# Common list elements

Common list elements 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.

Given an integer array nums, find the common list elements.

Example

nums = [2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9]

output = [3,6,9]

## 11l

```F cle(nums)
V r = Set(nums[0])
L(num) nums[1..]
r = r.intersection(Set(num))
R r

print(cle([[2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]]))```
Output:
```Set([3, 6, 9])
```

## Action!

```INCLUDE "D2:SORT.ACT" ;from the Action! Tool Kit

DEFINE PTR="CARD"

PROC PrintArray(BYTE ARRAY a BYTE len)
BYTE i

Print("  [")
IF len>0 THEN
FOR i=0 TO len-1
DO
PrintB(a(i))
IF i<len-1 THEN Put(' ) FI
OD
FI
PrintE("]")
RETURN

BYTE FUNC Contains(BYTE ARRAY a BYTE len,value)
BYTE i,count

count=0
IF len>0 THEN
FOR i=0 TO len-1
DO
IF a(i)=value THEN count==+1 FI
OD
FI
RETURN (count)

PROC CommonListElements(PTR ARRAY arrays
BYTE ARRAY lengths BYTE count
BYTE ARRAY res BYTE POINTER resLen)

BYTE ARRAY a
BYTE i,j,len,value,cnt,maxcnt

resLen^=0
IF count=0 THEN RETURN FI
FOR i=0 TO count-1
DO
IF lengths(i)=0 THEN RETURN FI
OD

a=arrays(0) len=lengths(0)
IF count=1 THEN
MoveBlock(res,a,len) RETURN
FI

FOR i=0 TO len-1
DO
value=a(i)
IF Contains(res,resLen^,value)=0 THEN
maxcnt=Contains(a,len,value)
FOR j=1 TO count-1
DO
cnt=Contains(arrays(j),lengths(j),value)
IF cnt<maxcnt THEN maxcnt=cnt FI
OD
IF maxcnt>0 THEN
FOR j=1 TO maxcnt
DO
res(resLen^)=value resLen^==+1
OD
FI
FI
OD
SortB(res,resLen^,0)
RETURN

PROC Test(PTR ARRAY arrays BYTE ARRAY lengths BYTE count)
BYTE ARRAY res(100)
BYTE len,i

CommonListElements(arrays,lengths,count,res,@len)
PrintE("Input:")
FOR i=0 TO count-1
DO
PrintArray(arrays(i),lengths(i))
OD
PrintE("Intersection:")
PrintArray(res,len) PutE()
RETURN

PROC Main()
PTR ARRAY arrays(3)
BYTE ARRAY
lengths(3)=[8 7 5],
a1(8)=[2 5 1 3 8 9 4 6],
a2(7)=[3 5 6 2 9 8 4],
a3(5)=[1 3 7 6 9],
a4(8)=[2 2 1 3 8 9 4 6],
a5(7)=[3 5 6 2 2 2 4],
a6(5)=[2 3 7 6 2],
a7(5)=[0 1 7 8 9]
BYTE len

Put(125) PutE() ;clear the screen

arrays(0)=a1 arrays(1)=a2 arrays(2)=a3
Test(arrays,lengths,3)

arrays(0)=a4 arrays(1)=a5 arrays(2)=a6
Test(arrays,lengths,3)

