# La La Land… no wait, Moo Moo Moonlight

This challenge is a tribute to the winner of Best Picture at the Oscars 2017, La La Land Moonlight!

Write a function/program that takes a string containing only letters [A-Za-z], the four symbols that are common in every day sentences .,'? and spaces, and outputs the string in the style of La La Land.

To be more specific, take the letters up to, and including, the first vowel group, and print/output it twice adding a space each time, then print/output the whole string. y is a vowel in this challenge. Punctuation and capitalization should be kept.

You may assume that all strings contain at least one vowel, and that all strings start with a letter.

Test cases:

Land
La La Land

Moonlight
Moo Moo Moonlight

quEueIng
quEueI quEueI quEueIng

This isn't a single word.
Thi Thi This isn't a single word.

It's fun to play golf
I I It's fun to play golf

Ooo
Ooo Ooo Ooo

I'm okay
I I I'm okay

Hmm, no. There will be at least one vowel, but it can be anywhere.
Hmm, no Hmm, no Hmm, no. There will be at least one vowel, but it can be anywhere.

Why is y a vowel?
Why Why Why is y a vowel?


This is so the shortest code in each language wins. Explanations are encouraged, also in mainstream languages.

• Test case for case insensitivity: MOONLIGHT. And just for fun: Why did the chicken cross the road? – Titus Mar 1 '17 at 11:05
• Challenge sponsored by: National Stuttering Association – sergiol Mar 1 '17 at 23:34
• Or Prof. Quirrell – Brian J Mar 2 '17 at 19:26
• 6th test case bought to you by Louis Prima and the Jungle Book. Joined just to add this (bad) pun. – Toby Mar 3 '17 at 16:23

## Sed 30 bytes

s/[^aeiouy]*[aeiouy]\+/& & &/i

• Could you use * instead of \+? – user41805 Jun 26 '18 at 18:37

# 05AB1E, 2319 18 bytes

Saved 1 byte thanks to Okx.

Dlð«žOsSåJTk>£D¹ðý


Try it online! or as a Test suite

Explanation

 Dl                  # create a lowercase copy of implicit input
ð«                # append a space
žO              # push the vowels
s             # swap lowercase input to the top of the stack
S            # split into a list of chars
å           # check each char for membership in the vowel-string
# (creates a list with 1 at the index of vowels and 0 for non-vowels)
J          # join to string
Tk        # find the index of 10
>       # increment
£      # take that many items from input
D     # duplicate this string
¹    # push input
ðý  # join the strings by space


# Jelly, 24 22 20 19 14 bytes

-5 bytes by utilising a trick from Emigna's brilliant answer (look for 10 in the isVowel list)

;⁶e€Øyw⁵ḣ@;⁶Ȯ;


Try it online! (not quite sure how to make a test suite for this full program)

15 byte alternative:

;⁶e€Øyw⁵ḣ@;⁶ẋ2;


Here is full test suite.

### How?

;⁶e€Øyw⁵ḣ@;⁶Ȯ; - Main link: string s
⁶             - space character
;              - concatenate to s (for all vowel edge case)
Øy         - vowels + y yield
e€           - exists in? for €ach (gives a list of isVowel identifiers)
⁵       - 10
w        - index of first sublist (with implicit decimalisation of 10 to [1,0])
ḣ@     - head with reversed @rguments (start of word up to & including vowel group)
⁶   - space character
;    - concatenate (start of word up to & including vowel group plus a space)
Ȯ  - print and yield (hence a full program...
-     ...the alternative ẋ2 repeats instead in order to return the result)
; - join with the input, s
- implicit print (of the second repetition and input string)


# Python, 61 bytes

import re;lambda x:re.sub('(.*?[aeiouy]+)',r'\1 \1 \1',x,1,2)


Here comes the first non-regex-based language (using regex).

Saved 1 byte thanks to Neil.

# JavaScript (ES6), 40 46

Edit 5+1 bytes saved thx @Arnauld

Overly long compared to other using the same trick (as usual)

x=>x.replace(/.*?[aeiouy]+/i,'$&$& $&')  let f= x=>x.replace(/.*?[aeiouy]+/i,'$& $&$&')

test=Land
La La Land

Moonlight
Moo Moo Moonlight

queueing
queuei queuei queueing

This isn't a single word.
Thi Thi This isn't a single word.

