Is string X a subsequence of string Y?

Question

Given strings X and Y, determine whether X is a subsequence of Y. The empty string is regarded as a subsequence of every string. (E.g., '' and 'anna' are subsequences of 'banana'.)

Input

• X, a possibly-empty case-sensitive alphanumeric string
• Y, a possibly-empty case-sensitive alphanumeric string

Output

• True or False (or equivalents), correctly indicating whether X is a subsequence of Y.

I/O examples

X      Y        output

''     'z00'    True
'z00'  'z00'    True 
'z00'  '00z0'   False
'aa'   'anna'   True
'anna' 'banana' True
'Anna' 'banana' False

Criteria

• The shortest program wins, as determined by the number of bytes of source code.

Example programs

• Several programs that could be adapted are in this related posting.

Answer 1

Perl 5, 17 bytes (+1?), full program

s//.*/g;$_=<>=~$_

Try it online!

Invoke with the p flag to the perl interpreter, as in perl -pe 's//.*/g;$_=<>=~$_'. Per the established scoring rules when this challenge was originally posted, this flag costs one extra byte. Under more recent rules, AFAICT, it may be free.

Either way, the input strings should be supplied on separate newline-terminated lines on stdin. Output (to stdout) will be 1 if the first input string is a substring of the second, or nothing at all if it's not.

Note that both input lines must have a newline at the end, or the program won't work correctly. Alternatively, you can add the l command line flag to the invocation to make perl strip the newlines; depending on the scoring rules in effect, this may or may not cost one extra byte. Note that using this flag will also append a newline to the output.

Original version (snippet, 18 bytes / chars)

$x=~s//.*/g,$y=~$x

Input is given in the variables $x and $y, result is the value of the expression (in scalar context). Note that $x is modified in the process. (Yes, I know using $_ instead of $x would let me save four chars, but doing that in a snippet that feels a bit too cheesy for me.)

How does it work?

The first part, $x=~s//.*/g, inserts the string .* between each character in $x. The second part, $y=~$x, treats $x as a regexp and matches $y against it. In Perl regexps, .* matches zero or more arbitrary characters, while all alphanumeric characters conveniently match themselves.

Answer 2

Ruby, 32 characters

s=->x,y{y=~/#{[*x.chars]*".*"}/}

This solution returns nil if x isn't a subsequence of y and a number otherwise (i.e. ruby equivalents to false and true). Examples:

p s['','z00']        # => 0   (i.e. true)
p s['z00','z00']     # => 0   (i.e. true)
p s['z00','00z0']    # => nil (i.e. false)
p s['anna','banana'] # => 1   (i.e. true)
p s['Anna','banana'] # => nil (i.e. false)

Answer 3

Haskell, 51 37

h@(f:l)%(g:m)=f==g&&l%m||h%m;x%y=x<=y

Thanks to Hammar for the substantial improvement. It's now an infix function, but there seems to be no reason why it shouldn't.

Demonstration:

GHCi> :{
GHCi| zipWith (%) [""   , "z00", "z00" , "anna"  , "Anna"]
GHCi|             ["z00", "z00", "00z0", "banana", "banana"]
GHCi| :}
[True,True,False,True,False]

Answer 4

Python (48 chars)

import re;s=lambda x,y:re.search('.*'.join(x),y)

Same approach as Howard's Ruby answer. Too bad about Python's need to import the regex package and its "verbose" lambda. :-)

Answer 5

Python, 59 characters

def s(x,y):
 for c in y:
  if x:x=x[c==x[0]:]
 return x==""

I figured my answer would be better expressed in Python.

Edit: Added r.e.s.'s suggestions.

Answer 6

GolfScript (22 chars)

X[0]+Y{1$(@={\}*;}/0=!

Assumes that input is taken as two predefined variables X and Y, although that is rather unusual in GolfScript. Leaves 1 for true or 0 for false on the stack.

Answer 7

C (52 chars)

s(char*x,char*y){return!*x||*y&&s(*x-*y?x:x+1,y+1);}

Test cases

Answer 8

Burlesque (6 chars)

6 chars in Burlesque: R@\/~[ (assuming x and y are on the stack. See here in action.)

