Detect rotated strings

Question

Read two strings from stdin.
Output Yes if one string is a rotated version of the other.
Otherwise output No

Testcases

Input

CodeGolf GolfCode

Output

Yes

Input

stackexchange changestackex

Output

Yes

Input

stackexchange changestack

Output

No

Input

Hello World

Output

No

Answer 1

Ruby 49 41

a,b=$*;puts (a*2).sub(b,'')==a ?:yes: :no

Edit: replaced gets.split by $*

Answer 2

APL (28)

Takes input on two lines.

'No' 'Yes'[1+(⊂⍞)∊⌽∘A¨⍳⍴A←⍞]

Explanation:

• A←⍞: read a line of input and store it in A
• ⌽∘A¨⍳⍴A: Rotate A by x, for each x in [1..length A]. Gives a list, i.e. estT stTe tTes Test
• (⊂⍞)∊: read another line of input, and see if it is in this list.
• 1+: add one to this, giving 1 if the strings were not rotated and 2 if they were
• 'No' 'Yes'[...]: Select either the first or second element from the list 'No' 'Yes' depending on whether the strings were rotated or not.
• This value is output automatically.

Answer 3

Python, 70 bytes

a,b=raw_input().split()
print ['No','Yes'][a in b*2and len(a)==len(b)]

Testing ...

Answer 4

Python 70 Characters

a,b=raw_input().split()
print'YNeos'[len(a)<>len(b)or a not in 2*b::2]

Thanks to gnibbler for the slice trick.

Answer 5

J, 47

y=:>2{ARGV
(>1{ARGV e.1|.^:(i.#y)y){'No',:'Yes'

Answer 6

According to the spec (same string lengths):

Perl, 42 43 chars

$.=pop;$_=(pop)x2;print+(qw'yes no')[!/$./]

If different sized strings are allowed, the solution would be:

Perl, 47 chars

$.=(pop)x8;$_=(pop)x9;print+(qw'yes no')[!/$./]

rbo

Answer 7

Golfscript, 31

' '/:)~,\,=)~.+\/,(&'Yes''No'if

This one check length first, so it should work as expected.

Answer 8

J, 57

{&('No';'Yes')@-:/@:((/:~@(|."0 _~i.&$))&.>)&.(;:&stdin)_

Sample use:

$ echo -n CodeGolf GolfCode | jconsole rotate.ijs
Yes
$ echo -n stackexchange changestackex | jconsole rotate.ijs
Yes
$ echo -n stackexchange changestack | jconsole rotate.ijs
No
$ echo -n Hello World | jconsole rotate.ijs
No

Answer 9

JavaScript, 51

function f(a,b)a&&(a+a).replace(b,"")==a?"Yes":"No"

JavaScript doesn't have a canonical host, so this answer is written as a function of two arguments. The score goes up to 60 if we disallow JS 1.7 features (expression closures).

In the SpiderMonkey shell this would be (for a score of 71):

[a,b]=readline().split(" ");print(a&&(a+a).replace(b,"")==a?"Yes":"No")

Answer 10

Windows PowerShell, 76

$a,$b=-split$input
('No','Yes')[+!($a.length-$b.length)*"$b$b".contains($a)]

Answer 11

Haskell (98 96 chars)

g x y@(t:r)(z:w)|x==y="Yes"|1>0=g x(r++[t])w
g _ _[]="No"
f(x:y:_)=g x y y
main=interact$f.words

Answer 12

Python, 66 63

a,b=raw_input().split()
print'YNeos'[a!=(2*a).replace(b,"")::2]

Another solution in 69 char

a,b=raw_input().split()
print['No','Yes'][a in b*2and len(a)==len(b)]

Answer 13

Haskell, 91 89 bytes

i[x,y]|x`elem`(take.length<*>iterate(tail<>take 1))y="Yes";i _="No"
main=interact$i.words

I would add a link to try it online, but unfortunately, tio.run seems to have a version of Haskell that doesn't include (<>) in the Prelude by default. I wrote this using GHC 9.4.8.

