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# Exactly three adjacent 3 in lists

Exactly three adjacent 3 in lists 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 5 lists of ints:
list[1] = [9,3,3,3,2,1,7,8,5]
list[2] = [5,2,9,3,3,7,8,4,1]
list[3] = [1,4,3,6,7,3,8,3,2]
list[4] = [1,2,3,4,5,6,7,8,9]
list[5] = [4,6,8,7,2,3,3,3,1]

For each list, print 'true' if the list contains exactly three '3's that form a consecutive subsequence, otherwise print 'false'.

## 11l

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

L(l) lists
print(l, end' ‘ -> ’)
L(i) 0 .< l.len - 2
I l[i] == l[i + 1] == l[i + 2] == 3
print(‘True’)
L.break
L.was_no_break
print(‘False’)
Output:
[9, 3, 3, 3, 2, 1, 7, 8, 5] -> True
[5, 2, 9, 3, 3, 7, 8, 4, 1] -> False
[1, 4, 3, 6, 7, 3, 8, 3, 2] -> False
[1, 2, 3, 4, 5, 6, 7, 8, 9] -> False
[4, 6, 8, 7, 2, 3, 3, 3, 1] -> True

## 8080 Assembly

org	100h
jmp demo
;;; See if the list at [HL] with length DE has three
;;; consecutive 3s.
;;; Returns with zero flag set if the list as three 3s,
;;; clear if not.
three3: lxi b,3 ; B = threes seen, C holds a 3
t_loop: mov a,m ; Get next element
inx h
cmp c ; A three?
jz three
mov a,b ; Not a three, not part of sequence
cmp c ; So we must have seen either three 3s,
jz t_next
ora a ; or none at all
rnz
t_next: dcx d ; Are we at the end yet?
mov a,d
ora e
rz
jmp t_loop ; If not, keep going
three: inr b ; A three - count it
mov a,c ; But see if we don't have too many 3s
cmp b
rc ; If too many 3s, stop
jmp t_next
;;; Test the given lists and print "true" or "false"
demo: lxi h,lists ; List pointer
d_loop: mov e,m ; Load pointer to next list
inx h
mov d,m
inx h
mov a,d ; If at the end, stop
ora e
rz
push h ; Otherwise, keep the pointer
xchg
lxi d,9 ; The lists are all of length 9
call three3 ; See if the list matches
mvi c,9 ; CP/M 'puts'
lxi d,true ; Print true or false
jz d_prn
lxi d,false
d_prn: call 5
pop h ; Get the list pointer back
jmp d_loop ; Next list
true: db "true \$"
false: db "false \$"
;;; Lists
lists: dw list1,list2,list3,list4,list5,0
list1: db 9,3,3,3,2,1,7,8,5
list2: db 5,2,9,3,3,7,8,4,1
list3: db 1,4,3,6,7,3,8,3,2
list4: db 1,2,3,4,5,6,7,8,9
list5: db 4,6,8,7,2,3,3,3,1
Output:
true false false false true

procedure Exactly_3 is

type List_Type is array (Positive range <>) of Integer;

function Has_3_Consecutive (List : List_Type) return Boolean is
Conseq : constant Natural := 3;
Match  : constant Integer := 3;
Count  : Natural := 0;
begin
for Element of List loop
if Element = Match then
Count := Count + 1;
else
if Count = Conseq then
return True;
else
Count := 0;
end if;
end if;
end loop;
return (Count = Conseq);
end Has_3_Consecutive;

procedure Put (List : List_Type) is
begin
Put ("[");
for Element of List loop
Put (Integer'Image (Element));
Put (" ");
end loop;
Put ("]");
end Put;

procedure Test (List : List_Type) is
Result : constant Boolean := Has_3_Consecutive (List);
begin
Put (List);
Put (" -> ");
Put (Boolean'Image (Result));
New_Line;
end Test;

begin
Test ((9,3,3,3,2,1,7,8,5));
Test ((5,2,9,3,3,7,8,4,1));
Test ((1,4,3,6,7,3,8,3,2));
Test ((1,2,3,4,5,6,7,8,9));
Test ((4,6,8,7,2,3,3,3,1));

Test ((4,6,8,7,2,3,3,3,3)); -- Four tailing
Test ((4,6,8,7,2,1,3,3,3)); -- Three tailing
Test ((1,3,3,3,3,4,5,8,9));

Test ((3,3,3,3));
Test ((3,3,3));
Test ((3,3));
Test ((1 => 3)); -- One element
Test ((1 .. 0 => <>)); -- No elements
end Exactly_3;
Output:
[ 9  3  3  3  2  1  7  8  5 ] -> TRUE
[ 5  2  9  3  3  7  8  4  1 ] -> FALSE
[ 1  4  3  6  7  3  8  3  2 ] -> FALSE
[ 1  2  3  4  5  6  7  8  9 ] -> FALSE
[ 4  6  8  7  2  3  3  3  1 ] -> TRUE
[ 4  6  8  7  2  3  3  3  3 ] -> FALSE
[ 4  6  8  7  2  1  3  3  3 ] -> TRUE
[ 1  3  3  3  3  4  5  8  9 ] -> FALSE
[ 3  3  3  3 ] -> FALSE
[ 3  3  3 ] -> TRUE
[ 3  3 ] -> FALSE
[ 3 ] -> FALSE
[] -> FALSE

## ALGOL 68

Including the extra test cases from the Raku and Wren samples.