Answer 9

C, 23:

while(*y)*x-*y++?0:x++;

result in *x

http://ideone.com/BpITZ

Answer 10

Python 3.8 (pre-release), 42 bytes

lambda a,b:''in[a:=a[a[:1]==c:]for c in b]

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Python 3.8 (pre-release), 48 bytes

lambda a,b,j=0:all((j:=1+b.find(c,j))for c in a)

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Python 2, 48 bytes

lambda a,b:re.search('.*'.join(a),b)>0
import re

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Copied from this answer of Lynn. The >0 can be omitted if just truthy/falsey output is OK.

Python 2, 50 bytes

f=lambda a,b:b and f(a[a[:1]==b[0]:],b[1:])or''==a

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Python 2, 50 bytes

lambda a,b:reduce(lambda s,c:s[c==s[:1]:],b,a)==''

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Answer 11

PHP, 90 characters

<?function s($x,$y){while($a<strlen($y))if($y[$a++]==$x[0])$x=substr($x,1);return $x=="";}

Answer 12

C #, 70 113 107 90 characters

static bool S(string x,string y){return y.Any(c=>x==""||(x=x.Remove(0,c==x[0]?1:0))=="");}

Answer 13

Scala 106:

def m(a:String,b:String):Boolean=(a.size==0)||((b.size!=0)&&((a(0)==b(0)&&m(a.tail,b.tail))||m(a,b.tail)))

Answer 14

CoffeeScript 112 100 95 89

My first attempt at code golf... hope I don't shame my family!

z=(x,y)->a=x.length;return 1if!a;b=y.indexOf x[0];return 0if!++b;z x[1..a],y[b..y.length]

Edit: turns out Coffeescript is more forgiving than I thought with whitespace.

Thanks to r.e.s. and Peter Taylor for some tips for making it a bit sleeker

Answer 15

Mathematica 19 17 27

LongestCommonSequence returns the longest non-contiguous subsequence of two strings. (Not to be confused with LongestCommonSubsequence, which returns the longest contiguous subsequence.

The following checks whether the longest contiguous subsequence is the first of the two strings. (So you must enter the shorter string followed by the larger string.)

LongestCommonSequence@##==#& 

Examples

LongestCommonSequence@## == # &["", "z00"]
LongestCommonSequence@## == # &["z00", "z00"]
LongestCommonSequence@## == # &["anna", "banana"]
LongestCommonSequence@## == # &["Anna", "banana"]

True True True False

The critical test is the third one, because "anna" is contained non contiguously in "banana".

Answer 16

PHP, 41 Bytes

prints 1 for true and nothing for false

<?=!levenshtein($argv[1],$argv[2],0,1,1);

If only insertions from word 1 to word 2 done the count is zero for true cases

levenshtein

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PHP, 57 Bytes

prints 1 for true and 0 for false

Creates a Regex

<?=preg_match(_.chunk_split($argv[1],1,".*")._,$argv[2]);

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Answer 17

Brachylog, 2 bytes

⊆ᵈ

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As with this answer, ⊆ is a built-in predicate that declares a relationship between the input and output variables, and ᵈ is a meta-predicate that modifies it to declare that same relationship between the first and second elements of the input variable instead (and unify the output variable with the second element but as this is a decision problem that doesn't end up mattering here). X⊆Y is an assertion that X is a subsequence of Y, therefore so is [X,Y]⊆ᵈ.

This predicate (which of course outputs through success or failure which prints true. or false. when it's run as a program) takes input as a list of two strings. If input is a bit more flexible...

Brachylog, 1 byte

⊆

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Takes string X as the input variable and string Y as the output variable. Outputs through success or failure, as before. If run as a full program, X is supplied as the input and Y is supplied as the first command line argument.

Answer 18

SWI-Prolog, SICStus

The built-in predicate sublist/2 of SICStus checks whether all the items in the first list also appear in the second list. This predicate is also available in SWI-Prolog via compatibility library, which can be loaded by the query [library(dialect/sicstus/lists)]..

Sample run:

25 ?- sublist("","z00").
true.

26 ?- sublist("z00","z00").
true .

