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.
- 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'.
That is, print false unless there is exactly one run of three '3's and there is no other occurrence of a '3'.
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
Ada
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
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
# 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
BASIC
Applesoft BASIC
The GW-BASIC solution works without any changes.
BASIC256
arraybase 1
dim list(5, 9)
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}}
for i = 1 to list[?][]
go = false
pass = true
c = 0
for j = 1 to list[][?]
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:
Similar to FreeBASIC entry.
Chipmunk Basic
100 cls
110 data 9,3,3,3,2,1,7,8,5
120 data 5,2,9,3,3,7,8,4,1
130 data 1,4,3,6,7,3,8,3,2
140 data 1,2,3,4,5,6,7,8,9
150 data 4,6,8,7,2,3,3,3,1
160 dim lista(5,9)
170 for i = 1 to ubound(lista)
180 for j = 1 to ubound(lista,2)
190 read lista(i,j)
200 next j
210 next i
220 for i = 1 to ubound(lista)
230 go = false
240 pass = true
250 c = 0
260 for j = 1 to ubound(lista,2)
270 if lista(i,j) = 3 then
280 c = c+1
290 go = true
300 else
310 if go = true and c <> 3 then pass = false
320 go = false
330 endif
340 next j
350 print i;" ";
360 if c = 3 and pass then print "True" else print "False"
370 next i
380 end
- Output:
Similar to FreeBASIC entry.
GW-BASIC
100 CLS : rem 100 HOME for Applesoft BASIC
110 LET f = 0
115 LET t = 1
120 DATA 9,3,3,3,2,1,7,8,5
130 DATA 5,2,9,3,3,7,8,4,1
140 DATA 1,4,3,6,7,3,8,3,2
150 DATA 1,2,3,4,5,6,7,8,9
160 DATA 4,6,8,7,2,3,3,3,1
170 DIM l(5,9)
180 FOR i = 1 TO 5
190 FOR j = 1 TO 9
200 READ l(i,j)
210 NEXT j
220 NEXT i
230 FOR i = 1 TO 5
240 LET g = f
250 LET p = t
260 LET c = 0
270 FOR j = t TO 9
280 IF l(i,j) = 3 THEN LET c = c+1
281 IF l(i,j) = 3 THEN LET g = t
282 IF l(i,j) <> 3 THEN GOSUB 340
283 IF l(i,j) <> 3 THEN LET g = f
290 NEXT j
300 PRINT i; " ";
310 IF c = 3 AND p = t THEN PRINT "true"
315 IF c <> 3 OR p <> t THEN PRINT "false"
320 NEXT i
330 END
340 IF g = t AND c <> 3 THEN LET p = f
350 RETURN
- Output:
Similar to FreeBASIC entry.
Minimal BASIC
100 LET F = 0
110 LET T = 1
120 DATA 9,3,3,3,2,1,7,8,5
130 DATA 5,2,9,3,3,7,8,4,1
140 DATA 1,4,3,6,7,3,8,3,2
150 DATA 1,2,3,4,5,6,7,8,9
160 DATA 4,6,8,7,2,3,3,3,1
170 DIM L(5,9)
180 FOR I = 1 TO 5
190 FOR J = 1 TO 9
200 READ L(I,J)
210 NEXT J
220 NEXT I
230 FOR I = 1 TO 5
240 LET G = F
250 LET P = T
260 LET C = 0
270 FOR J = T TO 9
280 IF L(I,J) = 3 THEN 300
290 IF L(I,J) <> 3 THEN 330
300 LET C = C + 1
310 LET G = T
320 GOTO 390
330 IF G = T THEN 360
340 LET G = F
350 GOTO 390
360 IF C <> 3 THEN 380
370 GOTO 390
380 LET P = F
390 NEXT J
400 PRINT I; " ";
410 IF C = 3 THEN 430
420 GOTO 440
430 IF P = T THEN 460
440 IF C <> 3 THEN 480
450 IF P <> T THEN 480
460 PRINT "TRUE"
470 GOTO 490
480 PRINT "FALSE"
490 NEXT I
500 END
- Output:
Similar to FreeBASIC entry.
