Exactly three adjacent 3 in lists

From Rosetta Code
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.
Task

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[edit]

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[edit]

	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

Ada[edit]

with Ada.Text_Io;  use Ada.Text_Io;

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[edit]

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[edit]

------- 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[edit]

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[edit]

# syntax: GAWK -f EXACTLY_THREE_ADJACENT_3_IN_LISTS.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[edit]

#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[edit]

% 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[edit]

proc nonrec three_adjacent([*]int arr) bool:
    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#[edit]

// 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[edit]

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[edit]

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)
                }
            }
            adjacent := false
            if len(indices) == d {
                adjacent = true
                for i := 1; i < len(indices); i++ {
                    if indices[i]-indices[i-1] != 1 {
                        adjacent = false
                        break
                    }
                }
            }
            fmt.Printf("%v -> %t\n", list, adjacent)
        }
        fmt.Println()
    }
}
Output:
Exactly 1 adjacent 1's:
[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 adjacent 2's:
[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 adjacent 3's:
[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 adjacent 4's:
[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

Haskell[edit]

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[edit]

(() => {
    "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[edit]

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);

The task

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]);

"Exactly three adjacent 3's:",
(lists[] 
 | "\(.) -> \(threeConsecutiveThrees)")
Output:

As for #Wren.

Julia[edit]

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[edit]

(# -> 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[edit]

Specific[edit]

#!/usr/bin/perl
 
use strict; # https://rosettacode.org/wiki/Exactly_three_adjacent_3_in_lists
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[edit]

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[edit]

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

printf(1,"\nExactly %d adjacent %d's:\n",3)
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)

Exactly 3 adjacent 3's:
{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[edit]

'''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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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++ {
        println("Exactly $d adjacent $d's:")
        for list in lists {
            mut indices := []int{}
            for i, e in list {
                if e == d {
                    indices << i
                }
            }
            mut adjacent := false
            if indices.len == d {
                adjacent = true
                for i in 1..indices.len {
                    if indices[i]-indices[i-1] != 1 {
                        adjacent = false
                        break
                    }
                }
            }
            println("$list -> $adjacent")
        }
        println('')
    }
}
Output:
Exactly three adjacent 3's:
[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[edit]

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]
]
System.print("Exactly three adjacent 3's:")
for (list in lists) {
    var condition = list.count { |n| n == 3 } == 3 && Lst.isSliceOf(list, [3, 3, 3])
    System.print("%(list) -> %(condition)")
}
Output:
Exactly three adjacent 3's:
[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) {
    System.print("Exactly %(d) adjacent %(d)'s:")
    for (list in lists) {
        var condition = list.count { |n| n == d } == d && Lst.isSliceOf(list, [d] * d)
        System.print("%(list) -> %(condition)")
    }
    System.print()
}
Output:
Exactly 1 adjacent 1's:
[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 adjacent 2's:
[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 adjacent 3's:
[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 adjacent 4's:
[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[edit]

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