One more LUL and I'm out

Question

Challenge description

On some channels on a popular streaming site twitch.tv a common message people tend to spam in chats to bait people into spamming "LUL" is

One more LUL and I'm out

LUL is a popular emote used to express that something funny happened on stream.

Soon dank memes showed their potential and a parody of the copy-pasta ensued:

One more "One more LUL and I'm out" and I'm out

Which is the same message nested in itself. Given a non-negative integer N, output the LUL-pasta nested N times in itself following the pattern below.

Standard code-golf rules apply, the shortest code in bytes wins.

Sample input / output

0: One more LUL and I'm out
1: One more "One more LUL and I'm out" and I'm out
2: One more "One more "One more LUL and I'm out" and I'm out" and I'm out
...
7: One more "One more "One more "One more "One more "One more "One more "One more LUL and I'm out" and I'm out" and I'm out" and I'm out" and I'm out" and I'm out" and I'm out" and I'm out

Notes

• Leading/trailing newlines are allowed
• Capitalization must be preserved
• Your code may be a full program or a function
• Instead of printing, you may return a string or its equivalent in your language of choice
• You may index from 1 instead of 0

Answer 1

Python 2, 56 bytes

lambda x:('"One more '*x+'LUL'+' and I\'m out"'*x)[1:-1]

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It is 1-indexed

Answer 2

JavaScript, 57 56 54 52 bytes

f=q=>`One more ${q?`"${f(q-1)}"`:"LUL"} and I'm out`

Test Snippet:

f=q=>`One more ${q?`"${f(q-1)}"`:"LUL"} and I'm out`

<input type=number min=0 oninput=o.textContent=f(+this.value)>

<p id=o>

For some reason the snack snippet is being buggy when the input is 0, but this works otherwise. Call it like f(4).

Explanation

f=q=>                      //declares a function f with argument q
`One more ... and I'm out` //hardcoded string
 ${q?`"${f(q-1)}"`:"LUL"}  // does recursion based on q
                           // if q is true, ie not 0, recurse
                           // else return "LUL"

Answer 3

Befunge, 91 bytes

&:00p10p>"tuo m'I dna "1v
09p00-1<^g09p01-1_v#:g0<<vg
>>00g:#^_$>:#,_@  >$"LUL">" erom enO"

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This is a breakdown of the source code with the various component parts highlighted.

We start by reading the repeat count N, and storing two duplicates of it in memory.
We then countdown the first N, pushing multiple copies of " and I'm out" onto the stack in reverse. Each additional copy is separated from the one before with a quote. The quote is generated with the sequence 90g (basically reading a copy from the first line of the source), since that's the shortest way to do so.
Once this first loop is complete, we push "LUL" onto the stack (technically this is in reverse, but it obviously makes no difference when it's a palindrome).
Then we have another loop, wrapping across the right border, over to the left of playfield, and then back again. This time we're counting down the second N, pushing multiple copies of "One more " onto the stack (again in reverse). And again, each additional copy is separated from the one before with a quote.
Once the second loop is complete, the entire phrase is now on the stack (in reverse), so we simply need to write it out.

Answer 4

C++, 118 + 16 = 134 bytes

auto L(int x){std::string k="One more LUL and I'm out",r=k;for(int i=0;i++<x;)r.replace(i*10-1,3,'"'+k+'"');return r;}

#include<string> - +16

replaces "LUL" to the whole string N times.

Anyone has better golfs?

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Massive thanks to Kritixi Lithos and hvd, for, uh, Massive help.

Answer 5

05AB1E, 30 29 bytes

…LULIF“"One€£ ÿ€ƒ I'm€Ä"“}}¦¨

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Different string-types doesn't seem to mix well, so for some reason I need to end the loop twice.

Answer 6

Javascript (ES6), 68 Bytes

f=(x,y="LUL")=>~x?f(--x,`"One more ${y} and I'm out"`):y.slice(1,-1)

Call like f(n).

You can also call it like f(n, "LUL") and replace LUL with any word you wish.

Answer 7

V, 39 37 bytes

Two bytes with the help of @KritixiLithos for coming up with the substitution method

iOne more LUL and I'm outÀñÓLUL/"."

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Hexdump:

00000000: 694f 6e65 206d 6f72 6520 4c55 4c20 616e  iOne more LUL an
00000010: 6420 4927 6d20 6f75 741b c0f1 d34c 554c  d I'm out....LUL
00000020: 2f22 122e 22                             /".."