27 ?- sublist("z00","00z0").
false.

28 ?- sublist("aa","anna").
true .

29 ?- sublist("anna","banana").
true .

30 ?- sublist("Anna","banana").
false.

The byte count can technically be 0, since all we are doing here is querying, much like how we run a program and supply input to it.

Answer 19

C - 74 71 64

This doesn't beat Peter Taylor's solution, but I think it's pretty fun (plus, this is a complete working program, not just a function)

main(int c,char**v){for(;*v[1]!=0;++v[1])v[2]+=*v[1]==*v[2];return*v[2];}

---

main(int c,char**v){for(;*v[1];++v[1])v[2]+=*v[1]==*v[2];return*v[2];}

---

main(c,v)char**v;{while(*v[1])v[2]+=*v[1]++==*v[2];return*v[2];}

And ungolfed:

main(int argc, char** argv){
   char * input = argv[1];
   char * test  = argv[2];

   // advance through the input string. Each time the current input
   // character is equal to the current test character, increment
   // the position in the test string.

   for(; *input!='\0'; ++input) test += *input == *test;

   // return the character that we got to in the test string.
   // if it is '\0' then we got to the end of the test string which
   // means that it is a subsequence, and the 0 (EXIT_SUCCESS) value is returned
   // otherwise something non-zero is returned, indicating failure.
   return *test;
}

To test it you can do something like:

./is_subsequence banana anna && echo "yes" || echo "nope"    
# yes
./is_subsequence banana foobar && echo "yes" || echo "nope"    
# nope

Answer 20

Ruby 32 30 28

f=->a,b{b.match a.tr'','.*'}

This will return MatchData instance if a is subsequence of b or nil otherwise.

Old version that find substring instead of subsequence

Ruby 15

f=->a,b{!!b[a]}

Using String#[](str) method that returns str if str is a substring of self and !! to return Boolean if returned value can be usable as boolean (and don't need to be true or false) then it can be only 13 chars:

f=->a,b{b[a]}

It will return nil if a is not a substring of b.

Answer 21

Python, 66 62 59 58 chars

Kind of a fun solution, definitely a neat problem.

def f(n,h,r=0):
 for c in h:r+=n[r:r+1]==c
 return r==len(n)

Answer 22

PHP, 75 65 64 bytes

for(;$p=@strpos(_.$argv[2],$c=$argv[1][$i++],$p+1););echo""==$c;

takes input from command line arguments; prints 1 for true, empty string for false. Run with -r.

explanation:

• strpos returns false if needle $c is not in the haystack $argv[2] (after position $p),
  causing the loop to break.
• strpos also returns false for an empty needle, breaking the loop at the end of $argv[1].
• If $argv[1] is a subsequence of $argv[2], $c will be empty when the loop breaks.
• strpos needs @ to suppress Empty needle warning.

Answer 23

Swift, 27

print(Y.range(of:X) != nil)

Answer 24

APL(NARS), chars 46, bytes 92

{(⊂,⍺)∊(⊂''),↑¨,/¨{⍵⊂w}¨{⍵⊤⍨k⍴2}¨⍳¯1+2*k←≢w←⍵}

test:

  h←{(⊂,⍺)∊(⊂''),↑¨,/¨{⍵⊂w}¨{⍵⊤⍨k⍴2}¨⍳¯1+2*k←≢w←⍵}
  '' h 'z00'
1
  'z00' h 'z00'
1
  'z00' h '00z0'
0
  'aa' h 'anna'
1
  'anna' h 'banana'
1
  'Anna' h 'banana'
0

comment:

{(⊂,⍺)∊(⊂''),↑¨,/¨{⍵⊂w}¨{⍵⊤⍨k⍴2}¨⍳¯1+2*k←≢w←⍵}
 ⍳¯1+2*k←≢w←⍵    this assign to w the argument and k argument lenght, it return 1..(2^k)-1 range
 {⍵⊤⍨k⍴2}¨      for each element of 1..(2^k)-1 convert in base 2 with lenght k (as the arg lenght)
 {⍵⊂w}¨         use the binary array above calculation for all partition argument
 ,/¨            concatenate each element of partition
 ↑¨             get the firs element of each element because they are all enclosed
 (⊂''),         add to the array the element (⊂'')
 (⊂,⍺)∊         see if (⊂,⍺) is one element of the array, and return 1 true o 0 false

How all you can see the comments are +- superfluous all is easy...