It's fun to play golf
I I It's fun to play golf

Ooo
Ooo Ooo Ooo

I'm okay
I I I'm okay

Hmm, no. There will be at least one vowel, but it can be anywhere.
Hmm, no Hmm, no Hmm, no. There will be at least one vowel, but it can be anywhere.
test.split(\n\n).forEach(z=>{
var [i,k]=z.split(\n),x=f(i);
console.log(k==x ? 'OK':'KO',i+'\n'+x);
})

• @Arnauld no, but I could use '$&$& $&' - I always forget the special dollar chars. Thank you. Unfortunately now it's really a port of Martin's retina answer. – edc65 Feb 27 '17 at 9:51 • The ^ is required in Retina which -- I think -- looks for all matches by default. But do we really need it here? – Arnauld Feb 27 '17 at 10:03 • @Arnauld you're right again – edc65 Feb 27 '17 at 13:09 • -2: x=>(y=/.*?[aeiouy]+/i.exec(x)+' ')+y+x – nderscore Feb 27 '17 at 16:31 • @ETHproductions in fact. Thanks for noticing. – edc65 Feb 27 '17 at 20:41 ## Retina, 24 bytes i^.*?[aeiouy]+$& $&$&


Try it online!

## Batch, 180 bytes

@echo off
set/ps=
set v=aeiouy
set g=c
set t=
:l
call set w=%%v:%s:~,1%=%%
if %v%==%w% goto %g%
set g=o
:c
set t=%t%%s:~,1%
set s=%s:~1%
goto l
:o
echo %t% %t% %t%%s%


Implements a state machine. g keeps track of whether we've ever seen a vowel, so if the current letter isn't a vowel we know whether to output or continue with the next letter.

# PowerShell, 46474139 38 bytes

$args-replace"^.*?[aeiouy]+",(,'$&'*3)


Try it online!

Thanks to Maarten Bamelis for the bugfix

Saved 6 bytes thanks to Rynant

Saved 2 3 bytes thanks to Joey

# Ruby, 3132 30 Bytes

->s{(s[/.*?[aeiouy]+/i]+' ')*2+s}


Two bytes saved thanks to G B and Cyoce.

# PHP, 55 54 bytes

Note: encoded version uses IBM-850 encoding.

echo preg_filter("/^(.*?[aeiouy]+)/i","$1$1 $0",$argn);
echo preg_filter(~ðíÎÐı└ñ×ÜûÉèåóÈÍðû,~█╬▀█╬▀█¤,$argn); # Encoded  Run like this: echo "This isn't a single word." | php -nR 'echo preg_filter(~ðíÎÐı└ñ×ÜûÉèåóÈÍðû,~█╬▀█╬▀█¤,$argn);'


# Explanation

Just a regex replace with non-eager matching any character at the start of the string, followed by any amount of vowels (use i option for case insensitivity). That capture group is then printed twice, followed by the entire string.

• Saved a byte by using -R to make $argn available (Thx Titus) # Javascript (ES6), 38 bytes x=>(y=/.*?[aeiouy]+/i.exec(x)+' ')+y+x  f= x=>(y=/.*?[aeiouy]+/i.exec(x)+' ')+y+x <!-- snippet demo: --> <input list=l oninput=console.log(f(this.value))> <datalist id=l><option value=Land> <option value=Moonlight> <option value=queueing> <option value="This isn't a single word."> <option value="It's fun to play golf"> <option value=Ooo> <option value="I'm okay."> <option value="Hmm, no. There will be at least one vowel, but it can be anywhere."> <option value="Why is y a vowel?"> # Perl, 25 + 1 (-p flag) s/.*?[aeiouy]+/$& $&$&/i


# Perl 6, 30 bytes

{S:i/.*?<[aeiouy]>+/{$/xx 3}/}  Try it online! • Interesting that this is the same length as {S:i/.*?<[aeiouy]>+/$/ $/$//} – Brad Gilbert b2gills Feb 27 '17 at 15:05

# C, 202196195193190 180

i,j,k,m,n;f(char*a){if((a[i/12]-"AEIOUY"[i++%6])%32==0)k=n=24-(i%12);else if(k&&!n--){m=j=(i-13)/12;for(i=0;i<j*2;)printf("%c%c",a[i%j],(i==j-1)*32),i++;printf(" %s", a);}m?:f(a);}

Try it online!