MSX Basic
100 CLS
110 false = 0 : true = 1
120 DATA 9,3,3,3,2,1,7,8,5
130 DATA 5,2,9,3,3,7,8,4,1
140 DATA 1,4,3,6,7,3,8,3,2
150 DATA 1,2,3,4,5,6,7,8,9
160 DATA 4,6,8,7,2,3,3,3,1
170 DIM lis(5,9)
180 FOR i = 1 TO 5
190 FOR j = 1 TO 9
200 READ lis(i,j)
210 NEXT j
220 NEXT i
230 FOR i = 1 TO 5
240 go = false
250 pass = true
260 c = 0
270 FOR j = true TO 9
280 IF lis(i,j) = 3 THEN c = c+1 : go = true ELSE GOSUB 340 : go = false
290 NEXT j
300 PRINT i;" ";
310 IF c = 3 AND pass = true THEN PRINT "true" ELSE PRINT "false"
320 NEXT i
330 END
340 IF go = true AND c <> 3 THEN pass = false
350 RETURN
- Output:
Similar to FreeBASIC entry.
PureBasic
OpenConsole()
Dim lista.i(5, 9)
Define.b go, pass
Define.i i, j, c
For i = 1 To ArraySize(lista())
For j = 1 To ArraySize(lista(),2)
Read.i lista(i,j)
Next j
Next i
For i = 1 To ArraySize(lista())
go = #False
pass = #True
c = 0
For j = 1 To ArraySize(lista(),2)
If lista(i, j) = 3:
c + 1
go = #True
Else
If go = #True And c <> 3:
pass = #False
EndIf
go = #False
EndIf
Next j
Print(Str(i) + #TAB$)
If c = 3 And pass = #True:
PrintN("True")
Else
PrintN("False")
EndIf
Next i
PrintN(#CRLF$ + "--- terminado, pulsa RETURN---"): Input()
CloseConsole()
DataSection
Data.i 9,3,3,3,2,1,7,8,5
Data.i 5,2,9,3,3,7,8,4,1
Data.i 1,4,3,6,7,3,8,3,2
Data.i 1,2,3,4,5,6,7,8,9
Data.i 4,6,8,7,2,3,3,3,1
EndDataSection
- Output:
Similar to FreeBASIC entry.
QBasic
CONST False = 0: True = NOT False
DATA 9,3,3,3,2,1,7,8,5
DATA 5,2,9,3,3,7,8,4,1
DATA 1,4,3,6,7,3,8,3,2
DATA 1,2,3,4,5,6,7,8,9
DATA 4,6,8,7,2,3,3,3,1
DIM lista(1 TO 5, 1 TO 9) AS INTEGER
FOR i = 1 TO UBOUND(lista)
FOR j = 1 TO UBOUND(lista, 2)
READ lista(i, j)
NEXT j
NEXT i
FOR i = 1 TO UBOUND(lista)
go = False
pass = True
c = 0
FOR j = 1 TO UBOUND(lista, 2)
IF lista(i, j) = 3 THEN
c = 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:
Similar to FreeBASIC entry.
QB64
The QBasic solution works without any changes.
Quite BASIC
The GW-BASIC solution works without any changes.
True BASIC
DIM lista(5, 9)
DATA 9, 3, 3, 3, 2, 1, 7, 8, 5
DATA 5, 2, 9, 3, 3, 7, 8, 4, 1
DATA 1, 4, 3, 6, 7, 3, 8, 3, 2
DATA 1, 2, 3, 4, 5, 6, 7, 8, 9
DATA 4, 6, 8, 7, 2, 3, 3, 3, 1
FOR i = 1 TO UBOUND(lista,1)
FOR j = 1 TO UBOUND(lista,2)
READ lista(i, j)
NEXT j
NEXT i
FOR i = 1 TO UBOUND(lista,1)
LET go = 0
LET pass = 1
LET c = 0
FOR j = 1 TO UBOUND(lista,2)
IF lista(i, j) = 3 THEN
LET c = c + 1
LET go = 1
ELSE
IF go = 1 AND c <> 3 THEN LET pass = 0
LET go = 0
END IF
NEXT j
PRINT i; " ";
IF c = 3 AND pass <> 0 THEN PRINT "True" ELSE PRINT "False"
NEXT i
END
- Output:
Similar to FreeBASIC entry.
