Is this a straight flush?

Related: Name the poker hand

A straight flush is a poker hand containing five cards of sequential rank, all of the same suit. As part of a straight flush, an ace can rank either above a king or below a two. An ace can rank either high (e.g. A♥ K♥ Q♥ J♥ 10♥ is an ace-high straight flush) or low (e.g. 5♦ 4♦ 3♦ 2♦ A♦ is a five-high straight flush), but cannot rank both high and low in the same hand (e.g. Q♣ K♣ A♣ 2♣ 3♣ is an ace-high flush, not a straight flush).

Challenge

Given N cards (in any reasonable format) output a truthy value if a straight flush is contained in the poker hand.

Input

• N numbers of cards. (In any reasonable format)

There are four suits; hearts, spades, diamonds, and clubs (H, S, D, C).

Each suit has one card for the numbers 2 to 10, plus 4 'picture' cards, Ace, Jack, Queen, and King (A, J, Q, K)

Note: You can take 10 as T

Output

• Truthy/Falsy value

Test case

["AS", "2S", "3S", "4S", "5S"] => true

["3D", "9C", "4S", "KH", "AD", "AC"] => false

["5D", "6D", "7D", "8H", "9D", "10D", "JD"] => false

["JC", "7C", "5D", "8C", "AC", "10C", "9C", "5S"] =>true

[] => false

["AS", "2S", "3S"] => false

["JC", "QC", "KC", "AC", "2C"] => false

[ "2H", "3H", "4H", "5H", "6H", "7H"] => true


Standard rules apply.

Winning criteria: Shortest code in each lang

• May we assume that there won't be two of the same card in the hand? – Jo King Jul 26 '18 at 23:52
• @JoKing yep, you wont have the same card twice or more – Luis felipe De jesus Munoz Jul 27 '18 at 0:48
• May we take 10 as T? – Kevin Cruijssen Jul 27 '18 at 7:13
• @JoKing I don't think that can happen IRL. ;-) – Erik the Outgolfer Jul 27 '18 at 9:08
• @EriktheOutgolfer I literally have about 5 packs of mixed cards less then a meter away from me – Jo King Jul 27 '18 at 10:04

Python 2, 95 bytes

lambda a:any(set('A234567891JQKA'[i/4:][:5])<={r['HCSD'[i%4]in r]for r in a}for i in range(40))


Try it online!

There are 40 possible straight flushes, and this simply checks them all. Chas Brown saved 2 bytes; Jo King saved 4 more.

• There are 40, you used A at both ends so I believe changing 36 to 40 should fix it. – Jonathan Allan Jul 26 '18 at 21:46
• Oops, aren't I good at counting. I fixed it! – Lynn Jul 27 '18 at 1:27
• – Chas Brown Jul 28 '18 at 20:36
• Swap the suit-value order and move the if condition to the index? – Jo King Jul 29 '18 at 12:48
• 95 bytes – Jo King Jul 29 '18 at 23:18

R, 128 12694 91 bytes

function(x,r=rle(outer(y<-chartr("J-X","A2-9TJQKAS",LETTERS),y,paste0)%in%x))any(r$l>4&r$v)


Try it online!

Original logic shortened considerably by @J.Doe.

Makes a 26 by 26 matrix with mostly nonsense but all the cards (with the Aces repeated at the bottom) contained in rows 10 to 23 of columns 3,4,8 and 24. The matrix is created by concatenating all combinations of the upper case alphabet with letters J through X replaced by A,2-9,T,J,Q,K,A,S via chartr. We get C, D, H for free!

The %in% flattens the matrix column-wise into a vector. Then see if run-length encoding is greater than 4 for any run of TRUE matches.

• Clever use of rle AND outer ! This saves two bytes – JayCe Jul 26 '18 at 21:06
• 94 bytes. Two changes: using a symmetric outer call that produces many invalid cards, and using the vector coercion of in to avoid apply. They both have to be in place for this to work! – J.Doe Sep 3 '18 at 8:54
• Very nice! Changed the answer and made it a community wiki. – ngm Sep 4 '18 at 13:24

JavaScript (ES6), 116 bytes

a=>[...'CDHS'].some(s=>a.map(c=>m|=c.match(s)&&2<<"234567891JQKA".search(c[0]),m=0)|(g=k=>k&&1+g(k&k/2))(m|m>>13)>4)


Try it online!