I have to say not understand why the last instruction is not "(,⍺)∊" in the place of "(⊂,⍺)∊" because for example in code

  q←{↑¨,/¨{⍵⊂w}¨{⍵⊤⍨k⍴2}¨⍳¯1+2*k←≢w←⍵}
  o q '123'
┌7──────────────────────────────────────┐
│┌1─┐ ┌1─┐ ┌2──┐ ┌1─┐ ┌2──┐ ┌2──┐ ┌3───┐│
││ 3│ │ 2│ │ 23│ │ 1│ │ 13│ │ 12│ │ 123││
│└──┘ └──┘ └───┘ └──┘ └───┘ └───┘ └────┘2
└∊──────────────────────────────────────┘
  o (,'3')
┌1─┐
│ 3│
└──┘

all you see array of 1 element (,'3') is the element of the set of instruction q '123', but

  o (,'3')∊q '123'
┌1─┐
│ 0│
└~─┘

return one array of unclosed 0 instead of number 1... for workaround that one has to write:

  o (⊂,'3')∊q '123'
1
~

that is the right result even if the element seems different because :

  o (⊂,'3')
┌────┐
│┌1─┐│
││ 3││
│└──┘2
└∊───┘

It is sure i make some error because i am not so smart... where is my error?

Answer 25

Perl 6, 34 28 bytes

-6 bytes thanks to nwellnhof

{$!=S:g/<(/.*/;&{?/<{$!}>/}}

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Anonymous code block that takes input curried, like f(X)(Y). This does the familiar join by .* and evaluate as a regex as other answers, but takes a couple of shortcuts.

Explanation:

{                          }  # Anonymous code block
 $!=            # Assign to $!
    S:g/<(/.*/  # Inserting .* between every character
              ;&{         }   # And return an anonymous code block
                 ?/      /    # That returns if the input matches
                   <{$!}>     # The $! regex

Answer 26

Haskell, 36 bytes

(a:b)%(c:d)=([a|a/=c]++b)%d
s%t=s<=t

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---

37 bytes

(.map concat.mapM(\c->["",[c]])).elem

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I think mapM postdates this challenge. If we can assume the characters are printable, we can do

36 bytes

(.map(filter(>' ')).mapM(:" ")).elem

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---

37 bytes

(null.).foldr(?)
c?(h:t)|c==h=t
c?s=s

Try it online!

Answer 27

Python - 72

def f(a,b):
 c=len(a)
 for i in b:a=a.replace(i,"",1)
 print len(a+b)==c

Answer 28

CoffeeScript 73

Here's an alternative CoffeeScript answer, using regexes instead of recursion:

z=(x,y)->a='.*';a+=c+'.*'for c in x;b=eval('/'+a+'/');(y.replace b,'')<y

If the haystack matches a very greedy regex constructed from the needle, it will be replaced with an empty string. If the haystack is shorter than it started, the needle was a subsequence.

Returns false when x and y are both empty strings. Think we need a philosopher to tell us if an empty string is a subsequence of itself!

(Posted as a separate answer from my previous because it feels different enough to justify it).

Answer 29

PowerShell, 38

$args[1]-clike($args[0]-replace'','*')

Of course, any such regex- or pattern-matching-based solution has severe performance problems with longer strings. But since shortness is the criterion ...

Answer 30

A sort of anti-solution generating all subsequences of Y:

Python 93

l=len(y)
print x in[''.join(c for i,c in zip(bin(n)[2:].rjust(l,'0'),y)if i=='1')for n in range(2**l)]

Answer 31

APL (31)

String handling is a bit lacking in APL.