Yabasic
dim lista(5, 9)
data 9,3,3,3,2,1,7,8,5
data 5,2,9,3,3,7,8,4,1
data 1,4,3,6,7,3,8,3,2
data 1,2,3,4,5,6,7,8,9
data 4,6,8,7,2,3,3,3,1
for i = 1 to arraysize(lista(),1)
for j = 1 to arraysize(lista(),2)
read lista(i,j)
next j
next i
for i = 1 to arraysize(lista(),1)
go = false
pass = true
c = 0
for j = 1 to arraysize(lista(),2)
if lista(i, j) = 3 then
c = c + 1
go = true
else
if go = true and c <> 3 pass = false
go = false
end if
next j
print i, " ";
if c = 3 and pass then print "True" else print "False" : fi
next i
- Output:
Similar to FreeBASIC entry.
BQN
"false"‿"true"⊏˜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:
⟨ "true" "false" "false" "false" "true" ⟩
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
C++
#include <algorithm>
#include <cstdint>
#include <iomanip>
#include <iostream>
#include <ranges>
#include <vector>
int main() {
std::vector<std::vector<int32_t>> 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 ( uint32_t i = 0; i < lists.size(); ++i ) {
bool count_started = false;
bool success = true;
int32_t count = 0;
for ( uint32_t j = 0; j < lists[i].size(); ++j ) {
if ( lists[i][j] == 3 ) {
count += 1;
count_started = true;
} else {
if ( count_started && count != 3 ) {
success = false;
}
count_started = false;
}
}
std::ranges::copy(lists[i], std::ostream_iterator<int32_t>(std::cout, " "));
std::cout << "=> " << std::boolalpha << success << std::endl;
}
}
- 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
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
Delphi
var ThreeList: array [0..4,0..8] of integer = (
(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));
function CountThrees(TA: array of integer): integer;
{Count the number threes in array}
var I,Cnt: integer;
begin
Result:=0;
for I:=0 to High(TA) do
if TA[I]=3 then
begin
Inc(Result);
if Result=3 then exit;
end
else Result:=0;
end;
procedure TestThreeArrays(Memo: TMemo);
var I,J: integer;
var B: boolean;
var S: string;
begin
for I:=0 to High(ThreeList) do
begin
S:='';
for J:=0 to High(ThreeList[I]) do
begin
if J>0 then S:=S+',';
S:=S+IntToStr(ThreeList[I][J])
end;
if CountThrees(ThreeList[I])=3 then S:=S+' True'
else S:=S+' False';
Memo.Lines.Add(S);
end;
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
Draco
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
DuckDB
# Exactly three occurrences of item, all in one run
# Assumption: the item contains no chr(0)
create or replace function threesome(lst, item) as (
select list_filter(lst, x -> x = item).length() = 3
and list_transform(lst, x -> x || chr(0))
.array_to_string('')
.regexp_matches( '(' || item::VARCHAR || chr(0) || '){3}')
);
### Examples
select lst, threesome(lst, 3)
from (select unnest( [
[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],
[3,3,3,4,3,5,6,7,8],
] ) as lst);
select ['x','y','z','a','a','a'] as lst, threesome(lst, 'a');
- Output:
┌─────────────────────────────┬───────────────────┐ │ lst │ threesome(lst, 3) │ │ int32[] │ boolean │ ├─────────────────────────────┼───────────────────┤ │ [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, 4, 3, 5, 6, 7, 8] │ false │ └─────────────────────────────┴───────────────────┘ ┌────────────────────┬─────────────────────┐ │ lst │ threesome(lst, 'a') │ │ varchar[] │ boolean │ ├────────────────────┼─────────────────────┤ │ [x, y, z, a, a, a] │ true │ └────────────────────┴─────────────────────┘
EasyLang
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 ] ]
func has3adj3 l[] .
for v in l[]
if v = 3
cnt += 1
else
if cnt = 3
break 1
.
cnt = 0
.