arrays(2)=a7
Test(arrays,lengths,3)
RETURN```
Output:
```Input:
[2 5 1 3 8 9 4 6]
[3 5 6 2 9 8 4]
[1 3 7 6 9]
Intersection:
[3 6 9]

Input:
[2 2 1 3 8 9 4 6]
[3 5 6 2 2 2 4]
[2 3 7 6 2]
Intersection:
[2 2 3 6]

Input:
[2 2 1 3 8 9 4 6]
[3 5 6 2 2 2 4]
[0 1 7 8 9]
Intersection:
[]
```

```with Ada.Text_Io;

procedure Common is

package Integer_Vectors is
Element_Type => Integer);
use Integer_Vectors;

function Common_Elements (Left, Right : Vector) return Vector is
Res : Vector;
begin
for E of Left loop
if Has_Element (Right.Find (E)) then
Res.Append (E);
end if;
end loop;
return Res;
end Common_Elements;

procedure Put (Vec : Vector) is
begin
Put ("[");
for E of Vec loop
Put (E'Image);  Put (" ");
end loop;
Put ("]");
New_Line;
end Put;

A : constant Vector := 2 & 5 & 1 & 3 & 8 & 9 & 4 & 6;
B : constant Vector := 3 & 5 & 6 & 2 & 9 & 8 & 4;
C : constant Vector := 1 & 3 & 7 & 6 & 9;
R : Vector;
begin
R := Common_Elements (A, B);
R := Common_Elements (R, C);
Put (R);
end Common;
```
Output:
`[ 3  9  6 ]`

## APL

APL has the built-in intersection function `∩`

```      ∩/ (2 5 1 3 8 9 4 6) (3 5 6 2 9 8 4) (1 3 7 9 6)
3 9 6
```

## AppleScript

### AppleScriptObjC

```use AppleScript version "2.4" -- OS X 10.10 (Yosemite) or later
use framework "Foundation"
use sorter : script "Insertion Sort" -- <https://www.rosettacode.org/wiki/Sorting_algorithms/Insertion_sort#AppleScript>

on commonListElements(listOfLists)
set mutableSet to current application's class "NSMutableSet"'s setWithArray:(beginning of listOfLists)
repeat with i from 2 to (count listOfLists)
tell mutableSet to intersectSet:(current application's class "NSSet"'s setWithArray:(item i of listOfLists))
end repeat

return (mutableSet's allObjects()) as list
end commonListElements

set commonElements to commonListElements({{2, 5, 1, 3, 8, 9, 4, 6}, {3, 5, 6, 2, 9, 8, 4}, {1, 3, 7, 6, 9}})
tell sorter to sort(commonElements, 1, -1)
return commonElements
```
Output:
```{3, 6, 9}
```

### Core language only

The requirement for AppleScript 2.3.1 is only for the 'use' command which loads the "Insertion Sort" script. If the sort's instead loaded with the older 'load script' command or copied into the code, this will work on systems as far back as Mac OS X 10.5 (Leopard) or earlier. Same output as above.

```use AppleScript version "2.3.1" -- Mac OS X 10.9 (Mavericks) or later.
use sorter : script "Insertion Sort" -- <https://www.rosettacode.org/wiki/Sorting_algorithms/Insertion_sort#AppleScript>

on commonListElements(listOfLIsts)
script o
property list1 : beginning of listOfLIsts
end script

set common to {}
set listCount to (count listOfLIsts)
repeat with i from 1 to (count o's list1)
set thisElement to {item i of o's list1}
if (thisElement is not in common) then
repeat with j from 2 to listCount
set OK to (item j of listOfLIsts contains thisElement)
if (not OK) then exit repeat
end repeat
if (OK) then set end of common to beginning of thisElement
end if
end repeat

return common
end commonListElements

set commonElements to commonListElements({{2, 5, 1, 3, 8, 9, 4, 6}, {3, 5, 6, 2, 9, 8, 4}, {1, 3, 7, 6, 9}})
tell sorter to sort(commonElements, 1, -1)
return commonElements
```

## Arturo

```commonElements: function [subsets][
if zero? size subsets -> return []
if 1 = size subsets -> return first subsets

result: first subsets

loop slice subsets 1 dec size subsets 'subset [
result: intersection result subset
]
return result
]

print commonElements [
[2 5 1 3 8 9 4 6]
[3 5 6 2 9 8 4]
[1 3 7 6 9]
]
```
Output:
`3 6 9`

## AutoHotkey

```Common_list_elements(nums){
counter := [], output := []
for i, num in nums
for j, d in num
if ((counter[d] := counter[d] ? counter[d]+1 : 1) = nums.count())
output.Push(d)
return output
}
```
Examples:
```nums := [[2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9]]
output := Common_list_elements(nums)
return
```
Output:
`[3, 6, 9]`