Thing left to golf:

•Collapse two printf into one.

•Printing my space char can be changed into %*c logic I'm sure.

•I'm using conditionals which can be removed somehow

j=(i-13)/12 can likely be shortened.

•[A-Y] conditional checks if ==0 which is usually un-necessary, though I'm currently stuck on that one (tried switching if-else and ditching the ==0 altogether but that requires adding more {brackets} and increase byte size)

Tricks I've used to golf this down:

•Combined a double for loop string search by using modulo for x-axis and integer-division for y-axis (input string vs vowel string). (X-axis is looped twice before iterating once on y-axis; the first time with [A-Z] and the second time with [a-z] using character value 32 differential.

•Bypassed having to use "[A-Y] and [a-y]" by just taking the distance between character sets and modulo 32. That way if the distance is 0 (A-A) or if the distance is 32 (a-A)

•Re-using integer variables that are no longer in use as boolean variables.

•Recursively calling a function with the same string to process through it and avoid a second for-loop.

•Set BOOL values to the logic of setting another variable. (e.g. bool = i=5;) to knock out both with one stone.

•Ternary empty-true exploit abuse. (GCC)

i,j,k,m,n;
f(char*a){
if((a[i/12]-"AEIOUY"[i++%6])%32==0)
k=n=24-(i%12);
else
if(k&&!n--){
m=j=(i-13)/12;
i=0;
for(;i<j*2;)
printf("%c%c",a[i%j],(i==j-1)?32:0),i++;
printf(" %s", a);
}
m?:f(a);
}


Knocked off 10 bytes thanks to Keyu Gan (in comments)

• Note to self: j=(i-13)/12 can likely be shortened. – Albert Renshaw Feb 27 '17 at 21:51
• Am I missing something, or could you start with i=j=k=m=n=0;? – Richard Irons Mar 1 '17 at 10:05
• @RichardIrons the variables have to be declared first. – Albert Renshaw Mar 1 '17 at 10:33
• You may use i,j,k,m,n; for initialization. – Keyu Gan Jun 7 '17 at 18:14
• @KeyuGan undefined behavior, not guaranteed to always be 0. (as far as I know?) – Albert Renshaw Jun 7 '17 at 22:20

# MATL, 33 bytes

'(^.*?[yY%s]+)'13Y2YD'$1$1 $1'YX  Try it at MATL Online Explanation  % Implicitly grab input as a string '(^.*?[yY%s]+)' % Push this string literal (regex pattern) 13Y2 % Push the string literal 'AEIUOaeiuo' YD % Replace the '%s' in the string with 'AEIUOaeiuo' '$1 $1$1'     % Push the string literal to use for replacement which repeats
% the first match 3 times
YX              % Perform regular expression matching and replacement
% Implicitly display the result

• '(^.*?[yY%s]+)'13Y2YD'$1 '8:)YX saves 2 bytes – Luis Mendo Jun 7 '17 at 17:00 • '(^.*?[%s]+)'19Y2YD'$1 '8:)YX saves another 2 – B. Mehta Sep 18 '17 at 22:48
• @B.Mehta 19Y2 didn't exist when this answer was submitted unfortunately – Suever Sep 18 '17 at 22:56
• Yeah, slightly expected that reply... I'll keep my comment so others can learn about inbuilt literal 'aeiouy' too. – B. Mehta Sep 18 '17 at 22:59
• @B.Mehta No worries. With MATL Online (matl.suever.net) you can select a specific version using the drop-down on the top right – Suever Sep 19 '17 at 0:32

# V, 21, 20 bytes

é /ã[aeiouy]«“.
3ä|<


Try it online!