How?

For each suit $s$, we convert all cards $c$ of suit $s$ into a 14-bit bitmask $m$, duplicate bit #13 (Ace) to bit #0 to handle the steel wheel (A,2,3,4,5) and count the number of consecutive bits. If it's greater than 4, we have a straight flush.

• I've gotten so used to your "currying notation" introductory line that I miss it when it's not needed. – ngm Jul 26 '18 at 20:56

Brachylog, 31 bytes

tᵍkᵐ²cᵐ{ps₅~s"A23456789TJQKA"}ᵉ


Try it online!

 ᵍ                    Group input by
t                     each element's "tail" (i.e. suit)
kᵐ²                   Knife off the suit character from each element in each array
cᵐ                    Concatenate the elements of each suit array into a string
{               }ᵉ    There exists at least one string in that such that
p                    it has a permutation
s₅                   which has a substring of length 5
~s                   which is also a substring of
"A23456789JQKA"


Retina 0.8.2, 66 bytes

J
11
Q
12
K
13
A
1$%'¶14 \d+(.)$1$&$*
O
^
¶
((?(1)\1.|¶.+)){5}\b


Try it online! Explanation:

J
11
Q
12
K
13


Convert the picture cards into their values.

A
1$%'¶14  A can be 1 or 14. \d+(.)$1$&$*
O


Convert the value to unary, and suffix it so that the cards sort properly.

^
¶
((?(1)\1.|¶.+)){5}\b


Match 5 cards that increase by 1 each time, and ensure that the last increase was exactly 1.

JavaScript (ES6), 106 bytes

h=>h.map(([r,s])=>[..."HSDCA23456789TJQKA"].map(c=>i+=c==s?i*15:c==r?d[i]=1:1,i=0),d=[])|/(,1){5}/.test(d)


Accepts an array of string representations of cards, replacing 10 with T. Try it online!

Explanation

Iterates over each card and sets a flag in an array of booleans using an index computed from the unique combination of its rank and suit. This array is then stringified to permit matching a pattern of 5 consecutive truthy values.

For example, a hand with a straight flush may produce the following as a substring of the full string representation of the boolean array: ,,,,1,1,1,1,1,,,,

Because the first rank value (i.e. A) is offset from the start of the string, there will always be empty values preceding all 1's in the array, ensuring the string representation will begin with a ,

h =>
h.map(([r, s]) =>                         // destructure card value, e.g. "JH" => ["J", "H"]
[..."HSDCA23456789TJQKA"].map(c =>    // mapping accounts for both positions of 'A'
i +=                              // increment index value
c == s                            // if found index of suit...
? i * 15                      // buffer so that cards from different suits cannot be confused
: c == r                          // if found index of rank...
? d[i] = 1                    // set flag to denote card is in hand
: 1,
i = 0
),
d = []
) |
/(,1){5}/.test(d)                         // implicitly converts to string joined with a ,

• Nice. This deserves more votes, but folks tend to lose interest in challenges a couple days after the initial posting. – Rick Hitchcock Jul 30 '18 at 20:22

Java 10, 189167165164160 157 bytes

s->{int i=10;for(;i-->0;)if(s.matches("AKQJT98765432A".substring(i,i+5).replaceAll(".","(?=.*$0\\\\1)").replaceFirst(".1","([HSDC])")+".*"))i=-2;return-1>i;}  Takes the input as a single space-delimited String, including trailing space (i.e. "AS 2S 3S 4S 5S "). -22 bytes thanks to @OlivierGrégoire. -1 byte thanks to @AlexRacer. Try it online. Golfed version of the code that I've used for Project Euler #54, which I primarily did with regexes (for fun and to learn more about regexes). Without regexes it would probably have been better for performance and easier (probably also applies for golfing this answer; will take a look later on). Explanation: s->{ // Method with String parameter and boolean return-type int i=10;for(;i-->0;) // Loop i in the range (10,0] if(s.matches( // If the input matches the following regex: "AKQJT98765432A".substring(i,i+5).replaceAll(".","(?=.*$0\\\\1)").replaceFirst(".1","([HSDC])")
+".*"))         //    With optionally zero or more other characters
i=-2;             //     Set i to -2,
//     which also stops the loops at the same time
return-1>i;}          //  Return whether i is not -2 (so whether the loop has finished)