## AWK

```# syntax: GAWK -f COMMON_LIST_ELEMENTS.AWK
BEGIN {
PROCINFO["sorted_in"] = "@ind_num_asc"
nums = "[2,5,1,3,8,9,4,6],[3,5,6,2,9,8,4],[1,3,7,6,9]"
printf("%s : ",nums)
n = split(nums,arr1,"],?") - 1
for (i=1; i<=n; i++) {
gsub(/[\[\]]/,"",arr1[i])
split(arr1[i],arr2,",")
for (j in arr2) {
arr3[arr2[j]]++
}
}
for (j in arr3) {
if (arr3[j] == n) {
printf("%s ",j)
}
}
printf("\n")
exit(0)
}
```
Output:
```[2,5,1,3,8,9,4,6],[3,5,6,2,9,8,4],[1,3,7,6,9] : 3 6 9
```

## CLU

```contains = proc [T: type] (a: array[T], v: T) returns (bool)
where T has equal: proctype (T,T) returns (bool)
for i: T in array[T]\$elements(a) do
if i=v then return(true) end
end
return(false)
end contains

common = proc [T: type] (lists: ssT) returns (sT)
where T has equal: proctype (T,T) returns (bool)
sT = sequence[T]
aT = array[T]
ssT = sequence[sequence[T]]

cur: aT := sT\$s2a(ssT\$bottom(lists))
for i: int in int\$from_to(2, ssT\$size(lists)) do
next: aT := aT\$[]
for e: T in sT\$elements(lists[i]) do
if contains[T](cur, e) then
end
end
cur := next
end
return(sT\$a2s(cur))
end common

start_up = proc ()
si = sequence[int]
ssi = sequence[sequence[int]]

nums: ssi := ssi\$[
si\$[2,5,1,3,8,9,4,6],
si\$[3,5,6,2,9,8,4],
si\$[1,3,7,6,9]
]

po: stream := stream\$primary_output()
for i: int in si\$elements(common[int](nums)) do
stream\$puts(po, int\$unparse(i) || " ")
end
end start_up```
Output:
`3 6 9`

## Excel

### LAMBDA

Binding the names INTERSECT and INTERSECTCOLS to a pair of lambda expressions in the Excel WorkBook Name Manager:

```INTERSECT
=LAMBDA(xs,
LAMBDA(ys,
FILTERP(
LAMBDA(x,
ELEM(x)(ys)
)
)(xs)
)
)

INTERSECTCOLS
=LAMBDA(xs,
IF(1 < COLUMNS(xs),
INTERSECT(
FIRSTCOL(xs)
)(
INTERSECTCOLS(
TAILCOLS(xs)
)
),
xs
)
)
```

and also assuming the following generic bindings in Name Manager:

```ELEM
=LAMBDA(x,
LAMBDA(xs,
ISNUMBER(MATCH(x, xs, 0))
)
)

FILTERP
=LAMBDA(p,
LAMBDA(xs,
FILTER(xs, p(xs))
)
)

FIRSTCOL
=LAMBDA(xs,
INDEX(
xs,
SEQUENCE(ROWS(xs), 1, 1, 1),
1
)
)

TAILCOLS
=LAMBDA(xs,
LET(
n, COLUMNS(xs) - 1,

IF(0 < n,
INDEX(
xs,
SEQUENCE(ROWS(xs), 1, 1, 1),
SEQUENCE(1, n, 2, 1)
),
NA()
)
)
)
```
Output:
 =INTERSECTCOLS(D2:F9) fx A B C D E F 1 Intersection of three columns 2 3 2 3 1 3 9 5 5 3 4 6 1 6 7 5 3 2 6 6 8 9 9 7 9 8 8 4 4 9 6

## Delphi

Works with: Delphi version 6.0

