 1 then ChOut(0, I);
]
Output:
156bgstz