BEGIN # test lists contain exactly 3 threes and that they are adjacent #
[]INT list1 = ( 9, 3, 3, 3, 2, 1, 7, 8, 5 ); # task test case #
[]INT list2 = ( 5, 2, 9, 3, 3, 7, 8, 4, 1 ); # " " " #
[]INT list3 = ( 1, 4, 3, 6, 7, 3, 8, 3, 2 ); # " " " #
[]INT list4 = ( 1, 2, 3, 4, 5, 6, 7, 8, 9 ); # " " " #
[]INT list5 = ( 4, 6, 8, 7, 2, 3, 3, 3, 1 ); # " " " #
[]INT list6 = ( 3, 3, 3, 1, 2, 4, 5, 1, 3 ); # additional test from the Raku/Wren sample #
[]INT list7 = ( 0, 3, 3, 3, 3, 7, 2, 2, 6 ); # additional test from the Raku/Wren sample #
[]INT list8 = ( 3, 3, 3, 3, 3, 4, 4, 4, 4 ); # additional test from the Raku/Wren sample #
[][]INT lists = ( list1, list2, list3, list4, list5, list6, list7, list8 );
FOR l pos FROM LWB lists TO UPB lists DO
[]INT list = lists[ l pos ];
INT threes := 0; # number of threes in the list #
INT three pos := 0; # position of the last three in the list #
BOOL list ok := FALSE;
FOR e pos FROM LWB list TO UPB list DO
IF list[ e pos ] = 3 THEN
threes +:= 1;
three pos := e pos
FI
OD;
IF threes = 3 THEN
# exactly 3 threes - check they are adjacent #
list ok := ( list[ three pos - 1 ] = 3 AND list[ three pos - 2 ] = 3 )
FI;
# show the result #
print( ( "[" ) );
FOR e pos FROM LWB list TO UPB list DO
print( ( " ", whole( list[ e pos ], 0 ) ) )
OD;
print( ( " ] -> ", IF list ok THEN "true" ELSE "false" FI, newline ) )
OD
END
Output:
[ 9 3 3 3 2 1 7 8 5 ] -> true
[ 5 2 9 3 3 7 8 4 1 ] -> false
[ 1 4 3 6 7 3 8 3 2 ] -> false
[ 1 2 3 4 5 6 7 8 9 ] -> false
[ 4 6 8 7 2 3 3 3 1 ] -> true
[ 3 3 3 1 2 4 5 1 3 ] -> false
[ 0 3 3 3 3 7 2 2 6 ] -> false
[ 3 3 3 3 3 4 4 4 4 ] -> false

## AppleScript

------- EXACTLY N INSTANCES OF N AND ALL CONTIGUOUS ------

-- nnPeers :: Int -> [Int] -> Bool
on nnPeers(n)
script p
on |λ|(x)
n = x
end |λ|
end script

script notP
on |λ|(x)
n ≠ x
end |λ|
end script

script
on |λ|(xs)
set {contiguous, residue} to ¬
span(p, dropWhile(notP, xs))

n = length of contiguous and ¬
all(notP, residue)
end |λ|
end script
end nnPeers

--------------------------- TEST -------------------------
on run
set xs to [¬
[9, 3, 3, 3, 2, 1, 7, 8, 5], ¬
[5, 2, 9, 3, 3, 7, 8, 4, 1], ¬
[1, 4, 3, 6, 7, 3, 8, 3, 2], ¬
[1, 2, 3, 4, 5, 6, 7, 8, 9], ¬
[4, 6, 8, 7, 2, 3, 3, 3, 1]]

set p to nnPeers(3)

script test
on |λ|(x)
showList(x) & " -> " & p's |λ|(x)
end |λ|
end script

unlines(map(test, xs))
end run

------------------------- GENERIC ------------------------

-- all :: (a -> Bool) -> [a] -> Bool
on all(p, xs)
-- True if p holds for every value in xs
tell mReturn(p)
set lng to length of xs
repeat with i from 1 to lng
if not |λ|(item i of xs, i, xs) then return false
end repeat
true
end tell
end all

-- dropWhile :: (a -> Bool) -> [a] -> [a]
-- dropWhile :: (Char -> Bool) -> String -> String
on dropWhile(p, xs)
set lng to length of xs
set i to 1
tell mReturn(p)
repeat while i ≤ lng and |λ|(item i of xs)
set i to i + 1
end repeat
end tell
items i thru -1 of xs
end dropWhile

-- intercalate :: String -> [String] -> String
on intercalate(delim, xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, delim}
set s to xs as text
set my text item delimiters to dlm
s
end intercalate

-- 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

-- showList :: [a] -> String
on showList(xs)
"[" & intercalate(", ", map(my str, xs)) & "]"
end showList

-- span :: (a -> Bool) -> [a] -> ([a], [a])
on span(p, xs)
-- The longest (possibly empty) prefix of xs
-- that contains only elements satisfying p,
-- tupled with the remainder of xs.
-- span(p, xs) eq (takeWhile(p, xs), dropWhile(p, xs))
script go
property mp : mReturn(p)
on |λ|(vs)
if {} ≠ vs then
set x to item 1 of vs
if |λ|(x) of mp then
set {ys, zs} to |λ|(rest of vs)
{{x} & ys, zs}
else
{{}, vs}
end if
else
{{}, {}}
end if
end |λ|
end script
|λ|(xs) of go
end span

-- str :: a -> String
on str(x)
x as string
end str

-- unlines :: [String] -> String
on unlines(xs)
-- A single string formed by the intercalation
-- of a list of strings with the newline character.
set {dlm, my text item delimiters} to ¬
{my text item delimiters, linefeed}
set s to xs as text
set my text item delimiters to dlm
s
end unlines
Output:
[9, 3, 3, 3, 2, 1, 7, 8, 5] -> true
[5, 2, 9, 3, 3, 7, 8, 4, 1] -> false
[1, 4, 3, 6, 7, 3, 8, 3, 2] -> false
[1, 2, 3, 4, 5, 6, 7, 8, 9] -> false
[4, 6, 8, 7, 2, 3, 3, 3, 1] -> true