```const Set1: set of byte = [2,5,1,3,8,9,4,6];
const Set2: set of byte = [3,5,6,2,9,8,4];
const Set3: set of byte = [1,3,7,6,9];

procedure CommonListElements(Memo: TMemo);
{Using Delphi "sets" to find common elements}
var I,Start,Stop: integer;
var Common: set of byte;
var S: string;
begin
{Uses "*" intersection set operator to}
{ find items common to all three sets}
Common:=Set1 * Set2 * Set3;
S:='';
{Display the common items}
for I:=0 to 9 do
if I in Common then S:=S+IntToStr(I)+',';
end;
```
Output:
```Common Elements in
[2,5,1,3,8,9,4,6]
[3,5,6,2,9,8,4]
[1,3,7,6,9]:
3,6,9,
Elapsed Time: 5.260 ms.
```

## EasyLang

```nums[][] = [ [ 2 5 1 3 8 9 4 6 ] [ 3 5 6 2 9 8 4 ] [ 1 3 7 6 9 ] ]
#
found = 1
for e in nums[1][]
for l = 2 to len nums[][]
found = 0
for x in nums[l][]
if e = x
found = 1
break 1
.
.
if found = 0
break 1
.
.
if found = 1
r[] &= e
.
.
print r[]```

## F#

Of course it is possible to use sets but I thought the idea was not to?

```// Common list elements. Nigel Galloway: February 25th., 2021
let nums=[|[2;5;1;3;8;9;4;6];[3;5;6;2;9;8;4];[1;3;7;6;9]|]
printfn "%A" (nums|>Array.reduce(fun n g->n@g)|>List.distinct|>List.filter(fun n->nums|>Array.forall(fun g->List.contains n g)));;
```
Output:
```[3; 9; 6]
```

## Factor

Note: in older versions of Factor, `intersect-all` was called `intersection`.

Works with: Factor version 0.99 2021-02-05
```USING: prettyprint sets ;

{ { 2 5 1 3 8 9 4 6 } { 3 5 6 2 9 8 4 } { 1 3 7 6 9 } } intersect-all .
```
Output:
```{ 3 6 9 }
```

## FreeBASIC

```dim as integer nums(1 to 3, 1 to 8) = {{2,5,1,3,8,9,4,6}, {3,5,6,2,9,8,4}, {1,3,7,6,9} }
redim as integer outp(0)
dim as integer i, j
dim as boolean found

function is_in( s() as integer, n as integer, r as integer ) as boolean
for i as uinteger = 1 to ubound(s,2)
if s(r,i)=n then return true
next i
return false
end function

for i = 1 to 8
found = true
for j = 2 to 3
if not is_in( nums(), nums(1,i), j ) then found = false
next j
if found then
redim preserve as integer outp(1 to 1+ubound(outp))
outp(ubound(outp)) = nums(1,i)
end if
next i

for i = 1 to ubound(outp)
print outp(i);" ";
next i```
Output:
`3 9 6`

## Go

Translation of: Wren
```package main

import "fmt"

func indexOf(l []int, n int) int {
for i := 0; i < len(l); i++ {
if l[i] == n {
return i
}
}
return -1
}

func common2(l1, l2 []int) []int {
// minimize number of lookups
c1, c2 := len(l1), len(l2)
shortest, longest := l1, l2
if c1 > c2 {
shortest, longest = l2, l1
}
longest2 := make([]int, len(longest))
copy(longest2, longest) // matching duplicates will be destructive
var res []int
for _, e := range shortest {
ix := indexOf(longest2, e)
if ix >= 0 {
res = append(res, e)
longest2 = append(longest2[:ix], longest2[ix+1:]...)
}
}
return res
}

func commonN(ll [][]int) []int {
n := len(ll)
if n == 0 {
return []int{}
}
if n == 1 {
return ll[0]
}
res := common2(ll[0], ll[1])
if n == 2 {
return res
}
for _, l := range ll[2:] {
res = common2(res, l)
}
return res
}

func main() {
lls := [][][]int{
{{2, 5, 1, 3, 8, 9, 4, 6}, {3, 5, 6, 2, 9, 8, 4}, {1, 3, 7, 6, 9}},
{{2, 2, 1, 3, 8, 9, 4, 6}, {3, 5, 6, 2, 2, 2, 4}, {2, 3, 7, 6, 2}},
}
for _, ll := range lls {
fmt.Println("Intersection of", ll, "is:")
fmt.Println(commonN(ll))
fmt.Println()
}
}
```
Output:
```Intersection of [[2 5 1 3 8 9 4 6] [3 5 6 2 9 8 4] [1 3 7 6 9]] is:
[3 6 9]