Answer 8

Java, 79 77 bytes

Golfed:

String f(int l){return"One more "+(l<1?"LUL":'"'+f(l-1)+'"')+" and I'm out";}

Ungolfed, with test:

public class OneMoreLulAndImOut {

  public static void main(String[] args) {
    OneMoreLulAndImOut obj = new OneMoreLulAndImOut();
    for (int i = 0; i < 8; ++i) {
      System.out.println(Integer.toString(i) + ": " + obj.f(i));
    }
  }

  String f(int l) {
    return "One more " + (l < 1 ? "LUL" : '"' + f(l - 1) + '"') + " and I'm out";
  }    
}

Program output:

0: One more LUL and I'm out
1: One more "One more LUL and I'm out" and I'm out
2: One more "One more "One more LUL and I'm out" and I'm out" and I'm out
3: One more "One more "One more "One more LUL and I'm out" and I'm out" and I'm out" and I'm out
4: One more "One more "One more "One more "One more LUL and I'm out" and I'm out" and I'm out" and I'm out" and I'm out
5: One more "One more "One more "One more "One more "One more LUL and I'm out" and I'm out" and I'm out" and I'm out" and I'm out" and I'm out
6: One more "One more "One more "One more "One more "One more "One more LUL and I'm out" and I'm out" and I'm out" and I'm out" and I'm out" and I'm out" and I'm out
7: One more "One more "One more "One more "One more "One more "One more "One more LUL and I'm out" and I'm out" and I'm out" and I'm out" and I'm out" and I'm out" and I'm out" and I'm out

Answer 9

Python, 79 bytes

I just wanted to do a recursive solution, even though it's longer than other answers.

x='"One more %s and I\'m out"'
f=lambda n,s=x:n and f(n-1,s%x)or(s%"LUL")[1:-1]

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Answer 10

C#, 125 bytes

n=>{string f="One more {0} and I'm out",s=f;for(int i=0;i++<n;)s=string.Format(s,$"\"{f}\"");return string.Format(s,"LUL");};

Answer 11

C#, 119 85 71 bytes

string m(int n)=>$"One more {(n<1?"LUL":$"\"{m(--n)}\"")} and I'm out";

Saved 14 bytes thanks to @Luc

Answer 12

Mathematica, 69 68 bytes

Thanks to Martin Ender for saving 1 hard-to-find byte!

""<>Nest[{q="\"",{"One more ",#," and I'm out"},q}&,"LUL",#+1][[2]]&

Unnamed function taking a nonnegative integer argument and returning a string. Nest applies a function repeatedly to an initial argument; in this case, the function is to surround its argument by the appropriate words and quotation marks. We start from "LUL" and iterate N+1 times; that results in unwanted quotation marks enclosing the entire phrase, but [[2]] keeps only the stuff between them. At the end, ""<> turns the resulting heavily nested list into a single string.

Previous submission:

""<>Nest[{o,q="\"",#,q,a}&,{o="One more ","LUL",a=" and I'm out"},#]&

Answer 13

C, 140 111 bytes

My first attempt at a golfing question.. Golfed:

#define F printf(
#define P 1&&putchar(34)
int a=0;void q(n){a=a?a:n,n?n>0?F"One more "),n-P:n?n+P,F" and I'm out"):0:F"LUL"),n+a?q(n-1):0;}

I have come to realise is the wrong output since q(0) just gives LUL. The next attempt:

#define o(Q) O(Q,n?34:0);
#define O printf
void q(int n){o("One more %c")n?q(n-1):O("LUL"),o("%c and I’m out")}

Example program (tested with GCC on OSX):

#define o(Q) O(Q,n?34:0);
#define O printf
void q(int n) {o("One more %c")n?q(n-1):O("LUL"),o("%c and I’m out")}

int main() {
    q(0),putchar('\n');
    q(1),putchar('\n');
    q(2),putchar('\n');
    q(3),putchar('\n');

    return 0;
}

Gives output

One more LUL and I’m out
One more "One more LUL and I’m out" and I’m out
One more "One more "One more LUL and I’m out" and I’m out" and I’m out
One more "One more "One more "One more LUL and I’m out" and I’m out" and I’m out" and I’m out

Answer 14

R, 100 97 92 bytes

"One more recursive function and I'm out"

f=function(n)paste("One more",`if`(n<1,"LUL",paste0('"',f(n-1),'"')),"and I'm out");cat(f(scan()))

Edit: Turns out that a non-recursive approach is slightly shorter:

x="One more ";y=" and I'm out";cat(x,rep(c('"',x),n<-scan()),"LUL",rep(c(y,'"'),n),y,sep="")