## AutoHotkey

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

L := []
for i, list in lists
{
c := cnsctv := 0
for j, v in list
{
cnsctv := (list[j] = 3 && list[j+1] = 3 && list[j+2] = 3) ? true : cnsctv
c += (v = 3) ? 1 : 0
L[i] .= (L[i] ? ", " : "" ) . v
}
result .= "[" L[i] "] : " (cnsctv && c=3 ? "true" : "false") "`n"
}
MsgBox % result
Output:
[9, 3, 3, 3, 2, 1, 7, 8, 5] : true
[5, 2, 9, 3, 3, 7, 8, 4, 1] : false
[1, 4, 3, 6, 7, 3, 8, 3, 2] : false
[1, 2, 3, 4, 5, 6, 7, 8, 9] : false
[4, 6, 8, 7, 2, 3, 3, 3, 1] : true

## AWK

BEGIN {
list[++n] = "9,3,3,3,2,1,7,8,5"
list[++n] = "5,2,9,3,3,7,8,4,1"
list[++n] = "1,4,3,6,7,3,8,3,2"
list[++n] = "1,2,3,4,5,6,7,8,9"
list[++n] = "4,6,8,7,2,3,3,3,1"
for (i=1; i<=n; i++) {
tmp = "," list[i] ","
printf("%s %s\n",sub(/,3,3,3,/,"",tmp)?"T":"F",list[i])
}
exit(0)
}

Output:
T 9,3,3,3,2,1,7,8,5
F 5,2,9,3,3,7,8,4,1
F 1,4,3,6,7,3,8,3,2
F 1,2,3,4,5,6,7,8,9
T 4,6,8,7,2,3,3,3,1

## C

#include <stdio.h>
#include <stdbool.h>

bool three_3s(const int *items, size_t len) {
int threes = 0;
while (len--)
if (*items++ == 3)
if (threes<3) threes++;
else return false;
else if (threes != 0 && threes != 3)
return false;
return true;
}

void print_list(const int *items, size_t len) {
while (len--) printf("%d ", *items++);
}

int main() {
int lists[][9] = {
{9,3,3,3,2,1,7,8,5},
{5,2,9,3,3,6,8,4,1},
{1,4,3,6,7,3,8,3,2},
{1,2,3,4,5,6,7,8,9},
{4,6,8,7,2,3,3,3,1}
};

size_t list_length = sizeof(lists[0]) / sizeof(int);
size_t n_lists = sizeof(lists) / sizeof(lists[0]);

for (size_t i=0; i<n_lists; i++) {
print_list(lists[i], list_length);
printf("-> %s\n", three_3s(lists[i], list_length) ? "true" : "false");
}

return 0;
}
Output:
9 3 3 3 2 1 7 8 5 -> true
5 2 9 3 3 6 8 4 1 -> false
1 4 3 6 7 3 8 3 2 -> false
1 2 3 4 5 6 7 8 9 -> false
4 6 8 7 2 3 3 3 1 -> true

## CLU

% See if a sequence has three consecutive 3s in it
% Works for any type that can be iterated over
three_3s = proc [T: type] (seq: T) returns (bool)
where T has elements: itertype (T) yields (int)
threes: int := 0

for n: int in T\$elements(seq) do
if n=3 then
if threes<3 then threes := threes + 1
else return(false)
end
else
if threes~=0 & threes~=3 then
return(false)
end
end
end
return(true)
end three_3s

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

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

po: stream := stream\$primary_output()
for list: si in ssi\$elements(lists) do
for i: int in si\$elements(list) do
stream\$puts(po, int\$unparse(i) || " ")
end
if three_3s[si](list) then
stream\$putl(po, "-> true")
else
stream\$putl(po, "-> false")
end
end
end start_up
Output:
9 3 3 3 2 1 7 8 5 -> true
5 2 9 3 3 6 8 4 1 -> false
1 4 3 6 7 3 8 3 2 -> false
1 2 3 4 5 6 7 8 9 -> false
4 6 8 7 2 3 3 3 1 -> true

## Draco

word i, n;
i := 0;
n := 0;
while i<dim(arr,1)
and (arr[i]=3 or n=0 or n=3)
and n<=3 do
if arr[i]=3 then n := n+1 fi;
i := i+1
od;
i=dim(arr,1) and n=3
corp

proc nonrec main() void:
[5][9]int list = (
(9,3,3,3,2,1,7,8,5),
(5,2,9,3,3,7,8,4,1),
(1,4,3,6,7,3,8,3,2),
(1,2,3,4,5,6,7,8,9),
(4,6,8,7,2,3,3,3,1)
);

word i, j;
for i from 0 upto 4 do
for j from 0 upto 8 do write(list[i][j]:2) od;
writeln(" -> ",
if three_adjacent(list[i]) then "true" else "false" fi)
od
corp
Output:
9 3 3 3 2 1 7 8 5 -> true
5 2 9 3 3 7 8 4 1 -> false
1 4 3 6 7 3 8 3 2 -> false
1 2 3 4 5 6 7 8 9 -> false
4 6 8 7 2 3 3 3 1 -> true

## F#

// Exactly three adjacent 3 in lists. Nigel Galloway: December 8th., 2021
let n=[[9;3;3;3;2;1;7;8;5];[5;2;9;3;3;7;8;4;1];[1;4;3;6;7;3;8;3;2];[1;2;3;4;5;6;7;8;9];[4;6;8;7;2;3;3;3;1]]
n|>List.iter(fun n->printfn "%A" (n|>List.windowed 3|>List.exists(fun(n::g::l::_)->n=3 && g=3 && l=3)))