{(⊂'')∊N←⍵↓⍨¨1,⍨=/⊃¨⍵:~×⍴⊃N⋄∇N}

usage:

      {(⊂'')∊N←⍵↓⍨¨1,⍨=/⊃¨⍵:~×⍴⊃N⋄∇N} 'anna' 'banana'
1
      {(⊂'')∊N←⍵↓⍨¨1,⍨=/⊃¨⍵:~×⍴⊃N⋄∇N} 'Anna' 'banana'
0
      {(⊂'')∊N←⍵↓⍨¨1,⍨=/⊃¨⍵:~×⍴⊃N⋄∇N} '' 'banana'
1

Answer 32

Python (75 52)

s=lambda a,b:a==''or b>''and s(a[a[0]==b[0]:],b[1:])

Simple recursive solution. First time golfing, so any tips on whittling this down are much appreciated :)

Tested with the following:

assert s('anna', 'banana') == True
assert s('oo0', 'oFopp0') == True
assert s 'this', 'this is a string') == True
assert s('that', 'this hat is large') == True
assert s('cba', 'abcdefg') == False

Thanks to @lirtosiast for some clever boolean tricks.

Answer 33

Python 132

Similar to Daniero's. Not the easiest solution, but it was fun to try. I'm new to Python, so I'm sure I could make it shorter if I knew a little bit more.

def f(i):
    s=x;j=0
    while j<len(s):t=~i%2;s=[s[:j]+s[j+1:],s][t];j+=t;i>>=1
    return s==y
print True in map(f,range(1,2**len(x)))

Answer 34

JavaScript (ES6), 42 bytes

Takes input n (needle) and h (haystack) in currying syntax (n)(h).

n=>h=>!!RegExp(n.split``.join`.*`).exec(h)

Test

let f =

n=>h=>!!RegExp(n.split``.join`.*`).exec(h)

console.log(f(''    )('z00'   )); // true
console.log(f('z00' )('z00'   )); // true 
console.log(f('z00' )('00z0'  )); // false
console.log(f('aa'  )('anna'  )); // true
console.log(f('anna')('banana')); // true
console.log(f('Anna')('banana')); // false

Answer 35

Java 8, 163 162 38 bytes

a->b->b.matches(a.replaceAll("",".*"))

-124 bytes by converting to Java 8, and pasting my answer from the duplicated challenge.

Try it online.

Explanation:

a->b->         // Method with two String parameters and boolean return-type
  b.matches(   //  Check if the second input matches the regex:
   a           //   The first input,
    .replaceAll("",".*"))
               //   where every character is surrounded with ".*"

For example:

a="anna"
b="banana"

Will do the check:

"banana".matches("^.*a.*n.*n.*a.*$")

Answer 36

Pyth, 5 bytes (non-competing)

}hQye

Explanation

}hQye
 hQ       The first input...
}         ... is an element of...
   y      ... the power set...
    eQ    ... of the second input.

Answer 37

REXX, 76 bytes

o=1
do while x>''
  parse var x a+1 x
  parse var y(a)b+1 y
  o=o&a=b
  end
return o

Note that x and y are consumed by this routine. Readability is impaired by skipping a lot of whitespace and parentheses.

Answer 38

APL (Dyalog Unicode), 18 bytes^SBCS

Full program. Prompts for Y then for X.

×≢(∊'.*'∘,¨⍞)⎕S⍬⊢⍞

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⍞ prompt (for Y)

⊢ yield that (separates ⍬ from it)

(…)⎕S⍬ search for occurrences of the following, yielding one empty list for each match:

 ⍞ prompt (for X)

 '.*'∘,¨ prepend .* to each character

 ∊ ϵnlist (flatten)

≢ tally the number of matches

× sign of that

Answer 39

Python 2, 47 bytes

x,y=map(list,input())
while x:y.remove(x.pop())

Try it online!

Takes two quote-delimited strings via STDIN as input.

Output is by presence or absence of an error, which is allowed per meta consensus.

Explanation:

             # get arguments from STDIN
             input()
    # convert each argument to a list
    map(list,.......)
# split arguments into two variables
x,y=..................

# while x still has elements
while x:
                 # remove the final element of x and return its value
                 x.pop()
        # remove the first matching item in y
        # this will error if there is no matching element
        y.remove(.......)