Explanation:

é               " Insert a space
/             " Jump forward too...
ã[aeiouy]«. "   The first non-vowel after a vowel
3ä              " Make three copies of
|             " Everything from the cursor to the first character
<            " Delete the space we inserted


Hexdump:

00000000: e920 2fe3 5b61 6569 6f75 795d ab93 2e0a  . /.[aeiouy]....
00000010: 33e4 7c3c                                3.|<


Alternate version (21 bytes):

Íã¨ƒ[aeiouy]«©/± ± &


Try it online!

This uses ridiculous regex compression, and it still manages to get it's butt kicked by the other golfing languages. For reference, this is about two/thirds the length of the regular "uncompressed" version, namely:

:%s/\v\c(.{-}[aeiou]).*/\1 \1 &


Explanation:

Í                               " Replace on every line:
ã                              "   Case-insensitive
<131>                        "   Any character, any number of times (non-greedy)
[aeiouy]«               "   A vowel, repeated once or more
<129>         "   Followed by anything
/        " Replaced with:
± ±     "   Capture group one twice, with spaces between
&   "   The whole matched pattern


Here is a hexdump:

00000000: cde3 a883 5b61 6569 6f75 795d aba9 812f  ....[aeiouy].../
00000010: b120 b120 26                             . . &

• +1 This has got to be the most impressive V regex submission I ever saw – user41805 Feb 27 '17 at 18:25
• Try it online! – nmjcman101 Feb 27 '17 at 18:35

# Python 3, 75 68 bytes

lambda s:(s[:[x in"aAeEiIoOuUyY"for x in s][1:].index(0)+1]+" ")*2+s


Try it online!

Explanation:

Works by generating a boolean value for every character in the input string based on whether or not it is a vowel, and finding the lowest index of 0, the first non-vowel (excluding the first character). It returns the substring to this index twice,separated by spaces, and the original string.

# Clojure, 192188 181 bytes

(fn[p](let[[f] p v #(#{\a \e \i \o \u \y}(Character/toLowerCase %))[q r](split-with(if(v f)v #(not(v %)))p)[w _](split-with v r)as #(apply str %)](str(as(repeat 2(str(as q)(as w) \ )))p)))


-4 bytes by inlining first-sp-pred (whoops).

-7 bytes by removing some missed spaces

This was far more challenging than I thought it would be going in! I'm manually parsing the string...since I still haven't gotten around to learning regex :/

See pre-golfed code for breakdown:

(defn repeat-prefix-cons [phrase]
(let [[first-letter] phrase ; Get first letter

; Function that checks if a lowercased character is a part of the vowel set
vowel? #(#{\a \e \i \o \u \y} (Character/toLowerCase %))

; cons(onant)? Negation of above
cons? #(not (vowel? %))

; Decide how to split it depending on if the first character is a vowel
first-sp-pred (if (vowel? first-letter) vowel? cons?)

; Split off the first chunk of cons/vowels
[pre1 r] (split-with first-sp-pred phrase)

; Split off the rest of the vowels
[pre2 r2] (split-with vowel? r)

; Shortcut function that turns a list into a string (Basically (join "" some-list-of-strings) )
as #(apply str %)]

(str ; ... then concat the prefix in front of the original phrase, and return
(as ; ...then turn it back into a string since "repeat" returns a list... ^
(repeat 2 ; ... then repeat it twice (shame Clojure doesn't have string multiplication)... ^
(str (as pre1) (as pre2) \ ))) ; Concat the 2 prefix parts together with an space at the end... ^
phrase)))


# Python 3, 101 96 bytes

s=input()
v=i=0
for c in s:
w=c in'aAeEiIoOuUyY'
if v*~-w:break
v=w;i+=1
print(s[:i],s[:i],s)


Try it online!

a non-regex solution

Commented:

s=input()
a='aAeEiIoOuUyY'
v=i=0
for c in s:          # for each character in the string
w=c in a            # w = True if the character is a vowel, else false
# true is equivalent to 1  and false to zero
# v*(w-1) evaluates only to true (-1 in this case) if v=1 (last character was a vowel) and w=0 (current character is not a vowel)
if v*(w-1):break    # if so, break the loop
v=w;i+=1            # increase the counter and set v to w
print(s[:i],s[:i],s)

• Why do you need a? Replace w=c in a with w=c in'aAeEiIoOuUyY' – sagiksp Mar 1 '17 at 10:14

# Ohm, 19 bytes (CP437), non-competing

New language, and as such, I had to add some new features to make this work, which sadly makes this non-competitive (because loopholes).