Output:
true
false
false
false
true

## FreeBASIC

dim as integer list(1 to 5, 1 to 9) = {_
{9,3,3,3,2,1,7,8,5}, {5,2,9,3,3,7,8,4,1},_
{1,4,3,6,7,3,8,3,2}, {1,2,3,4,5,6,7,8,9},_
{4,6,8,7,2,3,3,3,1}}

dim as boolean go, pass
dim as integer i, j, c

for i = 1 to 5
go = false
pass = true
c = 0
for j = 1 to 9
if list(i, j) = 3 then
c+=1
go = true
else
if go = true and c<>3 then pass=false
go = false
end if
next j
print i;" ";
if c = 3 and pass then print true else print false
next i
Output:

1   true
2   false
3   false
4   false
5   true

## Go

package main

import "fmt"

func main() {
lists := [][]int{
{9, 3, 3, 3, 2, 1, 7, 8, 5},
{5, 2, 9, 3, 3, 7, 8, 4, 1},
{1, 4, 3, 6, 7, 3, 8, 3, 2},
{1, 2, 3, 4, 5, 6, 7, 8, 9},
{4, 6, 8, 7, 2, 3, 3, 3, 1},
{3, 3, 3, 1, 2, 4, 5, 1, 3},
{0, 3, 3, 3, 3, 7, 2, 2, 6},
{3, 3, 3, 3, 3, 4, 4, 4, 4},
}
for d := 1; d <= 4; d++ {
fmt.Printf("Exactly %d adjacent %d's:\n", d, d)
for _, list := range lists {
var indices []int
for i, e := range list {
if e == d {
indices = append(indices, i)
}
}
if len(indices) == d {
for i := 1; i < len(indices); i++ {
if indices[i]-indices[i-1] != 1 {
break
}
}
}
}
fmt.Println()
}
}
Output:
[9 3 3 3 2 1 7 8 5] -> true
[5 2 9 3 3 7 8 4 1] -> true
[1 4 3 6 7 3 8 3 2] -> true
[1 2 3 4 5 6 7 8 9] -> true
[4 6 8 7 2 3 3 3 1] -> true
[3 3 3 1 2 4 5 1 3] -> false
[0 3 3 3 3 7 2 2 6] -> false
[3 3 3 3 3 4 4 4 4] -> false

[9 3 3 3 2 1 7 8 5] -> false
[5 2 9 3 3 7 8 4 1] -> false
[1 4 3 6 7 3 8 3 2] -> false
[1 2 3 4 5 6 7 8 9] -> false
[4 6 8 7 2 3 3 3 1] -> false
[3 3 3 1 2 4 5 1 3] -> false
[0 3 3 3 3 7 2 2 6] -> true
[3 3 3 3 3 4 4 4 4] -> false

[9 3 3 3 2 1 7 8 5] -> true
[5 2 9 3 3 7 8 4 1] -> false
[1 4 3 6 7 3 8 3 2] -> false
[1 2 3 4 5 6 7 8 9] -> false
[4 6 8 7 2 3 3 3 1] -> true
[3 3 3 1 2 4 5 1 3] -> false
[0 3 3 3 3 7 2 2 6] -> false
[3 3 3 3 3 4 4 4 4] -> false

[9 3 3 3 2 1 7 8 5] -> false
[5 2 9 3 3 7 8 4 1] -> false
[1 4 3 6 7 3 8 3 2] -> false
[1 2 3 4 5 6 7 8 9] -> false
[4 6 8 7 2 3 3 3 1] -> false
[3 3 3 1 2 4 5 1 3] -> false
[0 3 3 3 3 7 2 2 6] -> false
[3 3 3 3 3 4 4 4 4] -> true

import Data.Bifunctor (bimap)
import Data.List (span)

nnPeers :: Int -> [Int] -> Bool
nnPeers n xs =
let p x = n == x
in uncurry (&&) \$
bimap
(p . length)
(not . any p)
(span p \$ dropWhile (not . p) xs)

--------------------------- TEST -------------------------
main :: IO ()
main =
putStrLn \$
unlines \$
fmap
(\xs -> show xs <> " -> " <> show (nnPeers 3 xs))
[ [9, 3, 3, 3, 2, 1, 7, 8, 5],
[5, 2, 9, 3, 3, 7, 8, 4, 1],
[1, 4, 3, 6, 7, 3, 8, 3, 2],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[4, 6, 8, 7, 2, 3, 3, 3, 1]
]
Output:
[9,3,3,3,2,1,7,8,5] -> True
[5,2,9,3,3,7,8,4,1] -> False
[1,4,3,6,7,3,8,3,2] -> False
[1,2,3,4,5,6,7,8,9] -> False
[4,6,8,7,2,3,3,3,1] -> True

## JavaScript

(() => {
"use strict";

// ------- N INSTANCES OF N AND ALL CONTIGUOUS -------

// nnPeers :: Int -> [Int] -> Bool
const nnPeers = n =>
// True if xs contains exactly n instances of n
// and the instances are all contiguous.
xs => {
const
p = x => n === x,
mbi = xs.findIndex(p);

return -1 !== mbi ? (() => {
const
rest = xs.slice(mbi),
sample = rest.slice(0, n);

return n === sample.length && (
sample.every(p) && (
!rest.slice(n).some(p)
)
);
})() : false;
};

// ---------------------- TEST -----------------------
const main = () => [
[9, 3, 3, 3, 2, 1, 7, 8, 5],
[5, 2, 9, 3, 3, 7, 8, 4, 1],
[1, 4, 3, 6, 7, 3, 8, 3, 2],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[4, 6, 8, 7, 2, 3, 3, 3, 1]
]
.map(
xs => `\${JSON.stringify(xs)} -> \${nnPeers(3)(xs)}`
)
.join("\n");

return main();
})();
Output:
[9,3,3,3,2,1,7,8,5] -> true
[5,2,9,3,3,7,8,4,1] -> false
[1,4,3,6,7,3,8,3,2] -> false
[1,2,3,4,5,6,7,8,9] -> false
[4,6,8,7,2,3,3,3,1] -> true

## jq

Works with: jq

Works with gojq, the Go implementation of jq

The test cases, and the output, are exactly as for entry at #Wren.