• "AKQJT98765432A".substring(i,i+5) takes five adjacent cards based on i
• .replaceAll(".","(?=.*$0\\\\1)") replaces each of those cards with "(?=.*c\\1)" (where c is the card-character) • .replaceFirst(".1","([HSDC])") will then replace the first \\1 with ([HSDC]). I.e. the total regex to check the Straight Flush for cards in the value-range [9,5] will become: ^(?=.*9([HSDC]))(?=.*8\\1)(?=.*7\\1)(?=.*6\\1)(?=.*5\\1).*$
(NOTE: String#matches implicitly adds the trailing/leading ^...$ to check the entire String.) • 172 bytes. I only golfed the regex generation: it's still your algorithm. – Olivier Grégoire Jul 27 '18 at 14:48 • 167 bytes. I removed the unnecessary ".*"+ prefix. – Olivier Grégoire Jul 27 '18 at 14:53 • @OlivierGrégoire Thanks! Nice golfs. – Kevin Cruijssen Jul 27 '18 at 15:05 • -1 byte if you break out of loop instead of using f – AlexRacer Sep 1 '18 at 16:31 • @AlexRacer Smart, thanks! And been able to golf 2 more bytes by changing the break to i=-2 and the return to return-1>i; using your approach (and 2 more changing (.) to . and $1 to $0). :) – Kevin Cruijssen Sep 1 '18 at 17:30 Clean, 145 135 bytes import StdEnv,Data.List ?l=or[isInfixOf(map hd h)['A234567891JQKA']\\a<-l,b<-l,c<-l,d<-l,e<-l,h<-[[a,b,c,d,e]]|tl(nub(map last h))==[]]  Try it online! Simplified: ? l // function ? taking argument l = or [ // is at least one of these true isInfixOf (map hd h) ['A234567891JQKA'] // do the first characters of a hand appear in this string, in order \\ a <- l // loop level 1, assigns a , b <- l // loop level 2, assigns b , c <- l // loop level 3, assigns c , d <- l // loop level 4, assigns d , e <- l // loop level 5, assigns e , h <- [[a,b,c,d,e]] // trick to assign h, because it's cheaper than let .. in .. | tl (nub (map last h)) == [] // only take the loop iterations where all the suits are the same ]  Japt, 37 bytes Takes input as a 2D array. "AJQKA"i1Aò2 q)øUñÌòÏÌ¦XÌÃËmÎ¬ú5 á5Ãc  Try it Explanation "AJQKA" :String literal i1 :Insert at (0-based) index 1 Aò2 : Range [2,10] q : Join ) :End insert ø :Does that string contain any element in the following array? U :Input ñ :Sort Ì : By last element (grouping suits together) òÏ :Partition between X & Y where Ì : Last element of Y ¦ : Does not equal XÌ : Last element of X Ã :End partition Ë :Map m : Map Î : First elements (card values) ¬ : Join ú5 : Right pad with spaces to length 5 á5 : Permutations of length 5 Ã :End map c :Flatten  Jelly, 18 bytes Ṣœc5Uµ13R;1wṪ€ȧEµƇ  Try it online! Actual input format: list ([..., ...]) of lists of 2 integers each; the first is an integer in $[1,13]$, for $A23456789TJQK$ respectively, and the second is in $[1,4]$, for $CDHS$ respectively. The TIO link accepts input in the format of the test cases. Output format: empty list as falsy, non-empty list as truthy. • I don't see anything in the specs suggesting that integers can be substituted for the suits and picture cards - did I miss something? – Shaggy Jul 27 '18 at 11:37 • @Shaggy I assume that's within "any reasonable format", I don't think we have defaults regarding inputting playing cards. – Erik the Outgolfer Jul 27 '18 at 14:49 PHP, 264 bytes It echos 1 if it is a straight flush and 0 or null if not. If you name the file 1X then you can save 11 bytes as you don't need to change $argv[0]. Not sure at the moment why the filename can break it.