≡┬üC▓αy_ε;TF«u├DQüj


Explanation:

≡┬üC▓αy_ε;TF«u├DQüj     Main wire, arguments: s

≡                       Triplicate input
┬üC                    Push input, all lowercase with concatenated space character
▓    ;              Map string into an array with...
αy_ε                 Boolean: is element a vowel?
TF«u          Find first occurrence of [true, false]
├D        Slice input up to that index and duplicate it
Q       Reverse stack
üj     Join on spaces, implicitly print

• I'm curious to know which features you implemented...? – Stewie Griffin Feb 27 '17 at 21:04
• @StewieGriffin Stack reversal (Q), subarray searching (u), string/array slicing (├), and vowel constants (αv and αy). – Nick Clifford Feb 28 '17 at 0:02

# PHP, 69 65 53 bytes

<?=preg_filter("#.*?[aeiouy]+#i","$0$0 $0",$argn,1);


requires PHP 5.3 or later. Run as pipe with -F or try some versions online.

Saved 4 bytes (and fixed the code) with regex stolen from @aross;
10 more with preg_filter instead of preg_match and -F
and another two with an improved regex.

75 81 bytes for a non-regex version:

for(;$c=$argn[$i++];)($k+=$k^!trim($c,aeiouyAEIOUY))>1?:$w.=$c;echo"$w$w $argn";  requires PHP 5 or later; replace ?: with ?1: for older PHP. Run with -nR Breakdown for(;$c=$argn[$i++];)       // loop $c through input characters ($k+=$k^! // 2. !$k and vowel or $k and not vowel: increment$k
trim($c,aeiouyAEIOUY) // 1. strip vowels -> vowel=false, non-vowel=true )>1 // 3. if$k>1
?                           // do nothing
:$w.=$c;                    // else append $c to$w
echo"$w$w $argn"; // output  • Doesn't appear to work. Out put for This isn't a single word: T T This isn't a single word. – aross Feb 27 '17 at 12:20 • @aross seems it's checking just lowercase values? I may be wrong I don't know PHP that well – Albert Renshaw Mar 1 '17 at 10:53 • @AlbertRenshaw The regex version uses the i modifier that renders the regex case insensitive. The other one version did only check lowercase. Fixed. – Titus Mar 1 '17 at 11:02 ## R, 49bytes sub("(.*?[aeiouy]+)","\\1 \\1 \\1",scan(,""),T,T) Regex based replace, match everything until not a vowel, capture and replace it by itself 3 times. scan wait for a double type input, to tell it to use character type we have to give it two arguments, first is the default , emtpy string for stdin, and for the second the R evaluation allow to use only c as it's not ambiguous for character in this context. T stands for TRUE and save some char as 4th and 5th parameter to sub to tell it to ignore case and use PCRE (the greedyness is not the same with R regex syntax) 4 bytes saved courtesy of Sumner18 along with the Tio link to running code # Java 8, 147 140 bytes Golfed: import java.util.regex.*;s->{Matcher m=Pattern.compile("([^aeiouy]*[aeiouy]+)",2).matcher(s);m.find();return m.group()+" "+m.group()+" "+s;}  Ungolfed: import java.util.regex.*; public class LaLaLandNoWaitMooMooMoonlight { public static void main(String[] args) { for (String[] strings : new String[][] { { "Land", "La La Land" }, { "Moonlight", "Moo Moo Moonlight" }, { "queueing", "queuei queuei queueing" }, { "This isn't a single word.", "Thi Thi This isn't a single word." }, { "It's fun to play golf", "I I It's fun to play golf" }, { "Ooo", "Ooo Ooo Ooo" }, { "I'm okay", "I I I'm okay" }, { "Hmm, no. There will be at least one vowel, but it can be anywhere.", "Hmm, no Hmm, no Hmm, no. There will be at least one vowel, but it can be anywhere." } }) { final String input = strings[0]; final String expected = strings[1]; final String actual = f(s -> { java.util.regex.Matcher m = java.util.regex.Pattern.compile("([^aeiouy]*[aeiouy]+)", 2).matcher(s); m.find(); return m.group() + " " + m.group() + " " + s; } , input); System.out.println("Input: " + input); System.out.println("Expected: " + expected); System.out.println("Actual: " + actual); System.out.println(); } } private static String f(java.util.function.Function<String, String> function, String input) { return function.apply(input); } }  Note: the literal 2 in the code is the value of java.util.regex.Pattern.CASE_INSENSITIVE. • I think you can use import java.util.regex.*; to save some bytes. – Roman Gräf Feb 27 '17 at 20:14 • @RomanGräf you are correct. I had the packages spelled out because in an earlier version of the code (did not work) it was shorter not to use imports. I did not reevaluate after fixing the code. – user18932 Feb 27 '17 at 21:00 # C, 123 bytes #define v(x)while(x strchr("AEIOUY",*s&95))++s; a;f(s,t)char*s,*t;{t=s;v(!)v()a=*s;*s=0;printf("%s %s ",t,t);*s=a;puts(t);}  Call as: main(){char s[] = "queueing"; f(s);}  • This is nice! You knocked my C solution out of the park lol. – Albert Renshaw Mar 1 '17 at 10:54 # Pyke, 22 bytes l1~V\y+L{$0R+f2<ssDQdJ