Preliminaries

def count(s): reduce s as \$x (0; .+1);

def lists : [
[9,3,3,3,2,1,7,8,5],
[5,2,9,3,3,7,8,4,1],
[1,4,3,6,7,3,8,3,2],
[1,2,3,4,5,6,7,8,9],
[4,6,8,7,2,3,3,3,1],
[3,3,3,1,2,4,5,1,3],
[0,3,3,3,3,7,2,2,6],
[3,3,3,3,3,4,4,4,4]
];

def threeConsecutiveThrees:
count(.[] == 3 // empty) == 3
and index([3,3,3]);

(lists[]
| "\(.) -> \(threeConsecutiveThrees)")

Output:

As for #Wren.

## Julia

function onlyconsecutivein(a::Vector{T}, lis::Vector{T}) where T
return any(i -> a == lis[i:i+length(a)-1], 1:length(lis)-length(a)+1) &&
all(count(x -> x == a[i], lis) == count(x -> x == a[i], a) for i in eachindex(a))
end

needle = [3, 3, 3]
for haystack in [
[9,3,3,3,2,1,7,8,5],
[5,2,9,3,3,7,8,4,1],
[1,4,3,3,3,3,8,3,2],
[1,2,3,4,5,6,7,8,9],
[4,6,8,7,2,3,3,3,1]]
println("\$needle in \$haystack: ", onlyconsecutivein(needle, haystack))
end

needle = [3, 2, 3]
for haystack in [
[9,3,3,3,2,3,7,8,5],
[5,6,9,1,3,2,3,4,1],
[1,4,3,6,7,3,8,3,2],
[1,2,3,4,5,6,7,8,9],
[4,6,8,7,2,3,2,3,1]]
println("\$needle in \$haystack: ", onlyconsecutivein(needle, haystack))
end

Output:
[3, 3, 3] in [9, 3, 3, 3, 2, 1, 7, 8, 5]: true
[3, 3, 3] in [5, 2, 9, 3, 3, 7, 8, 4, 1]: false
[3, 3, 3] in [1, 4, 3, 3, 3, 3, 8, 3, 2]: false
[3, 3, 3] in [1, 2, 3, 4, 5, 6, 7, 8, 9]: false
[3, 3, 3] in [4, 6, 8, 7, 2, 3, 3, 3, 1]: true
[3, 2, 3] in [9, 3, 3, 3, 2, 3, 7, 8, 5]: false
[3, 2, 3] in [5, 6, 9, 1, 3, 2, 3, 4, 1]: true
[3, 2, 3] in [1, 4, 3, 6, 7, 3, 8, 3, 2]: false
[3, 2, 3] in [1, 2, 3, 4, 5, 6, 7, 8, 9]: false
[3, 2, 3] in [4, 6, 8, 7, 2, 3, 2, 3, 1]: false

## Mathematica / Wolfram Language

(# -> MemberQ[Partition[#, 3, 1], {3, 3, 3}]) & /@ {{9, 3, 3, 3, 2, 1,
7, 8, 5}, {5, 2, 9, 3, 3, 7, 8, 4, 1}, {1, 4, 3, 6, 7, 3, 8, 3,
2}, {1, 2, 3, 4, 5, 6, 7, 8, 9}, {4, 6, 8, 7, 2, 3, 3, 3,
1}} // TableForm
Output:

{9,3,3,3,2,1,7,8,5}->True {5,2,9,3,3,7,8,4,1}->False {1,4,3,6,7,3,8,3,2}->False {1,2,3,4,5,6,7,8,9}->False {4,6,8,7,2,3,3,3,1}->True

## Perl

### Specific

#!/usr/bin/perl

use warnings;

my @lists = (
[9,3,3,3,2,1,7,8,5],
[5,2,9,3,3,7,8,4,1],
[1,4,3,6,7,3,8,3,2],
[1,2,3,4,5,6,7,8,9],
[4,6,8,7,2,3,3,3,1]);

for my \$ref ( @lists )
{
my @n = grep \$ref->[\$_] == 3, 0 .. \$#\$ref;
print "@\$ref => ",
@n == 3 && \$n[0] == \$n[1] - 1 && \$n[1] == \$n[2] - 1 ? 'true' : 'false',
"\n";
}
Output:
9 3 3 3 2 1 7 8 5 => true
5 2 9 3 3 7 8 4 1 => false
1 4 3 6 7 3 8 3 2 => false
1 2 3 4 5 6 7 8 9 => false
4 6 8 7 2 3 3 3 1 => true

### General

use strict;
use warnings;

my @lists = (
[ < 9 3 3 3 2 1 7 8 5 > ],
[ < 5 2 9 3 3 7 8 4 1 > ],
[ < 1 4 3 6 7 3 8 3 2 > ],
[ < 1 2 3 4 5 6 7 8 9 > ],
[ < 4 6 8 7 2 3 3 3 1 > ],
[ < 3 3 3 1 2 4 5 1 3 > ],
[ < 0 3 9 3 3 7 2 2 6 > ],
[ < 3 3 3 3 3 4 4 4 4 > ],
);

print ' 'x21 . '0x0 1x1 2x2 3x3 4x4' . "\n";
for my \$ref ( @lists ) {
print "@\$ref: ";
for my \$n (0..4) {
my @i = grep \$ref->[\$_] == \$n, 0 .. \$#\$ref;
print ' ', \$n==0 && !@i || @i == \$n && (\$n==1 || (\$n-1 == grep \$i[\$_-1]+1 == \$i[\$_], 1..\$n-1)) ? 'Y' : 'N';
}
print "\n";
}
Output:
0x0 1x1 2x2 3x3 4x4
9 3 3 3 2 1 7 8 5:    Y   Y   N   Y   N
5 2 9 3 3 7 8 4 1:    Y   Y   N   N   N
1 4 3 6 7 3 8 3 2:    Y   Y   N   N   N
1 2 3 4 5 6 7 8 9:    Y   Y   N   N   N
4 6 8 7 2 3 3 3 1:    Y   Y   N   Y   N
3 3 3 1 2 4 5 1 3:    Y   N   N   N   N
0 3 9 3 3 7 2 2 6:    N   N   Y   N   N
3 3 3 3 3 4 4 4 4:    Y   N   N   N   Y