Answer 15

Retina, 51 bytes

.+
$*00LUL1$&$*
0
"One more 
1
 and I'm out"
^"|"$

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Answer 16

R, 97 bytes

function(n){s="One more LUL and I'm out";while(n){s=sub("LUL",paste0('"',s,'"'),s);n=n-1};cat(s)}

Ungolfed:

function(n) {
  s = "One more LUL and I'm out";
  while(n) {
    s = sub("LUL", paste0('"', s, '"'), s);
    n = n - 1
  };
  cat(s)
}

Answer 17

PowerShell, 72 67 bytes

"$('"One more '*($c=1+"$args"))LUL$(' and I''m out"'*$c)".Trim('"')

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Answer 18

Japt, 39 bytes

Thank you @ETHproductions for helping.

`"O Ú `p°U +"LUL"+` d I'm t"`pU)s1J

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Answer 19

Ruby, 51 bytes

One-indexed. Try it online

->n{['One more ']*n*?"+'LUL'+[" and I'm out"]*n*?"}

Answer 20

Lua, 101 bytes

i,f,g='"One more ',' and I\'m out"',io.read()+1 print((i:rep(g).."LUL"..f:rep(g)):sub(2,g*24-(g-2)))

Obvious string attempt. Repeats "One more and and I'm out" exactly input + 1 times, with a LUL inbetween, then removes first and last quote.

Answer 21

Stacked, 54 bytes

('"One more ' ' and I''m out"')*'LUL'join'^.|.$'εrepl

Try it here! Example usage of "function":

1
('"One more ' ' and I''m out"')*'LUL'join'^.|.$'εrepl
out

---

One for 56 bytes:

@n'One more LUL and I''m out':@o['LUL' '"'o'"'+ +repl]n*

Answer 22

PHP, 69 66 bytes

Note: uses IBM-850 encoding

for($o=LUL;~$argn--;$d=~¦)$o="One more $d$o$d and I'm out";echo$o;

Run like this:

echo 2 | php -nR 'for($o=LUL;~$argn--;$d=~¦)$o="One more $d$o$d and I'"'"'m out";echo$o;';echo
> One more "One more "One more LUL and I'm out" and I'm out" and I'm out

Explanation

for(
  $o=LUL;                           # Start with output string "LUL".
  ~$argn--;                         # Loop until N is -1.
  $d=~¦                             # Set delimiter to `"` after first
                                    #   iteration.
)
  $o="One more $d$o$d and I'm out"; # Add text with previous output
                                    #   embedded, between delims.

echo$o;                             # Output resulting string.

Answer 23

Haskell, 51 bytes

Indexes from 1.

f 0="LUL";f n="One more \""++f(n-1)++"\" and I'm out"

Answer 24

CJam, 51 49 bytes

" and I'm out\"""\"One more "li1+_@*"LUL"+1>@@*W<

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Ungolfed:

" and I'm out\""   "\"One more " // Push two strings to the stack
     l i 1 +                     // Read a number and add 1
     _ @                         // Copy number and rise '"One more ' to the top
     *                           // Multiply '"One more ' by a number
     "LUL" +                     // Add "LUL"
     1>                          // Chop the first quote
     @@                          // Push the result down
     *                           // Multiply ' and I\'m out"' by a number
     W<                          // Chop the last quote

Answer 25

Python 3, 68 Bytes

def f(a):return('"One more '*a+'LUL'+(' and I%sm out"'%"'")*a)[1:-1]

Answer 26

Ruby, 70 bytes

def l x,t="LUL";x.times{t='"One more %s and I\'m out"'%t};t[1..~1];end

Simply loops for the amount it's given, surrounding the last string via a format string each time.

Index starts at one.

Answer 27

Mathematica, 65 63 Bytes

Nest["\"One more "<>#<>" and I'm out\""&,"LUL",#]~StringTrim~_&

Two bytes off by noticing the challenge allows 1-indexing.

Answer 28

Groovy, 54 bytes

{x->('"One more '*x+'LUL'+' and I\'m out\"'*x)[1..-2]}

Pretty straightforward, same as the python answer but 2 bytes shorter. It is also 1-indexed.

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Answer 29

PHP

Hello, i found so far two ways for doing this.