De-golfed version:

-- Because of the partial function `tail`,
-- `rotateOnce` throws an error for empty lists.
rotateOnce :: [a] -> [a]
rotateOnce xs = tail xs ++ take 1 xs

-- `allRotationsOf` never calls `rotateOnce` on an empty list, so it's okay.
allRotationsOf :: [a] -> [[a]]
allRotationsOf xs = take (length xs) (iterate rotateOnce xs)

-- Note: String is the same as [Char].
answer :: [String] -> String
answer [x, y]
  | elem x (allRotationsOf y) = "Yes"
answer _ = "No"
main = interact (answer . words)

The function rotateOnce rotates a list leftwards once. This comment explains how it's essentially equivalent to rotateOnce xs = tail xs ++ take 1 xs, which means that it appends its first element to the very end. tail essentially works like drop 1 except that an exception is thrown on an empty list, but it's shorter. It returns the list it was given, but with the first element removed.

The function iterate is from the standard library. iterate f x evaluates to an infinite list [x, f x, f (f x), ...], and so on. So, iterate rotateOnce xs just keeps generating rotations of the list xs.

To ensure the program terminates, it stops after the infinite list has generated as many elements as there are in the original list, because a list of n elements can only be rotated n times before repetitions are encountered.

Using the Monad instance for functions, \x -> f (g x) (h x) can be rewritten liftM2 f g h, which is equivalent to f <$> g <*> h. The operator (<$>) is a synonym for fmap, and for functions, fmap is just the function composition operator, so it can be rewritten f . g <*> h to save characters. That's why take (length xs) (iterate rotateOnce xs) can be rewritten like that in the golfed version.

So anyway, using the functions described, the program just checks if y is some rotation of x. If so, the string "Yes" is returned, otherwise "No" is returned. And that's it.

Answer 14

J, 28 22 21 bytes

(No`Yes{::~e.)#\|."{]

Input taken as function arguments, -1 thanks to ovs

Attempt This Online!

(No`Yes{::~e.)#\|."{]
                    ]  NB. ‎⁡right arg
              #\       NB. ‎⁢take the right arg then get the length of each prefix [1..len(y)]
                |."{   NB. ‎⁣rotate the right by each element of the left, obtain all rotations
(No`Yes{::~e.)         NB. single unit to create a hook
           e.          NB. ‎⁤is the left arg a member of the resulting table?
 No`Yes                NB. a lesser known trick, yields a boxed array but requires valid identifiers
       {::~            NB. ‎⁢⁢fetch using the boolen result of e., unlike {, this unboxes the result

Answer 15

J, 84

y=:(>1{ARGV),:(>2{ARGV)
((0{y)e.(y&((]$0{[),(]-~[:}.[:$[)$1{[)/.i.}.$y)){'No',:'Yes'

Answer 16

JavaScript (120 chars)

function f(a,b) {for (i=0,A=a.split("");A.join("")!=b&&i++<a.length;A.push(A.shift()));return A.join("")==b?'Yes':'No';}

Output:

f('CodeGolf','GolfCode'); //Yes
f('stackexchange','changestackex'); //Yes
f('stackexchange','changestack'); //No
f('Hello','World'); //No
f('nn','nBn'); //No

Answer 17

Ruby, 58 (62) characters

a,b=gets.split;$><<(a.size==b.size&&/#{a}/=~b*2?:Yes: :No)

This solution assumes the input contains only alphanumeric characters (actually everything that doesn't have a special meaning inside a regular expression is ok).

A solution that doesn't have this constraint is 4 characters longer

a,b=gets.split;$><<(a.size==b.size&&(b*2).index(a)?:Yes: :No)

Answer 18

Python, 71

a,b=raw_input().split()
print'Yes'if a in b*2and len(a)==len(b)else'No'

Answer 19

PHP, 61

<?echo preg_match('/^(.+)(.*) \\2\\1$/',fgets(STDIN))?Yes:No;

Answer 20

Ruby, 41

puts gets =~ /^(.+)(.*) \2\1$/ ?:Yes: :No

Answer 21

Q (50 43 chars)

{`No`Yes x in((!)(#)y)rotate\:y}." "vs(0:)0

Answer 22

Scala 78

val b=readLine split " "
print(b(0).size==b(1).size&&(b(0)+b(0)contains b(1)))

It's a shame about the size check, without it the count drops to 54

val a=readLine split " "
print(a(0)+a(0)contains a(1))

