Flippign Lettesr Aroudn

Question

In chat, we are often fast-typers and don't really look at the order of letters before posting a message. Since we are lazy, we need a program that automatically swaps the last two letters in our words, but since we don't want to respond too late, the code must be short.

Your task, if you wish to accept it, is to write a program that flips the last two letters of each word in a given string (so the word Thansk turns into Thanks). A word is a sequence of two or more letters in the English alphabet delimited by a single space.

• The string / list of characters you receive as input is guaranteed to only contain alphabetic characters and spaces (ASCII [97 - 122], [65 - 90] and 32).

• You can take input and provide output through any standard method, in any programming language, while taking note that these loopholes are forbidden by default.

• The output may have one trailing space and / or one trailing newline.

• The input will always contain words only (and the corresponding whitespace) and will consist of at least one word.

This is code-golf, so the shortest submission (scored in bytes), in each language wins!

Test cases

Note that the strings are surrounded with quotes for readability.

Input -> Output

"Thansk"                                 -> "Thanks"
"Youer welcoem"                          -> "Youre welcome"
"This is an apple"                       -> "Thsi si na appel"
"Flippign Lettesr Aroudn"                -> "Flipping Letters Around"
"tHe oDd chALlEneg wiht swappde lettesR" -> "teH odD chALlEnge with swapped letteRs"

Or, for test suite convenience, here are the inputs and their corresponding outputs separately:

Thansk
Youer welcoem
This is an apple
Flippign Lettesr Aroudn
tHe oDd chALlEneg wiht swappde lettesR

Thanks
Youre welcome
Thsi si na appel
Flipping Letters Around
teH odD chALlEnge with swapped letteRs

Thanks to DJMcMayhem for the title. This was originally a CMC.

Answer 1

V, 4 5 bytes

òeXp

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|| denotes the cursor

The buffer starts with |w|ord and more words and the cursor being on the first character.

Recursively ò

go to the end of a word

wor|d| and more words

remove X the character to the left of the cursor

wo|d| and more words

paste it over the next character

wod|r| and more words

Implicit ending ò, repeat the same process for other words until the end of the buffer is reached

Answer 2

Jelly, 7 bytes

Ḳœ?@€2K

A monadic link taking and returning lists of characters

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How?

Ḳœ?@€2K - Link: list of characters
Ḳ       - split at spaces
     2  - literal two
    €   - for €ach:
   @    -   with sw@pped arguments:
 œ?     -     nth permutation (the 2nd permutation has the rightmost elements swapped)
      K - join with spaces

Answer 3

Brain-Flak, 122 bytes

{(({})[((((()()){}){}){}){}])((){[()](<{}>)}{}){{}<>(({}({}))[({}[{}])])(<>)}{}({}<>)<>}<>(({}({}))[({}[{}])]){({}<>)<>}<>

Try it online!

The worst language for the job :)

Readable Slightly more readable version:

{
    (({})[((((()()){}){}){}){}])((){[()](<{}>)}{})

    {
        {}
        <>

        (({}({}))[({}[{}])])

        (<>)
    }
    {}

    ({}<>)<>

}<>

(({}({}))[({}[{}])])

{

    ({}<>)
    <>
}<>

Answer 4

Python 3, 50 bytes

print(*(w[:-2]+w[:-3:-1]for w in input().split()))

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This answer abuses Python 3's behavior of print: Multiple arguments are printed with a single space between them. Of course, we can't just give it multiple arguments because we don't know how many words will be in the input. So we use the splat operator. Basically

print(*[a,b,c])

is exactly the same thing as

print(a,b,c)

Abusing that makes a full program turn out shorter than a function/lambda where we'd have to use ' '.join or something similar.

Answer 5

Haskell, 40 bytes

(f=<<).words
f[a,b]=b:a:" "
f(x:r)=x:f r

Try it online! Usage example: (f=<<).words $ "abc xyz" yields "acb xzy ".

Answer 6

Retina, 13 bytes

(.)(.)\b
$2$1

Try it online! Link includes test cases.

Answer 7

Matlab (R2016b), 51 50 bytes

Saved 49 50 (!) bytes thanks to @Giuseppe.

function s(a),regexprep(a,'(\w)(\w)( |$)','$2$1 ')

And my previous answer:

Matlab (R2016b), 100 bytes

(Just for the fun of it :P)

function s(a),a=regexp(a,' ','split');for i=1:length(a),fprintf('%s ',a{i}([1:end-2 end end-1])),end

Explanation:

function s(a) % Defining as a function...
a=regexp(a,' ','split'); % Splits the input string at the spaces
for i=1:length(a) % Loops through each word
    fprintf('%s ',a{i}([1:end-2 end end-1])) % And prints everything with the last two characters swapped.
end

Answer 8

C,  62   58  54 bytes

Thanks to @Dennis for saving  four  eight bytes!

f(char*s){s[1]>32||(*s^=s[-1]^=*s^=s[-1]);*++s&&f(s);}

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Answer 9

Prolog (SWI), 60 bytes

[A,B]+[B,A].
[A,B,32|U]+[B,A,32|Y]:-U+Y,!.
[A|U]+[A|Y]:-U+Y.