Try it online!

This is 4 bytes longer than it should really be had I implemented a shorter way of getting vowels including y.

# Retina, 24 bytes

i1.*?[aeiouy]+
$0$0 $0  Try it online • Quite similar to this – Emigna Feb 27 '17 at 14:49 • Yeah, I know. But I made the answer independently. Even so, it's been decided that duplicate answers are allowed, if the work is not plagiarized. – mbomb007 Feb 27 '17 at 14:56 # Python 3, 130 102 bytes w=input();a='';v=0 for i in w: if i in 'aeiouyAEIOUY': v=1 elif v: break a+=i a+=' ';print(a*2+w)  Try it online! Uses no function of any kind and no external libraries! (Unless print and input count as functions, which they do). Works by seeing if it gets out of the consonants in the start of the title into the 'vowel zone'. If it is in the 'vowel zone' and detects a consonant, then it prints the title. Saved 28 bytes thanks to @LliwTelracs # MATLAB / Octave, 58 51 bytes 7 Bytes saved thanks to @HughNolan @(x)regexprep(x,'(^.*?[aeiouyAEIOUY]+)','$1 $1$1')


Creates an anonymous function named ans which can be called by passing a string to it: ans('Land')

Online Demo

For MATLAB compatibility, $0 should be used in place of $1 in the above function.

• Was thinking about this and then saw you had it done already. Save a few bytes: @(x)regexprep(x,'^.*?[aeiouyAEIOUY]+','$0$0 $0 '); - also Matlab seems to use$0 rather than \$1 strangely – Hugh Nolan Mar 1 '17 at 12:13
• @HughNolan Great point, thanks! – Suever Mar 1 '17 at 14:27

# C (gcc), 111 110 bytes

*d="AEIOUYaeiouy";b;f(char*a){b=strcspn(a,d);write(printf(" "),a,write(1,a,b+strspn(a+b,d)));printf(" %s",a);}


Try it online!

This just uses library functions strspn() and strcspn() and exploits the order in which gcc evaluates function parameters. Slightly less golfed

*d="AEIOUYaeiouy";b;
f(char*a){
b=strcspn(a,d);
write(printf(" "),a,write(1,a,b+strspn(a+b,d)));
printf(" %s",a);
}


Thanks to @gastropner for -1.

# Pyth - 24 bytes

jd[Ke:Q"^.*?[aeiou]+"1KQ

• I think you maybe forgot y? – Stewie Griffin Feb 27 '17 at 18:04