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

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 5

Retina, 13 bytes

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

Try it online! Link includes test cases.

Answer 6

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 7

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 8

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 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

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 11

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 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

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 14

Funky, 34 bytes

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

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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

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

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Answer 18

QuadR, 8 bytes

..\b
⌽⍵M

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Answer 19

05AB1E, 7 bytes

#vy`sðJ

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-1 thanks to Magic Octopus Urn.

Prints one trailing space.

Answer 20

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 21

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

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

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Answer 22

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 23

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 24

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 25

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

Answer 26

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 27

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 28

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 29

Jelly, 10 bytes

ḲṪ,¥\€Ḋ€UK

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Answer 30

SNOBOL4 (CSNOBOL4), 136 119 bytes

	I =INPUT
B	I SPAN(&LCASE &UCASE) . Y ARBNO(' ') =:F(O)
	Y RPOS(2) REM . Z =REVERSE(Z)
	O =O Y ' '	:(B)
O	OUTPUT =O
END

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Prints with a trailing space. You know you've done something wrong when a language is a backronym for StriNg Oriented and symBOlic Language and your code is longer than Brain-Flak :( now it's slightly better.

Line B takes I and replaces (alphabetic characters saved as Y)(some number of spaces) with the empty string.

The following line extracts the last 2 characters of Y as Z and replaces them as Z reversed, then the next line concatenates O, Y, and a single space character.

Finally, it prints when I no longer matches the required pattern in line B.

Answer 31

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

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

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Answer 32

Clean, 75 59 bytes

-16 from Laikoni

@[a,b]=[b,a]
@[a,b,' ':c]=[b,a,' ': @c]
@[a:b]=[a: @b]
@e=e

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Answer 33

Standard ML (MLton), 92 77 bytes

implode o foldr(fn(a,b:: #" "::r)=>b::a:: #" "::r|(a,r)=>a::r)[#" "]o explode

Try it online! Based on xnor's Haskell solution.

---

Previous 92 byte version

fun f(a::b:: #" "::r)=b::a:: #" "::f r|f(x::r)=x::f r|f r=r
fun$s=implode(f(explode(s^" ")))

Try it online! Example usage: $ "Flippign Lettesr Aroudn" yields "Flipping Letters Around ".

Ungolfed

fun f (a :: b :: #" " :: r) = b :: a :: #" " :: f r
  | f              (x :: r) = x :: f r
  | f                    r  = r

fun g s = implode (f (explode (s ^ " ")))

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Answer 34

Japt, 7 bytes

A port of Jonathan's Jelly solution.

®á g1}S

Try it

---

Explanation

® }S splits the input on spaces, maps over the array and rejoins with spaces, á generates an array of all permutations of the current element and g1 gets the second element of that array.

Answer 35

Wolfram Language (Mathematica), 72 bytes

x""<>(Characters[StringSplit@x][[;;,#]]&/@{;;-3,-1,-2})~Riffle~" "

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Another approach using manipulation of character list and take parts instead of string regex.

The two Unicode characters are \[Function] and \[Transpose] in private use area of Mathematica, take 3 bytes each.

74 bytes

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Answer 36

Excel VBA, 78 Bytes

Anonymous VBE immediate window that takes input from range A1 and outputs to the VBE immediate window.

For Each s in Split([A1]):l=Len(s):?Left(s,l-2)Right(s,1)Mid(s,l-1,1)" ";:Next

Answer 37

Ruby, 53 bytes

$*.each{|x|print x[0,x.length-2]+x[-2,2].reverse," "}

The program takes input through command line arguments (formatted as a regular sentence, without quotes).

$* is a shorthand for ARGV, therefore i can iterate over each of the words, already in an array.

Only thing that bugs me is that awful x.length-2 snippet. I would prefer to use x.last(-2), but i believe that is only in RoR.

Called like:

ruby file.rb Helol Wordl

So that each word becomes a parameter. (Not wrapped in quotes)

Answer 38

Clojure, 122 117 bytes

-5 bytes by shortcutting the indices of the characters to swap

(fn[p](map #(let[l(count %)v(vec %)y(dec l)z(- l 2)](apply str(assoc v y(v z)z(v y))))(clojure.string/split p #" ")))

It's a shame that most of Clojure's String function are tucked away inside of Clojure.string, and that there are no good ways of manipulating the characters of Strings directly. clojure.string/split is atrociously long, and the conversion to-and-from vectors added quite a lot. I might be able to shave off a couple bytes by shortcutting the trailing indices, but this is the only decent algorithm I could think of.

(defn flip-last [phrase]
  (let [words (clojure.string/split phrase #" ")]

    ; For each word...
    (map #(let [len (count %)
                ; Need to turn the string into a vector so we can manipulate it using indices
                vec-word (vec %)]

            ; ... then turn the vector of characters back into a String.
            (apply str
               ; Associate the element at the second last index with the element at the last index,
               ;  and vice-versa...
              (assoc vec-word (dec len) (vec-word (- len 2))
                              (- len 2) (vec-word (dec len)))))

         words)))

Answer 39

QuadR, 8 bytes

..\b
⌽⍵M

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..\b any two characters followed by a word boundary

gets replaced with

⌽⍵M the reverse of the Match

---

Equivalent to the following 18-byte Dyalog APL function:

'..\b'⎕R{⌽⍵.Match}