.
return if cnt = 3
.
for i to len lists[][]
write has3adj3 lists[i][] & " "
.
- Output:
1 0 0 0 1
F#
let has_adjacent n x =
let rec loop c = function
h :: t when h = x -> loop (c+1) t
|_ :: t -> c = n || loop 0 t
|_ -> c = n
in loop 0
[[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]]|>List.iter((has_adjacent 3 3)>>printfn "%A")
- Output:
true false false false true
Factor
USING: formatting generalizations kernel math.statistics
sequences.extras ;
: adjacent? ( seq -- ? )
[ 3 = ] arg-where differences V{ 1 1 } = ;
{ 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 }
[ dup adjacent? "%u -> %u\n" printf ] 5 napply
- Output:
{ 9 3 3 3 2 1 7 8 5 } -> t
{ 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 } -> f
{ 4 6 8 7 2 3 3 3 1 } -> t
A somewhat simpler implementation without the fancy statistics and generalizations vocabs.
USING: io kernel sequences ;
{
{ 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 = ] count 3 = ]
[ { 3 3 3 } subseq-of? ]
bi and "true" "false" ? print
] each
- 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
FutureBasic
include "NSLog.incl"
local fn ThreeAdjacentThrees
NSUInteger i, j
CFMutableArrayRef lists = fn MutableArrayNew
MutableArrayInsertObjectAtIndex( lists, @"9,3,3,3,2,1,7,8,5", 0 )
MutableArrayInsertObjectAtIndex( lists, @"5,2,9,3,3,7,8,4,1", 1 )
MutableArrayInsertObjectAtIndex( lists, @"1,4,3,6,7,3,8,3,2", 2 )
MutableArrayInsertObjectAtIndex( lists, @"1,2,3,4,5,6,7,8,9", 3 )
MutableArrayInsertObjectAtIndex( lists, @"4,6,8,7,2,3,3,3,1", 4 )
for i = 0 to len(lists) -1
CFArrayRef tempArr = fn StringComponentsSeparatedByString( lists[i], @"," )
NSUInteger counter = 0, elements = len(tempArr) -1
for j = 0 to elements
if ( counter == 3 ) then NSLog( @"%@: TRUE — contains 3 adjacent 3s.", lists[i] )
if ( counter != 3 ) and ( j == elements )
NSLog( @"%@: FALSE — doesn't contain 3 adjacent 3s.", lists[i] )
end if
if fn StringIsEqual( tempArr[j], @"3" ) == NO then counter = 0 : continue
if fn StringIsEqual( tempArr[j], @"3" ) == YES then counter++ : continue
next
next
end fn
fn ThreeAdjacentThrees
HandleEvents- Output:
9,3,3,3,2,1,7,8,5: TRUE — contains 3 adjacent 3s. 9,3,3,3,2,1,7,8,5: FALSE — doesn't contain 3 adjacent 3s. 5,2,9,3,3,7,8,4,1: FALSE — doesn't contain 3 adjacent 3s. 1,4,3,6,7,3,8,3,2: FALSE — doesn't contain 3 adjacent 3s. 1,2,3,4,5,6,7,8,9: FALSE — doesn't contain 3 adjacent 3s. 4,6,8,7,2,3,3,3,1: TRUE — contains 3 adjacent 3s.
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)
}
}
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
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]
]
or, in terms of groupBy:
import Data.List (groupBy)
nnPeers :: Int -> [Int] -> Bool
nnPeers n = const (any p . groupBy bothN) n
where
p g@(h:_) = n == h && n == length g
bothN x y = n == x && x == y
--------------------------- 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]
]
Or with less generality:
threeThree :: [Int] -> Bool
threeThree [] = False
threeThree (3:3:3:_) = True
threeThree (_:xs) = threeThree xs
--------------------------- TEST -------------------------
main :: IO ()
main =
putStrLn $
unlines $
fmap
(\xs -> show xs <> " -> " <> show (threeThree 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
J
For the given test cases:
lists=: >cutLF{{)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
}}
(,.~ (;:'false true')>@{~'3 3 3' +./@E.&".]) lists
true 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
true 4 6 8 7 2 3 3 3 1
However, for example, it's not clear what the result should be for an argument of 3 3 3 3 3 3 3 3 3.