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Explanation

First we define the base case:

p([A,B],[B,A]).

This means that the last two letters will always be swapped.

Then we define what happens if we are right next to a space:

p([A,B,32|U],[B,A,32|Y]):-p(U,Y),!.

Two strings match if right before a space the letters before the space are swapped and the remainder if the strings match. We then use ! to cut.

Our last case is if we are not next to a space the first two letters need to match.

p([A|U],[A|Y]):-p(U,Y).

Answer 10

Wolfram Language, 117 bytes

StringReplace[RegularExpression["\\b[[:alpha:]]{2,}\\b"]:>StringDrop[StringInsert["$0",StringTake["$0",{-1}],-3],-1]]

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Applied to the test strings.

StringReplace[
  RegularExpression["\\b[[:alpha:]]{2,}\\b"] :> 
   StringDrop[StringInsert["$0", StringTake["$0", {-1}], -3], -1]] /@
 {"Thansk", "Youer welcoem", "This is an apple", 
  "Flippign Lettesr Aroudn", "tHe oDd chALlEneg wiht swappde lettesR"} // Column

Thanks
Youre welcome
Thsi si na appel
Flipping Letters Around
teH odD chALlEnge with swapped letteRs

Answer 11

R, 111 51 41 bytes

Courtesy of @Giuseppe, a regex approach which blows my old method out of the water.

cat(gsub("(.)(.)\\b",'\\2\\1',scan(,"")))

Answer 12

APL (Dyalog Classic), 28 bytes

1↓∊((¯2↓⊢),2↑⌽)¨' '(,⊂⍨⊣=,)⍞

⎕ML and ⎕IO are both 1,

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Explanation

• ... (,⊂⍨⊣=,) ... Split (while keeping borders, and appending a border to the beginning) ...
• ... ⍞ ... the input ...
• ... ' ' ... ... at spaces.
• ... ( ... )¨ ... Then, to each element of that:
  • ... , ... Concatenate ...
  • ... (¯2↓⊢) ... ... every item except the last two ...
  • ... 2↑⌽ ... ... with the reverse of the last two elements.
• 1↓∊ ... Finally, return all but the first element of the flattened result.

Answer 13

Funky, 34 bytes

s=>s::gsub("(.)(.)( |$)","$2$1$3")

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Answer 14

Haskell, 45 bytes

-2 bytes thanks to H.PWiz.

(r.g.r=<<).words
g(x:y:z)=' ':y:x:z
r=reverse

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Answer 15

J, 20 19 11 bytes

Credit to @Bolce Bussiere

1&A.&.>&.;:

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       &.;:      on words
    &.>          on each
  A.             apply the permutation
1&               number 1, swap the last two elements

Answer 16

Alice, 24 bytes

/0RR'.%$1\' o
\ix*o ne@/

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Explanation

/...\' o
\.../

This forms a loop where the loop body is a linear Ordinal snippet and we execute ' o in Cardinal mode between every two loop iterations. The latter just prints a space.

Unfolding the zigzag structure of the Ordinal code, the linear loop body actually looks like this:

iR*' %e10xRo.n$@

Breaking this down:

i     Read all input. On subsequent iterations, this will push an empty string.
R     Reverse.
*     Join. On the first iteration, this joins the input to an implicit empty string,
      which does nothing. On subsequent iterations, it will join the empty string to
      the word on top of the string, thereby getting rid of the empty string.
' %   Split around spaces. On the first iteration, this will split the input
      into individual words. On subsequent iterations, this does nothing.
e10   Push "10".
x     Use this to permute the (reversed) word on top of the stack. In
      particular, the word is rearranged with the same permutation that is
      required to sort the string "10", which means the first two letters
      get swapped (which correspond to the last two letters of the actual
      word).
R     Reverse the swapped word.
o     Print it.
.n$@  If there are no words left on the stack, terminate the program.

Answer 17

brainfuck, 109 100 bytes

Edit: don’t have to handle one letter words

,[>++++[-<-------->],]>+[-<[>++++[<++++++++>-]<[->>+<<]<]<<[->>+<<]>[[-<+>]>]<<[>+>+>]-<]>>>>>>>[.>]

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Prints a trailing space

How It Works

,[>++++[-<-------->],] Puts input on the tape and subtracts 32 from each character
                       This separates each word

>+[- Start the loop
   <[>++++[<++++++++>-]<[->>+<<]<] Add 32 to each letter of the word
                                   Skip this on the first iteration for the last word

   <<[->>+<<]>[[-<+>]>] Swaps the last two letters of the word
   <<[>+>+>]- If there is another word to the left continue loop
              Also set up to add a space to the end of the word
 <] End loop
 >>>>>>>[.>] Print the modified string

Previous version, 109 bytes

,[>++++[-<-------->],]>+[-<[>++++[<++++++++>-]<[->>+<<]<]<<[[->>+<<]>[[-<+>]>]<<[<]]>[>]<[>+>+>]-<]>>>>>>[.>]

Try it online!