## Phix

with javascript_semantics
procedure test(integer n, sequence s)
sequence f = find_all(n,s)
printf(1,"%v: %t\n",{s,length(f)=n and f[\$]-f[1]=n-1})
end procedure

papply(true,test,{3,{{9, 3, 3, 3, 2, 1, 7, 8, 5},
{5, 2, 9, 3, 3, 7, 8, 4, 1},
{1, 4, 3, 6, 7, 3, 8, 3, 2},
{1, 2, 3, 4, 5, 6, 7, 8, 9},
{4, 6, 8, 7, 2, 3, 3, 3, 1}}})
Output:

(Agrees with Raku and Wren with a for loop and the three extra tests)

{9,3,3,3,2,1,7,8,5}: true
{5,2,9,3,3,7,8,4,1}: false
{1,4,3,6,7,3,8,3,2}: false
{1,2,3,4,5,6,7,8,9}: false
{4,6,8,7,2,3,3,3,1}: true

## Python

'''N instances of N and all contiguous'''

from itertools import dropwhile, takewhile

# nnPeers :: Int -> [Int] -> Bool
def nnPeers(n):
'''True if xs contains exactly n instances of n
and all instances are contiguous.
'''

def p(x):
return n == x

def go(xs):
fromFirstMatch = list(dropwhile(
lambda v: not p(v),
xs
))
ns = list(takewhile(p, fromFirstMatch))
rest = fromFirstMatch[len(ns):]

return p(len(ns)) and (
not any(p(x) for x in rest)
)

return go

# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''Tests for N=3'''
print(
'\n'.join([
f'{xs} -> {nnPeers(3)(xs)}' for xs in [
[9, 3, 3, 3, 2, 1, 7, 8, 5],
[5, 2, 9, 3, 3, 7, 8, 4, 1],
[1, 4, 3, 6, 7, 3, 8, 3, 2],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[4, 6, 8, 7, 2, 3, 3, 3, 1]
]
])
)

# MAIN ---
if __name__ == '__main__':
main()
Output:
[9, 3, 3, 3, 2, 1, 7, 8, 5] -> True
[5, 2, 9, 3, 3, 7, 8, 4, 1] -> False
[1, 4, 3, 6, 7, 3, 8, 3, 2] -> False
[1, 2, 3, 4, 5, 6, 7, 8, 9] -> False
[4, 6, 8, 7, 2, 3, 3, 3, 1] -> True

## Raku

Generalized

for 1 .. 4 -> \$n {

say "\nExactly \$n {\$n}s, and they are consecutive:";

say .gist, ' ', lc (.Bag{\$n} == \$n) && ( so .rotor(\$n=>-(\$n - 1)).grep: *.all == \$n ) for
[9,3,3,3,2,1,7,8,5],
[5,2,9,3,3,7,8,4,1],
[1,4,3,6,7,3,8,3,2],
[1,2,3,4,5,6,7,8,9],
[4,6,8,7,2,3,3,3,1],
[3,3,3,1,2,4,5,1,3],
[0,3,3,3,3,7,2,2,6],
[3,3,3,3,3,4,4,4,4]
}
Output:
Exactly 1 1s, and they are consecutive:
[9 3 3 3 2 1 7 8 5] true
[5 2 9 3 3 7 8 4 1] true
[1 4 3 6 7 3 8 3 2] true
[1 2 3 4 5 6 7 8 9] true
[4 6 8 7 2 3 3 3 1] true
[3 3 3 1 2 4 5 1 3] false
[0 3 3 3 3 7 2 2 6] false
[3 3 3 3 3 4 4 4 4] false

Exactly 2 2s, and they are consecutive:
[9 3 3 3 2 1 7 8 5] false
[5 2 9 3 3 7 8 4 1] false
[1 4 3 6 7 3 8 3 2] false
[1 2 3 4 5 6 7 8 9] false
[4 6 8 7 2 3 3 3 1] false
[3 3 3 1 2 4 5 1 3] false
[0 3 3 3 3 7 2 2 6] true
[3 3 3 3 3 4 4 4 4] false

Exactly 3 3s, and they are consecutive:
[9 3 3 3 2 1 7 8 5] true
[5 2 9 3 3 7 8 4 1] false
[1 4 3 6 7 3 8 3 2] false
[1 2 3 4 5 6 7 8 9] false
[4 6 8 7 2 3 3 3 1] true
[3 3 3 1 2 4 5 1 3] false
[0 3 3 3 3 7 2 2 6] false
[3 3 3 3 3 4 4 4 4] false

Exactly 4 4s, and they are consecutive:
[9 3 3 3 2 1 7 8 5] false
[5 2 9 3 3 7 8 4 1] false
[1 4 3 6 7 3 8 3 2] false
[1 2 3 4 5 6 7 8 9] false
[4 6 8 7 2 3 3 3 1] false
[3 3 3 1 2 4 5 1 3] false
[0 3 3 3 3 7 2 2 6] false
[3 3 3 3 3 4 4 4 4] true