Java
import java.util.Collections;
import java.util.List;
public final class ExactlyThreeAdjacent3InLists {
public static void main(String[] args) {
List<List<Integer>> lists = List.of( List.of( 9, 3, 3, 3, 2, 1, 7, 8, 5 ),
List.of( 5, 2, 9, 3, 3, 7, 8, 4, 1 ), List.of( 1, 4, 3, 6, 7, 3, 8, 3, 2 ),
List.of( 1, 2, 3, 4, 5, 6, 7, 8, 9 ), List.of( 4, 6, 8, 7, 2, 3, 3, 3, 1 )
);
lists.forEach( list -> {
final boolean result = Collections.indexOfSubList(list, List.of( 3, 3, 3 )) != -1;
System.out.println(list + " => " + 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
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)} -> ${nnPeers(3)(xs)}`
)
.join("\n");
return main();
})();
Or, recursively:
(() => {
"use strict";
// ------- 3 INSTANCES OF 3 AND ALL CONTIGUOUS ------
// nnPeers :: Int -> [Int] -> Bool
const nnPeers = n =>
xs => {
const go = ns =>
ns.slice(0, n).every(x => n === x) || (
n < ns.length && go(ns.slice(1))
);
return go(xs);
}
const threeThree = nnPeers(3);
// ---------------------- TEST -----------------------
return [
[9, 3, 3, 3, 2, 1, 7, 8, 5],
[5, 2, 9, 3, 3, 7, 8, 4, 1],
[1, 4, 3, 6, 7, 3, 8, 3, 2],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[4, 6, 8, 7, 2, 3, 3, 3, 1]
]
.map(
xs => `${JSON.stringify(xs)} -> ${threeThree(xs)}`
)
.join("\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
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
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
Nim
const 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]]
func contains3Adjacent3(s: openArray[int]): bool =
if s.len < 3: return false
var count = 0
for i in 0..s.high:
if s[i] == 3:
inc count
if count == 3: return true
else:
count = 0
for list in Lists:
echo list, ": ", list.contains3Adjacent3()
- 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
Nu
[
[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]
]
| each {
window 3 | each { $in == [3 3 3] } | into int | str join | $in =~ '(?<!1)1(?!1)'
}- Output:
╭───┬───────╮ │ 0 │ true │ │ 1 │ false │ │ 2 │ false │ │ 3 │ false │ │ 4 │ true │ ╰───┴───────╯
OCaml
let has_adjacent n x =
let rec loop c = function
| h :: t when h = x -> loop (succ c) t
| _ :: t -> c = n || loop 0 t
| _ -> c = n
in loop 0
let 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]]
let () =
List.iter (fun l -> Printf.printf " %B" (has_adjacent 3 3 l)) list
- Output:
true false false false true
Perl
Specific
#!/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
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 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
'''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
Quackery
Includes the extra test cases from the Raku and Wren solutions.
[ [] swap witheach
[ 3 = if
[ i^ join ] ]
dup size 3 != iff
[ drop false ]
done
unpack
1 - over != iff
[ 2drop false ]
done
1 - = ] is three-threes ( [ --> b )
' [ [ 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 ] ]
witheach
[ dup echo sp
three-threes iff
[ say "true" ]
else [ say "false" ]
cr ]- 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
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...
RPL
The program below creates a list of the positions of the number 3, then calculates the list of first differences, which must be equal to { 1 1 } if there are exactly 3 adjacent 3 in the input list.
