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;$_=<>=~$_

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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, 45 bytes

lambda a,b:{1,l:=iter(b)}>{c in l for c in a}

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Builds on @qwatry's method used in @xnor's answer: iter consumes elements as they are searched for, so c in l "uses up" the character if it is found.

Some trickery lets us combine the all and the iterator assignment:

• {c in l for c in a} is a set comprehension which effectively deduplicates its contents, so it's always one of {}, {False}, {True}, or {False, True}
• 1 and 0 are equivalent to True and False
• The > operator tests if {True, iter(b)} is a strict superset of the result, which is true only if that result is {} or {True}, i.e. all the c in l expressions were true
• By including l:=iter(b) in the set, we can sequence the evaluation order correctly with minimal parentheses by reusing the set's {} grouping
• It also lets us use > instead of >= because the iterator will never be present in the boolean set
• Otherwise, >= would be necessary because {True} is not a strict superset of {True}

Answer 5

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 6

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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In Python 3.10+, we don't need parens around the walrus expression for 46 bytes.

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, 49 bytes

def f(a,b):l=iter(b);print all(c in l for c in a)

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A dazzling new method introduced to me by qwatry. The idea is that turning b into the iterator l will consume values whenever they are read. So, checking c in l will consume values left to right until it finds one that equals c or it reaches the end.

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 7

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 8

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 9

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 10

C (52 chars)

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

Test cases

Answer 11

Burlesque (6 chars)

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

Answer 12

C, 23:

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

result in *x

http://ideone.com/BpITZ

Answer 13

PHP, 90 characters

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

Answer 14

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 15

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 16

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 17

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 18

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 19

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

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Answer 20

Charcoal, 18 17 bytes

ＦηＦ⁼ι∧θ§θ⁰≔Φθμθ¬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.

Ｆ⁼ι∧θ§θ⁰

See if this character is the next character of X.

≔Φθμθ

If so then remove that character from X.

¬θ

Were all the characters of X consumed?

Answer 21

Factor + math.combinatorics, 19 bytes

[ all-subsets in? ]

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Answer 22

JavaScript (no regex), 45 bytes

y=>x=>(i=0,[...y].map(c=>c==x[i]&&i++),!x[i])

Should be called like

f(haystack)(needle)

Answer 23

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 24

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 25

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 26

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 27

Red, 82 bytes

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

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Answer 28

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 29

Zsh -o glob_subst, 14 bytes

>:$1>${2///*}*

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Explanation:

• >:$1: copy stdin to a file named : and the first argument. The colon ensures that we don't try to create a file with an empty name in the case of an empty string as input. Since we don't take any input from stdin, this just creates an empty file.
• >${2///*}*: try to copy stdin (again) to a file that matches the pattern resulting from the expansion of ${2///*}* If no such file exists, an error will occur. The -o glob_subst ensures that the expanded value is treated as a pattern. Otherwise, it will be treated as a literal string. Alternatively, this option can be removed and ${~2///*}* used instead for +1 byte.
  • ${2}: take the second argument
  • ///*: replace every empty string with an asterisk; this results in abc -> *a*b*c
  • append another asterisk. Asterisks simply mean "match any sequence

Answer 30

Java 8, 163 162 38 35 bytes

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

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

NOTE: Doesn't work if the input contains special regex-characters, but this would invalidate a lot of existing answers as well.

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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,
    .replace("",".*"))
               //   where every character is surrounded with ".*"

For example:

a="anna"
b="banana"

Will do the check:

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