## Ring

see "working..." + nl

list = List(5)
list[1] = [9,3,3,3,2,1,7,8,5]
list[2] = [5,2,9,3,3,7,8,4,1]
list[3] = [1,4,3,6,7,3,8,3,2]
list[4] = [1,2,3,4,5,6,7,8,9]
list[5] = [4,6,8,7,2,3,3,3,1]

for n = 1 to 5
good = 0
cnt = 0
len = len(list[n])
for p = 1 to len
if list[n][p] = 3
good++
ok
next
if good = 3
for m = 1 to len-2
if list[n][m] = 3 and list[n][m+1] = 3 and list[n][m+2] = 3
cnt++
ok
next
ok
showarray(list[n])
if cnt = 1
see " > " + "true" + nl
else
see " > " + "false" + nl
ok
next

see "done..." + nl

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:
working...
[9,3,3,3,2,1,7,8,5] > true
[5,2,9,3,3,7,8,4,1] > false
[1,4,3,6,7,3,8,3,2] > false
[1,2,3,4,5,6,7,8,9] > false
[4,6,8,7,2,3,3,3,1] > true
done...

## Ruby

Using the Raku/Wren testset:

tests = [[9,3,3,3,2,1,7,8,5],
[5,2,9,3,3,7,8,4,1],
[1,4,3,6,7,3,8,3,2],
[1,2,3,4,5,6,7,8,9],
[4,6,8,7,2,3,3,3,1],
[3,3,3,1,2,4,5,1,3],
[0,3,3,3,3,7,2,2,6],
[3,3,3,3,3,4,4,4,4]]

(1..4).each do |n|
c = [n]*n
puts "Contains exactly #{n} #{n}s, consecutive:"
tests.each { |t| puts "#{t.inspect} : #{t.count(n)==n && t.each_cons(n).any?{|chunk| chunk == c }}" }
end

Output:
Contains exactly 1 1s, consecutive:
[9, 3, 3, 3, 2, 1, 7, 8, 5] : true
[5, 2, 9, 3, 3, 7, 8, 4, 1] : true
[1, 4, 3, 6, 7, 3, 8, 3, 2] : true
[1, 2, 3, 4, 5, 6, 7, 8, 9] : true
[4, 6, 8, 7, 2, 3, 3, 3, 1] : true
[3, 3, 3, 1, 2, 4, 5, 1, 3] : false
[0, 3, 3, 3, 3, 7, 2, 2, 6] : false
[3, 3, 3, 3, 3, 4, 4, 4, 4] : false
Contains exactly 2 2s, consecutive:
[9, 3, 3, 3, 2, 1, 7, 8, 5] : false
[5, 2, 9, 3, 3, 7, 8, 4, 1] : false
[1, 4, 3, 6, 7, 3, 8, 3, 2] : false
[1, 2, 3, 4, 5, 6, 7, 8, 9] : false
[4, 6, 8, 7, 2, 3, 3, 3, 1] : false
[3, 3, 3, 1, 2, 4, 5, 1, 3] : false
[0, 3, 3, 3, 3, 7, 2, 2, 6] : true
[3, 3, 3, 3, 3, 4, 4, 4, 4] : false
Contains exactly 3 3s, consecutive:
[9, 3, 3, 3, 2, 1, 7, 8, 5] : true
[5, 2, 9, 3, 3, 7, 8, 4, 1] : false
[1, 4, 3, 6, 7, 3, 8, 3, 2] : false
[1, 2, 3, 4, 5, 6, 7, 8, 9] : false
[4, 6, 8, 7, 2, 3, 3, 3, 1] : true
[3, 3, 3, 1, 2, 4, 5, 1, 3] : false
[0, 3, 3, 3, 3, 7, 2, 2, 6] : false
[3, 3, 3, 3, 3, 4, 4, 4, 4] : false
Contains exactly 4 4s, consecutive:
[9, 3, 3, 3, 2, 1, 7, 8, 5] : false
[5, 2, 9, 3, 3, 7, 8, 4, 1] : false
[1, 4, 3, 6, 7, 3, 8, 3, 2] : false
[1, 2, 3, 4, 5, 6, 7, 8, 9] : false
[4, 6, 8, 7, 2, 3, 3, 3, 1] : false
[3, 3, 3, 1, 2, 4, 5, 1, 3] : false
[0, 3, 3, 3, 3, 7, 2, 2, 6] : false
[3, 3, 3, 3, 3, 4, 4, 4, 4] : true

## Sidef

func contains_n_consecutive_objs(arr, n, obj) {

# In Sidef >= 3.99, we can also say:
# arr.contains(n.of(obj)...)

arr.each_cons(n, {|*a|
if (a.all { _ == obj }) {
return true
}
})

return false
}

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

lists.each {|list|
say (list, " => ", contains_n_consecutive_objs(list, 3, 3))
}
Output:
[9, 3, 3, 3, 2, 1, 7, 8, 5] => true
[5, 2, 9, 3, 3, 7, 8, 4, 1] => false
[1, 4, 3, 6, 7, 3, 8, 3, 2] => false
[1, 2, 3, 4, 5, 6, 7, 8, 9] => false
[4, 6, 8, 7, 2, 3, 3, 3, 1] => true