≪ → list
≪ { } 1 list SIZE FOR j
IF list j GET 3 == THEN j + END NEXT
IFERR ΔLIST { 1 1 } == THEN DROP 0 END
≫ ≫ 'ADJ3?' STO
≪ {{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}} → cases
≪ { } 1 cases SIZE FOR j cases j GET ADJ3?’ + NEXT ≫ ≫ ‘TASK’ STO
- Output:
1: { 1 0 0 0 1 0 0 0 }
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
Rust
Using both Julia and Wren examples' test sets.
fn only_consecutive_in(a: &[i32], lis: &[i32]) -> bool {
return (0..lis.len() - a.len() + 1).any(|i| a == &lis[i..i + a.len()])
&& (0..a.len()).all(|i| {
lis.iter().filter(|x| **x == a[i]).count() == a.iter().filter(|x| **x == a[i]).count()
});
}
fn main() {
let needles = [[3, 3, 3], [3, 2, 3]];
let pair_of_haystacks = [
[
[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],
],
[
[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],
],
];
for idx in [0_usize, 1] {
let (haystacks, needle) = (pair_of_haystacks[idx], needles[idx]);
for haystack in haystacks {
println!(
"{:?} in {:?}: {:?}",
needle,
haystack,
only_consecutive_in(&needle, &haystack)
);
}
}
let haystacks = [
[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 i in 1_i32..=4 {
let needle = vec![i; i as usize];
for haystack in haystacks {
println!(
"{:?} in {:?}: {:?}",
needle,
haystack,
only_consecutive_in(&needle.as_slice(), &haystack)
);
}
}
}
- 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 [1] in [9, 3, 3, 3, 2, 1, 7, 8, 5]: true [1] in [5, 2, 9, 3, 3, 7, 8, 4, 1]: true [1] in [1, 4, 3, 6, 7, 3, 8, 3, 2]: true [1] in [1, 2, 3, 4, 5, 6, 7, 8, 9]: true [1] in [4, 6, 8, 7, 2, 3, 3, 3, 1]: true [1] in [3, 3, 3, 1, 2, 4, 5, 1, 3]: false [1] in [0, 3, 3, 3, 3, 7, 2, 2, 6]: false [1] in [3, 3, 3, 3, 3, 4, 4, 4, 4]: false [2, 2] in [9, 3, 3, 3, 2, 1, 7, 8, 5]: false [2, 2] in [5, 2, 9, 3, 3, 7, 8, 4, 1]: false [2, 2] in [1, 4, 3, 6, 7, 3, 8, 3, 2]: false [2, 2] in [1, 2, 3, 4, 5, 6, 7, 8, 9]: false [2, 2] in [4, 6, 8, 7, 2, 3, 3, 3, 1]: false [2, 2] in [3, 3, 3, 1, 2, 4, 5, 1, 3]: false [2, 2] in [0, 3, 3, 3, 3, 7, 2, 2, 6]: true [2, 2] in [3, 3, 3, 3, 3, 4, 4, 4, 4]: false [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, 6, 7, 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, 3, 3] in [3, 3, 3, 1, 2, 4, 5, 1, 3]: false [3, 3, 3] in [0, 3, 3, 3, 3, 7, 2, 2, 6]: false [3, 3, 3] in [3, 3, 3, 3, 3, 4, 4, 4, 4]: false [4, 4, 4, 4] in [9, 3, 3, 3, 2, 1, 7, 8, 5]: false [4, 4, 4, 4] in [5, 2, 9, 3, 3, 7, 8, 4, 1]: false [4, 4, 4, 4] in [1, 4, 3, 6, 7, 3, 8, 3, 2]: false [4, 4, 4, 4] in [1, 2, 3, 4, 5, 6, 7, 8, 9]: false [4, 4, 4, 4] in [4, 6, 8, 7, 2, 3, 3, 3, 1]: false [4, 4, 4, 4] in [3, 3, 3, 1, 2, 4, 5, 1, 3]: false [4, 4, 4, 4] in [0, 3, 3, 3, 3, 7, 2, 2, 6]: false [4, 4, 4, 4] in [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
V (Vlang)
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
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
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
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