Intersection of [[2 2 1 3 8 9 4 6] [3 5 6 2 2 2 4] [2 3 7 6 2]] is:
[3 6 2 2]
```

```import qualified Data.Set as Set

task :: Ord a => [[a]] -> [a]
task xs = Set.toAscList . foldl1 Set.intersection . map Set.fromList \$ xs

main = print \$ task  [[2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9]]
```
Output:
`[3,6,9]`

## J

```   2 5 1 3 8 9 4 6([-.-.)3 5 6 2 9 8 4([-.-.)1 3 7 6 9
3 9 6
```

Or,

```   ;([-.-.)&.>/2 5 1 3 8 9 4 6;3 5 6 2 9 8 4;1 3 7 6 9
3 9 6
```

## jq

Works with: jq

Works with gojq, the Go implementation of jq

The following definition of `intersection` does not place any restrictions on the arrays whose intersection is sought. The helper function, `ios`, might be independently useful and so is defined as a top-level filter.

```# If a and b are sorted lists, and if all the elements respectively of a and b are distinct,
# then [a,b] | ios will emit the stream of elements in the set-intersection of a and b.
def ios:
.[0] as \$a | .[1] as \$b
| if 0 == (\$a|length) or 0 == (\$b|length) then empty
elif \$a[0] == \$b[0] then \$a[0], ([\$a[1:], \$b[1:]] | ios)
elif \$a[0]  < \$b[0] then [\$a[1:], \$b] | ios
else [\$a, \$b[1:]] | ios
end ;

# input: an array of arbitrary JSON arrays
# output: their intersection as sets
def intersection:
def go:
if length == 1 then (.[0]|unique)
else [(.[0]|unique), (.[1:] | go)] | [ios]
end;
if length == 0 then []
elif any(.[]; length == 0) then []
else sort_by(length) | go
end;```
```# The task:
[[2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9]]```
Output:
```[3,6,9]
```

## K

Works with: ngn/k
```{x^x^y}/(2 5 1 3 8 9 4 6;1 3 7 6 9;3 5 6 2 9 8 4)
3 9 6
```

## Ksh

```#!/bin/ksh

# Find the common list elements in an integer array nums[][]

#	# Variables:
#
typeset -a nums=( (2 5 1 3 8 9 4 6) (3 5 6 2 9 8 4) (1 3 7 6 9) )

#	# Functions:
#

#	# Function _flatten(arr[][] arr[]) - flatten input matrix
#
function _flatten {
typeset _inarr ; nameref _inarr="\$1"
typeset _outarr ; nameref _outarr="\$2"
typeset _i ; integer _i

_oldIFS=\$IFS ; IFS=\|
for ((_i=1; _i<\${#_inarr[*]}; _i++)); do
_outarr[_i]=\${_inarr[_i][*]}
done
IFS=\$oldIFS
}

######
# main #
######

typeset -a flatarr output

_flatten nums flatarr

integer i j
for ((i=0; i<\${#nums[0][*]}; i++)); do
integer cnt=0
for ((j=1; j<=\${#flatarr[*]}; j++)); do
[[ \${nums[0][i]} == @(\${flatarr[j]%\|}) ]] && (( cnt++ ))
done
(( cnt == 2 )) && output+=( \${nums[0][i]} )
done

print "( \${output[@]} )"
```
Output:
`( 3 9 6 )`

## Julia

```julia> intersect([2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9])
3-element Array{Int64,1}:
3
9
6
```

## Lambdatalk

```{def intersection

{def intersection.r
{lambda {:a :b :c :d}
{if {A.empty? :a}
then :d
else {intersection.r {A.rest :a} :b :c
{if {and {> {A.in? {A.first :a} :b} -1}
{> {A.in? {A.first :a} :c} -1}}
else :d} }}}}