## Vlang

Translation of: go

fn main() {
lists := [
[9, 3, 3, 3, 2, 1, 7, 8, 5],
[5, 2, 9, 3, 3, 7, 8, 4, 1],
[1, 4, 3, 6, 7, 3, 8, 3, 2],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[4, 6, 8, 7, 2, 3, 3, 3, 1],
[3, 3, 3, 1, 2, 4, 5, 1, 3],
[0, 3, 3, 3, 3, 7, 2, 2, 6],
[3, 3, 3, 3, 3, 4, 4, 4, 4],
]
for d := 1; d <= 4; d++ {
for list in lists {
mut indices := []int{}
for i, e in list {
if e == d {
indices << i
}
}
if indices.len == d {
for i in 1..indices.len {
if indices[i]-indices[i-1] != 1 {
break
}
}
}
}
println('')
}
}
Output:
[9, 3, 3, 3, 2, 1, 7, 8, 5] -> true
[5, 2, 9, 3, 3, 7, 8, 4, 1] -> false
[1, 4, 3, 6, 7, 3, 8, 3, 2] -> false
[1, 2, 3, 4, 5, 6, 7, 8, 9] -> false
[4, 6, 8, 7, 2, 3, 3, 3, 1] -> true
[3, 3, 3, 1, 2, 4, 5, 1, 3] -> false
[0, 3, 3, 3, 3, 7, 2, 2, 6] -> false
[3, 3, 3, 3, 3, 4, 4, 4, 4] -> false

## Wren

Library: Wren-seq
import "./seq" for Lst

var lists = [
[9,3,3,3,2,1,7,8,5],
[5,2,9,3,3,7,8,4,1],
[1,4,3,6,7,3,8,3,2],
[1,2,3,4,5,6,7,8,9],
[4,6,8,7,2,3,3,3,1],
[3,3,3,1,2,4,5,1,3],
[0,3,3,3,3,7,2,2,6],
[3,3,3,3,3,4,4,4,4]
]
for (list in lists) {
var condition = list.count { |n| n == 3 } == 3 && Lst.isSliceOf(list, [3, 3, 3])
System.print("%(list) -> %(condition)")
}
Output:
[9, 3, 3, 3, 2, 1, 7, 8, 5] -> true
[5, 2, 9, 3, 3, 7, 8, 4, 1] -> false
[1, 4, 3, 6, 7, 3, 8, 3, 2] -> false
[1, 2, 3, 4, 5, 6, 7, 8, 9] -> false
[4, 6, 8, 7, 2, 3, 3, 3, 1] -> true
[3, 3, 3, 1, 2, 4, 5, 1, 3] -> false
[0, 3, 3, 3, 3, 7, 2, 2, 6] -> false
[3, 3, 3, 3, 3, 4, 4, 4, 4] -> false

Or, more generally, replacing everything after 'lists' with the following:

for (d in 1..4) {
for (list in lists) {
var condition = list.count { |n| n == d } == d && Lst.isSliceOf(list, [d] * d)
System.print("%(list) -> %(condition)")
}
System.print()
}
Output:
[9, 3, 3, 3, 2, 1, 7, 8, 5] -> true
[5, 2, 9, 3, 3, 7, 8, 4, 1] -> true
[1, 4, 3, 6, 7, 3, 8, 3, 2] -> true
[1, 2, 3, 4, 5, 6, 7, 8, 9] -> true
[4, 6, 8, 7, 2, 3, 3, 3, 1] -> true
[3, 3, 3, 1, 2, 4, 5, 1, 3] -> false
[0, 3, 3, 3, 3, 7, 2, 2, 6] -> false
[3, 3, 3, 3, 3, 4, 4, 4, 4] -> false

[9, 3, 3, 3, 2, 1, 7, 8, 5] -> false
[5, 2, 9, 3, 3, 7, 8, 4, 1] -> false
[1, 4, 3, 6, 7, 3, 8, 3, 2] -> false
[1, 2, 3, 4, 5, 6, 7, 8, 9] -> false
[4, 6, 8, 7, 2, 3, 3, 3, 1] -> false
[3, 3, 3, 1, 2, 4, 5, 1, 3] -> false
[0, 3, 3, 3, 3, 7, 2, 2, 6] -> true
[3, 3, 3, 3, 3, 4, 4, 4, 4] -> false

[9, 3, 3, 3, 2, 1, 7, 8, 5] -> true
[5, 2, 9, 3, 3, 7, 8, 4, 1] -> false
[1, 4, 3, 6, 7, 3, 8, 3, 2] -> false
[1, 2, 3, 4, 5, 6, 7, 8, 9] -> false
[4, 6, 8, 7, 2, 3, 3, 3, 1] -> true
[3, 3, 3, 1, 2, 4, 5, 1, 3] -> false
[0, 3, 3, 3, 3, 7, 2, 2, 6] -> false
[3, 3, 3, 3, 3, 4, 4, 4, 4] -> false

[9, 3, 3, 3, 2, 1, 7, 8, 5] -> false
[5, 2, 9, 3, 3, 7, 8, 4, 1] -> false
[1, 4, 3, 6, 7, 3, 8, 3, 2] -> false
[1, 2, 3, 4, 5, 6, 7, 8, 9] -> false
[4, 6, 8, 7, 2, 3, 3, 3, 1] -> false
[3, 3, 3, 1, 2, 4, 5, 1, 3] -> false
[0, 3, 3, 3, 3, 7, 2, 2, 6] -> false
[3, 3, 3, 3, 3, 4, 4, 4, 4] -> true

## XPL0

func Check(L);  \Return 'true' if three adjacent 3's
int L, C, I, J;
def Size = 9; \number of items in each List
[C:= 0;
for I:= 0 to Size-1 do
if L(I) = 3 then [C:= C+1; J:= I];
if C # 3 then return false; \must have exactly three 3's
return L(J-1)=3 & L(J-2)=3; \the 3's must be adjacent
];

int List(5+1), I;
[List(1):= [9,3,3,3,2,1,7,8,5];
List(2):= [5,2,9,3,3,7,8,4,1];
List(3):= [1,4,3,6,7,3,8,3,2];
List(4):= [1,2,3,4,5,6,7,8,9];
List(5):= [4,6,8,7,2,3,3,3,1];
for I:= 1 to 5 do
[IntOut(0, I);
Text(0, if Check(List(I)) then " true" else " false");
CrLf(0);
];
]
Output:
1 true
2 false
3 false
4 false
5 true