Answer 18

QuadR, 8 bytes

..\b
⌽⍵M

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Answer 19

Google Sheets, 33 Bytes

Anonymous worksheet function that takes input from cell A1 and outputs to the calling cell

=RegExReplace(A1,"(.)(.)\b","$2$1

-2 Bytes Thanks to @KevinCruijssen for the use of (.) over (\w)

Answer 20

Julia 1.0, 30 bytes

!s=replace(s,r"..\b"=>reverse)

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detect end of words with a regex r"..\b" and apply reverse on the matches

Answer 21

Curry, 33 bytes

Tested in both KiCS2 and PAKCS.

f(u++[x,y])=u++[y,x]
(>>=f).words

Curry's powerful patterns let it beat the Haskell answer!

To test it you can use Smap. Just select KiCS2 2.2.0 or PAKCS 2.2.0 and paste the following complete code:

f(u++[x,y])=u++[y,x]
g=(>>=f).words
main=f "Hello world an happy day"

Answer 22

Perl 5, 19 + 1 (-p) = 20 bytes

s/(\w)(\w)\b/$2$1/g

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Answer 23

PHP, 119 107 bytes

Edit: thanks to totallyhuman

<?php foreach(explode(" ",trim(fgets(STDIN)))as$w)echo substr($w,0,strlen($w)-2).strrev(substr($w,-2))," ";

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Answer 24

Haskell, 41 bytes

foldr(%)" "
a%(b:' ':r)=b:a:' ':r
a%s=a:s

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Outputs with a trailing space.

The repeated ' ':r looks wasteful. But a%(b:t@(' ':r))=b:a:t is the same length and a%(b:t)|' ':_<-t=b:a:t is one byte longer.

---

Haskell, 41 bytes

f(a:b:t)|t<"A"=b:a:f t|1>0=a:f(b:t)
f e=e

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Answer 25

Haskell, 53 45 bytes

u[a,b]=[b,a]
u(a:b)=a:u b
unwords.map u.words

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Explanation

u is a function that swaps the last two letters of a word. To apply it to all the words we use words to split the list, map it across all of the words and then use unwords to put it back together.

Answer 26

sed, 20 17+1 (-r) = 18 bytes

s/(.)(.)\b/\2\1/g

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Answer 27

PHP, 65 bytes

requires PHP 7.1 (or later)

for(;$s=$argv[++$i];$s[-1]=$s[-2],$s[-2]=$c,print"$s ")$c=$s[-1];

takes sentence as separate command line arguments. Run with -nr.

---

working on a single string, 77+1 bytes:

foreach(explode(" ",$argn)as$s){$c=$s[-1];$s[-1]=$s[-2];$s[-2]=$c;echo"$s ";}

Run as pipe with -nR.

---

... or try them online.

Answer 28

Java 8, 35 bytes

s->s.replaceAll("(.)(.)\\b","$2$1")

Port of @TaylorScott's Google Sheets answer, after I golfed two bytes. EDIT: I see it's now a port of Neil's Retina answer after my two golfed bytes.

Explanation:

Try it online.

s->                           // Method with String as both parameter and return-type
   s.replaceAll("(.)(.)       //  Replace any two characters,
                       \\b",  //  with a word-boundary after it (space or end of String)
                "$2$1")       //  With the two characters swapped

Answer 29

JavaScript (Node.js), 38 36 32 bytes

s=>s.replace(/(.)(.)( |$)/g,"$2$1 ") 

s=>s.replace(/(.)(.)\b/g,"$2$1")

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RegExp approach courtesy @Giuseppe (although I thought of this independently), assuming words separated by only one space

-2 for only considering 1 space and add trailing space

-4 Thanks @Shaggy

Answer 30

K (oK), 23 22 bytes

" "/{x@prm[!#x]1}'" "\

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Example:

" "/{x@prm[!#x]1}'" "\"Hello World"
"Helol Wordl"

Explanation:

Port of FrownyFrog's solution to save 1 byte.

I'll come back to this.

" "/{prm[x]1}'" "\ / the solution
              " "\ / split input on " "
    {       }'     / apply lambda to each
     prm[x]        / permute input x
           1       / and take the 2nd result
" "/               / join with " "

Previous solution:

• " "/-2{(x_y),|x#y}'" "\ 23 bytes