Answer 23

bash 56

read a b
[[ $a$a =~ $b&&$b$b =~ $a ]]&&echo Yes||echo No

Answer 24

GolfScript, 25 bytes

' '/~.2*@/''+='Yes''No'if

How it works

             # STACK: "CodeGolf GolfCode"
' '/         # Split input string by spaces.
             # STACK: [ "CodeGolf" "GolfCode" ]
~            # Dump the array.
             # STACK: "CodeGolf" "GolfCode"
.            # Duplicate the topmost string.
             # STACK: "CodeGolf" "GolfCode" "GolfCode"
2*           # Repeat the topmost string.
             # STACK: "CodeGolf" "GolfCode" "GolfCodeGolfCode"
@            # Rotate the three topmost strings.
             # STACK: "GolfCode" "GolfCodeGolfCode" "CodeGolf"
/            # Split the second topmost string around the topmost one.
             # STACK: "GolfCode" [ "Golf" "Code" ]
''+          # Flatten the array of strings.
             # STACK: "GolfCode" "GolfCode"
=            # Check for equality.
             # STACK: 1
'Yes''No'if  # Push 'Yes' for 1, 'No' for 0.
             # STACK: "Yes"

Answer 25

CJam, 21 bytes

r]r_,,\f{\{(+}*}_@+^!

Try it online!

Answer 26

Husk, 14 bytes

Not sure about the rules here, Husk doesn't do IO at all. The closest alternative is a function:

!w¨Ye∫No¨€U¡ṙ1

Try it online!

Explanation

!w¨Ye∫No¨€U¡ṙ1
           ¡    -- iterate the following for ever:
            ṙ1  --   rotate string by 1
          U     -- only keep the longest prefix with unique elements
         €      -- is the argument in that list?
  ¨Ye∫No¨       -- compressed string: "Yes No"
 w              -- split on space
!               -- modular index (1-based)

Answer 27

jq -Rn, 62 + 4 = 65 bytes

len('if[inputs|explode|sort|implode]|.[0]==.[1]then"Yes"else"No"end')

As in the accepted answer, the two strings should be on separate lines. -R is used to avoid having to quote the input strings, and -n is used to disable the default input processing because inputs takes care of that.

Answer 28

C (gcc), 118 113 111 bytes

-5 bytes thanks to ceilingcat

i,l,r;f(s,t)char*s,*t;{for(l=r=0,i=strlen(s);i--;)r=strncmp(s+i,t,++l)|strncmp(s,t+l,i)?r:3;puts(r+"No\0Yes");}

Try it online!

Answer 29

05AB1E, 13 bytes

ā._Êßi„Noë”…Ü

Try it online or verify all test cases.

Explanation:

ā         # Push a list in the range [0, length] for the (implicit) input-string,
          # without popping this input-string itself
          #  i.e. "CodeGolf" → [1,2,3,4,5,6,7,8]
 ._       # Rotate the string that many times towards the left (vectorized)
          #  "CodeGolf" and [1,2,3,4,5,6,7,8] → ["olfCodeG","lfCodeGo","fCodeGol","CodeGolf",
          #                                      "odeGolfC","deGolfCo","eGolfCod","GolfCode"]
   Ê      # Check for each whether it's NOT equal to the second (implicit) input-string
          #  "GolfCode" → [1,1,1,0,1,1,1,1]
    ß     # Pop and push the minimum (to check if any are falsey)
          #  [1,1,1,0,1,1,1,1] → 0 (falsey)
     i    # If this minimum is truthy (so all rotations were unequal to the second input):
      „No #  Push string "No"
     ë    # Else:
      ”…Ü #  Push dictionary string "Yes"
          # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why ”…Ü is "Yes".

Answer 30

Jelly, 17 13 bytes

ɠṙJɠeị“Yes“No

Try it online!

How it works

ɠṙJɠeị“Yes“No - Main link. Takes no arguments
ɠ             - Read a line from STDIN
  J           - Indices; [1, 2, ..., len(S)]
 ṙ            - All rotations of S
   ɠe         - Read a line from STDIN; is that in the rotations?
     ị“Yes“No - Index that into ["Yes", "No"]