{lambda {:a :b :c}
{A.sort! < {intersection.r :a :b :c {A.new}}} }}
-> intersection

{intersection
{A.new 2 5 1 3 8 9 4 6}
{A.new 3 5 6 2 9 8 4}
{A.new 1 3 7 6 9}
}
-> [3,6,9]
```

## Mathematica/Wolfram Language

```Intersection[{2, 5, 1, 3, 8, 9, 4, 6}, {3, 5, 6, 2, 9, 8, 4}, {1, 3, 7, 6, 9}]
```
Output:
`{3, 6, 9}`

## Maxima

```common_elems(lst):=block(map(setify,lst),apply(intersection,%%),listify(%%))\$
nums:[[2,5,1,3,8,9,4,6],[3,5,6,2,9,8,4],[1,3,7,6,9]]\$
common_elems(nums);
```
Output:
```[3,6,9]
```

## Nim

```import algorithm, sequtils

proc commonElements(list: openArray[seq[int]]): seq[int] =
for val in list[0].deduplicate():     # Check values only once.
block checkVal:
for i in 1..list.high:
if val notin list[i]:
break checkVal

echo commonElements([@[2,5,1,3,8,9,4,6], @[3,5,6,2,9,8,4], @[1,3,7,6,9]])
```
Output:
`@[3, 6, 9]`

## Perl

```@nums = ([2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9]);
map { print "\$_ " if @nums == ++\$c{\$_} } @\$_ for @nums;
```
Output:
`3 6 9`

## Phix

```function intersection(sequence s)
sequence res = {}
if length(s) then
for i=1 to length(s[1]) do
integer candidate = s[1][i]
for j=2 to length(s) do
if not find(candidate,s[j]) then
exit
end if
end for
if bAdd then res &= candidate end if
end for
end if
return res
end function
?intersection({{2,5,1,3,8,9,4,6},{3,5,6,2,9,8,4},{1,3,7,6,9}})
?intersection({{2,2,1,3,8,9,4,6},{3,5,6,2,2,2,4},{2,3,7,6,2}})
```

Note that a (slightly more flexible) intersection() function is also defined in sets.e, so you could just include that instead, and use it the same way.

Output:
```{3,9,6}
{2,3,6}
```

## Python

### Without Duplicates

```"""Find distinct common list elements using set.intersection."""

def common_list_elements(*lists):
return list(set.intersection(*(set(list_) for list_ in lists)))

if __name__ == "__main__":
test_cases = [
([2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]),
([2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]),
]

for case in test_cases:
result = common_list_elements(*case)
print(f"Intersection of {case} is {result}")
```
Output:
```intersection of ([2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]) is [9, 3, 6]
intersection of ([2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]) is [2, 3, 6]
```

### With Duplicates

```"""Find common list elements using collections.Counter (multiset)."""

from collections import Counter
from functools import reduce
from itertools import chain

def common_list_elements(*lists):
counts = (Counter(list_) for list_ in lists)
intersection = reduce(lambda x, y: x & y, counts)
return list(chain.from_iterable([elem] * cnt for elem, cnt in intersection.items()))

if __name__ == "__main__":
test_cases = [
([2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]),
([2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]),
]

for case in test_cases:
result = common_list_elements(*case)
print(f"Intersection of {case} is {result}")
```
Output:
```intersection of ([2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]) is [9, 3, 6]
intersection of ([2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]) is [2, 2, 3, 6]
```

## Quackery

```  [ behead sort swap witheach
[ sort [] temp put
[ over [] !=
over [] != and while
over 0 peek
over 0 peek = iff
swap join temp put
over 0 peek
over 0 peek < if swap
2drop temp take ] ]          is common ( [ [ --> [ )

' [ [ 2 5 1 3 8 9 4 6 ] [ 3 5 6 2 9 8 4 ] [ 1 3 7 6 9 ] ] common echo```
Output:
`[ 3 6 9 ]`