For some reason the strings :;<=> get sorted before the strings 0123456789 by asort in TIO even though :;<=> have ASCII values 58-62 and 0123456789 have ASCII values 48-57. So if you take the code from the TIO link or below and use PHPTester with the following test suite it works.

$argb[0] = [".code.tio", "AS", "2S", "3S", "4S", "5S"]; // => true$argb[1] = [".code.tio", "3D", "9C", "4S", "KH", "AD", "AC"]; // => false
$argb[2] = [".code.tio", "5D", "6D", "7D", "8H", "9D", "TD", "JD"]; // => false$argb[3] = [".code.tio", "JC", "7C", "5D", "8C", "AC", "TC", "9C", "5S"]; // => true
$argb[4] = [".code.tio", ]; // => false$argb[5] = [".code.tio", "AS", "2S", "3S"]; // => false
$argb[6] = [".code.tio", "JC", "QC", "KC", "AC", "2C"]; // => false$argb[7] = [".code.tio", "TC", "JC", "QC", "KC", "AC", "2C"]; // => true
$argb[8] = [".code.tio", "2H", "3H", "4H", "5H", "6H", "7H"]; // => true for ($z=0; $z<9;$z++){
$argv=$argb[$z]; array_shift($argv);
unset($a,$b,$c,$d,$e,$f,$g,$h,$i);$f=false; // not needed, just removes several notices

// TIO code here

echo "<br>";


TIO Code

for($b=count($a=$argv);$b;){$a[0]='1X';$a[--$b]=strtr($a[$b],'ATJQK','1:;<=');$a[]=($a[$b][0]==1?">".$a[$b][1]:1);}asort($a);foreach($a as$c){$d[$c[1]][]=$c[0];}foreach($d as$e){if(4<$g=count($e)){for($h=0;$g>$i=4+$h;){$f|=(ord($e[$i])-ord($e[$h++])==4);}}}echo$f;


Try it online!

Kotlin, 226 bytes

Used T for 10 so all cards are 2 characters in length.

{h:List<String>->val m=List(4){mutableSetOf<Int>()}
var r=0>1
for(i in 0..9)r=b.containsAll((i..i+4).toList())||r}
r}


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Pascal (FPC), 223216210 209 bytes

var a,b:char;c:set of byte;i:byte;begin repeat readln(a,b);i:=pos(b,'HDC')*14+pos(a,'23456789TJQK');c:=c+[i];if a='A'then c:=c+[i+13]until eof;i:=0;while not([i..i+4]<=c)or(i mod 14>9)do i:=i+1;write(i<52)end.


Try it online!

Uses T for 10. Input contains 1 card per line.

Now I golfed it so much that I don't know how it works anymore...

Explanation:

var a,b:char; //for reading cards
c:set of byte; //this set is for remembering which cards are present in the input
//14 numbers used for each suit
i:byte;
begin
repeat
i:=pos(b,'HDC')*14+pos(a,'23456789TJQK');
//temporary use i to calculate corresponding number for the card
//1st pos() is for the group of numbers for that suit, 2nd pos() is for offset
c:=c+[i];                //include i into set
if a='A'then c:=c+[i+13] //if rank is A, include the number at the end of group as well
until eof;
i:=0;
while not(
([i..i+4]<=c) //if NOT 5 cards in a row are present...
and           //while the check is started from 10 (T)...
(i mod 14<10) //(otherwise, it is checking across 2 different suits)
)do i:=i+1;     //increment i, otherwise stop
write(i<52) //if i<=51, there is a straight flush starting at the card corresponding to i
//(if there isn't a straight flush, i stops at 252 due to i..i+4, I don't know why)
end.