# if the loop is exited without error, all elements in x were in y

Answer 40

x86-16 Assembly, 11 bytes

Binary:

00000000: 41ac f2ae e303 4a75 f83b ca              A.....Ju.;.

Unassembled listing:

41          INC  CX                 ; Loop counter is 1 greater than string length 
        SCAN_LOOP: 
AC          LODSB                   ; load next char of acronym into AL 
F2/ AE      REPNZ SCASB             ; search until char AL is found 
E3 03       JCXZ DONE               ; stop if end of first string reached 
4A          DEC  DX                 ; decrement second string counter 
75 F8       JNZ  SCAN_LOOP          ; stop if end of second string reached 
        DONE: 
3B CA       CMP  CX, DX             ; which string ended first?

Input: Y string pointer in SI, length in CX. X string pointer in DI, length in DX. Output is Falsey if CF.

Example test program:

This test program uses additional PC DOS API I/O routines to take multi-line input from STDIN.

Download and test ACRON.COM.

Answer 41

JavaScript, 32 bytes

Repost of a port of Kevin's Java solution to a duplicate challenge, modified in case my choices for I/O weren't standards at the time this challenge was posted.

x=>y=>!!y.match([...x].join`.*`)

Try it online! (will update with this challenge's test cases when I get back to a computer)

Answer 42

Japt, 4 bytes

Repost from a duplicate challenge.

Takes input in reverse order (Y, then X).

à øV

Try it

Answer 43

Red, 82 bytes

func[x y][parse/case y collect[foreach c x[keep reduce['to c 'skip]]keep[to end]]]

Try it online!

Answer 44

Charcoal, 18 bytes

Ｆη≔✂θ⁼ι∧θ§θ⁰Ｌθ¹θ¬θ

Try it online! Link is to verbose version of code. Uses Charcoal's default Boolean output of - for true, nothing for false. Explanation:

Ｆη

Loop over the characters of Y.

⁼ι∧θ§θ⁰

See if this character is the next character of X.

≔✂θ...Ｌθ¹θ

If so then remove that character from X.

¬θ

Were all the characters of X consumed?

Answer 45

Julia 1.0, 53 bytes

f(x,y)=(x==""||[x=x[2:end] for c=y if c==x[1]];x=="")

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Answer 46

05AB1E, 3 bytes

æQà

Try it online or verify all test cases.

Or alternatively:

æså

Try it online or verify all test cases.

With both programs the first input is \$Y\$ and the second input is \$X\$.

Explanation:

æ    # Get the powerset of the (implicit) input-string `Y`
 Q   # Check for each if it's equal to the (implicit) input-String `X`
  à  # And check if any are truthy by taking the maximum
     # (after which this result is output implicitly)

æ    # Get the powerset of the (implicit) input-String `Y`
 s   # Swap to take the (implicit) input-String `X`
     # (could also be `I` or `²` to simply take input)
  å  # Check if this string is in the powerset-list of strings
     # (after which this result is output implicitly)

Answer 47

J (20 chars)

(<x)e.(#:i.2^#y)<@#y

The input is given in the variables x and y. It makes a list of all subsequences of y, so don't use it for very big strings.

Answer 48

Python (72)

import itertools as I;any(tuple(X)==Z for Z in I.permutations(Y,len(X)))

Answer 49

C, 120

main(i,c){char x[99],y[99];c=0;gets(y),gets(x);for(i=0;y[i]!='\0';)c+=y[i++]==x[c]?1:0;puts(x[c]=='\0'?"True":"False");}

Answer 50

Retina, 26 bytes (not competing)

The language is newer than the challenge. Byte count assumes ISO 8859-1 encoding. Input is taken on two lines with Y first.

+`^(.)(.*¶)(?(\1).|)
$2
¶$

Try it online

Answer 51

Javascript, 104 chars

function _(x,y){f=true;for(c in x){if(y.indexOf(x[c])>-1)y=y.replace(x[c],"");else{f=!f;break}}return f}

Ungolfed

function _(x,y){
f=true;
    for(c in x)
    {
        if(y.indexOf(x[c])>-1)
           y=y.replace(x[c],"");
        else {
            f=!f;
            break;
        }
    }
    return f;
}