## Raku

```put [∩] [2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9,3];
```
Output:
`6 9 3`

## REXX

This REXX version properly handles the case of duplicate entries in a list   (which shouldn't happen in a true list).

```/*REXX program finds and displays the  common list elements  from a collection of sets. */
parse arg a                                      /*obtain optional arguments from the CL*/
if a='' | a=","  then a= '[2,5,1,3,8,9,4,6]  [3,5,6,2,9,8,4]  [1,3,7,6,9]'   /*defaults.*/
#= words(a)                                      /*the number of sets that are specified*/
do j=1  for #                        /*process each set  in  a list of sets.*/
@.j= translate( word(a, j), ,'],[')  /*extract   a   "  from "   "   "   "  */
end   /*j*/
\$=                                               /*the list of common elements (so far).*/
do k=1  for #-1                               /*use the last set as the base compare.*/
do c=1  for words(@.#);  x= word(@.#, c)   /*obtain an element from a set.        */
do f=1  for #-1                         /*verify that the element is in the set*/
if wordpos(x, @.f)==0  then iterate c   /*Is in the set?  No, then skip element*/
end   /*f*/
if wordpos(x, \$)==0  then \$= \$ x           /*Not already in the set?  Add common. */
end      /*c*/
end         /*k*/
/*stick a fork in it,  we're all done. */
say 'the list of common elements in all sets: '       "["translate(space(\$), ',', " ")']'
```
output   when using the default inputs:
```the list of common elements in all sets:  [3,6,9]
```

## Ring

```nums = [[2,5,1,3,8,9,4,6],[3,5,6,2,9,8,4],[1,3,7,6,9]]
sumNums = []
result = []

for n = 1 to len(nums)
for m = 1 to len(nums[n])
next
next

sumNums = sort(sumNums)
for n = len(sumNums) to 2 step -1
if sumNums[n] = sumNums[n-1]
del(sumNums,n)
ok
next

for n = 1 to len(sumNums)
flag = list(len(nums))
for m = 1 to len(nums)
flag[m] = 1
ind = find(nums[m],sumNums[n])
if ind < 1
flag[m] = 0
ok
next
flagn = 1
for p = 1 to len(nums)
if flag[p] = 1
flagn = 1
else
flagn = 0
exit
ok
next
if flagn = 1
ok
next

see "common list elements are: "
showArray(result)

func showArray(array)
txt = ""
see "["
for n = 1 to len(array)
txt = txt + array[n] + ","
next
txt = left(txt,len(txt)-1)
txt = txt + "]"
see txt```
Output:
```common list elements are: [3,6,9]
```

## Ruby

```nums = [2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9]
p nums.inject(&:intersection)  # or nums.inject(:&)
```
Output:
```[3, 9, 6]
```

## V (Vlang)

Translation of: go
```fn index_of(l []int, n int) int {
for i in 0..l.len {
if l[i] == n {
return i
}
}
return -1
}

fn common2(l1 []int, l2 []int) []int {
// minimize number of lookups
c1, c2 := l1.len, l2.len
mut shortest, mut longest := l1.clone(), l2.clone()
if c1 > c2 {
shortest, longest = l2.clone(), l1.clone()
}
mut longest2 := longest.clone()
mut res := []int{}
for e in shortest {
ix := index_of(longest2, e)
if ix >= 0 {
res << e
longest2 << longest2[ix+1..]
}
}
return res
}

fn common_n(ll [][]int) []int {
n := ll.len
if n == 0 {
return []int{}
}
if n == 1 {
return ll[0]
}
mut res := common2(ll[0], ll[1])
if n == 2 {
return res
}
for l in ll[2..] {
res = common2(res, l)
}
return res
}

fn main() {
lls := [
[[2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]],
[[2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]],
]
for ll in lls {
println("Intersection of \$ll is:")
println(common_n(ll))
println('')
}
}```
Output:
```Intersection of [[2, 5, 1, 3, 8, 9, 4, 6] [3, 5, 6, 2, 9, 8, 4] [1, 3, 7, 6, 9]] is:
[3, 6, 9]

Intersection of [[2 2 1 3 8 9 4 6] [3 5 6 2 2 2 4] [2, 3, 7, 6, 2]] is:
[3, 6, 2, 2]```

Alternative:

```fn main()
{
lls := [
[[2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]],
[[2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]],
]
for ll in lls {
println("Intersection of \$ll is:")
println(common_list_elements(ll))
println('')
}

}

fn common_list_elements(md_arr [][]int) []int {
mut counter := map[int]int{}
mut output := []int{}

for sd_arr in md_arr {
for value in sd_arr {
if counter[value] == counter[value] {counter[value] = counter[value] + 1} else { 1 } {
if counter[value] >= md_arr.len && output.any(it == value) == false {
output << value
}
}
}
}
return output
}```
Output:
```Intersection of [[2, 5, 1, 3, 8, 9, 4, 6] [3, 5, 6, 2, 9, 8, 4] [1, 3, 7, 6, 9]] is:
[3, 6, 9]

Intersection of [[2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]] is:
[2, 3, 6]```

## Wren

Library: Wren-seq

As we're dealing here with lists rather than sets, some guidance is needed on how to deal with duplicates in each list in the general case. A drastic solution would be to remove all duplicates from the result. Instead, the following matches duplicates - so if List A contains 2 'a's and List B contains 3 'a's, there would be 2 'a's in the result.

```import "./seq" for Lst

var common2 = Fn.new { |l1, l2|
// minimize number of lookups
var c1 = l1.count
var c2 = l2.count
var shortest = (c1 < c2) ? l1 : l2
var longest = (l1 == shortest) ? l2 : l1
longest = longest.toList // matching duplicates will be destructive
var res = []
for (e in shortest) {
var ix = Lst.indexOf(longest, e)
if (ix >= 0) {
longest.removeAt(ix)
}
}
return res
}

var commonN = Fn.new { |ll|
var n = ll.count
if (n == 0) return ll
if (n == 1) return ll[0]
var first2 = common2.call(ll[0], ll[1])
if (n == 2) return first2
return ll.skip(2).reduce(first2) { |acc, l| common2.call(acc, l) }
}

var lls = [
[[2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]],
[[2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]]
]

for (ll in lls) {
System.print("Intersection of %(ll) is:")
System.print(commonN.call(ll))
System.print()
}
```
Output:
```Intersection of [[2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]] is:
[3, 6, 9]

Intersection of [[2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]] is:
[2, 3, 6, 2]
```

Since the above was written, we can also now offer a library based solution.

```import "./seq" for Lst

var lls = [
[[2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]],
[[2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]]
]

for (ll in lls) {
System.print(Lst.intersect(ll[0], Lst.intersect(ll[1], ll[2])))
}
```
Output:
```[3, 9, 6]
[2, 2, 3, 6]
```

## XPL0

A 32-bit integer is used to specify a set of values 0..31. The [-1] terminator helps determine the number of lists.

```int IntSize, Nums, Sets, Ans, N, ListSize, Set, I;
[IntSize:= @Nums - @IntSize;    \number of bytes in an integer
Nums:= [[2,5,1,3,8,9,4,6], [3,5,6,2,9,8,4], [1,3,7,6,9], [-1]];
Sets:= 0;       \find the number of lists = number of Sets
while Nums(Sets, 0) > 0 do Sets:= Sets+1;
Ans:= -1;
for N:= 0 to Sets-1 do
[ListSize:= (Nums(N+1) - Nums(N)) / IntSize;
Set:= 0;
for I:= 0 to ListSize-1 do  \Set = union (or) of list elements
Set:= Set or 1<<Nums(N, I);
Ans:= Ans & Set;            \Answer is intersection (&) of Sets
];
I:= 0;
while Ans do                    \show common list elements
[if Ans & 1 then
[IntOut(0, I);  ChOut(0, ^ )];
Ans:= Ans>>1;
I:= I+1;
];
]```
Output:
```3 6 9
```