The replacement way 1-indexed (121 bytes).

function f($x){$v='One more LUL and i\'m out';$t=$v;for($i=1;$i<=$x;$t=str_replace('LUL','"'.$t.'"',$v),$i++);return $t;}

The recursive way (86 bytes).

function r($n){$v=($n==0)?'LUL':'"'.r($n-1).'"';return'One more '.$v.' and i\'m out';}

Answer 30

C++, 80 + 16 = 96 bytes

std::string L(int c){return"One more "+(c?'"'+L(--c)+'"':"LUL")+" and I'm out";}

#include<string> - +16

Ungolfed:

std::string LUL(int count) {
    return "One more " + (count? ('"' + LUL(--count) + '"') : "LUL") + " and I'm out";
}

Calls itself recursively and uses string addition. Pretty straight forward. I mean what else can I say? Even the ungolfed version is essentially a one liner.

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Answer 31

Common Lisp, 90 89 108 105 103 bytes

(set's"One more LUL and I'm out")(dotimes(n(read)(princ s))(set's(format()"One more ~a and I'm out"s)))

Other solution (91 bytes)

Using Winny's solution in Racket (my is not exact copy)

(defun l(n)(if(< n 0)'LUL(format()"One more ~:[~a~;\"~a\"~] and I'm out"(> n 0)(l(1- n)))))

usage:

(format t"~a"(l 2))

Ideas for improvement are welcomed.

Answer 32

Cheddar, 71 bytes

i->('One more "'*i).slice(0,-1)+'LUL '+('and I\'m out" '*i).slice(0,-2)

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Answer 33

Jelly, 39 bytes

First Jelly submission! Can probably slice off a few by encoding the strings, but I couldn't get it working with the spaces yet.

“"One more ”;;“ and I'm out"”
“LUL”Ç¡ḊṖ

A 0-indexed version needs an additional Ç after the Ç¡.

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Explanation will follow soon!

Answer 34

Haskell - 73 bytes

n!s=[1..n]>>s
f n=init(n!"One more \"")++" LUL"++tail(n!"\" and I'm out")

I wonder if there something I could use besides init and tail to save a few bytes.

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Answer 35

Perl 6, 57 bytes

{substr "\"One more "x$_~"LUL"~" and I'm out\""x$_,1,*-1}

Slightly more interesting, but 58 bytes:

{join "LUL",("One more "," and I'm out"Xxx$_)».join('"')}

Answer 36

Clojure, 84 bytes

#(nth(iterate(fn[s](str"One more \""s"\" and I'm out"))"One more LUL and I'm out")%)

Argh why isn't the first LUL in quotes? :( Almost correct in 63 bytes:

#(nth(iterate(fn[s](str"One more \""s"\" and I'm out"))"LUL")%)

Answer 37

GameMaker Language, 106 bytes

1-indexed

s='LUL'for(a=0,a<argument0;a++)s='"One more '+s+" and I'm out\""return string_copy(s,2,string_length(s)-2)

Answer 38

Clojure, 90 bytes

(fn c[n](if(= n -1)"LUL"(let[q(if(> n 0) \")](str"One more "q(c(dec n))q" and I'm out"))))

Written and golfed on my phone on a soft keyboard using the "Clojure REPL" app. That was sure a treat.

Explicitly recursive solution. Inserts the results of recursing into the "template string".

Pregolfed:

(defn lul [n]
  (if (= n -1)
    "LUL"
    (let [quote? (if (> n 0) \")]
      (str "One more "
            quote? (lul (dec n)) quote?
           " and I'm out" ))))

Answer 39

Octave, 72 70 bytes

@(n)[{'"One more '}{o=!(1:n)+1} "LUL " {"and I'm out\" "}{o}](2:end-2)

1 based index

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Answer 40

PowerShell, 88 bytes

for($d=($s="One more LUL and I'm out");$i++-lt$args[0]){$d=$d.Replace("LUL","`"$s`"")}$d

Answer 41

Batch, 95 bytes

@set s=LUL
@for /l %%i in (0,1,%1)do @call set s="One more %%s%% and I'm out"
@echo %s:~1,-1%

0-indexed, because that means it works even if you forget to pass a parameter. (The 1-indexed version would just print U in that case, which looks silly.)

Answer 42

C#, 102 90 bytes

string l(string s,int n){return n<0?s.Substring(1,s.Length-2):l($"\"One more {s} and I'm out\"",--n);}

Recursive, call with l("LUL",7);

EDIT

Thank you for pointing out that it is illegal to pass input not specified in the challenge. Here is the fixed version that only takes an integer N:

string l(int N){return N<0?"":$"One more \"{l(--N)}\" and I'm out".Replace("\"\"","LUL");}

Answer 43

Racket, 88 87 bytes

(define(f n)(format"One more ~a and I'm out"(if(= n 0)'LUL(format"\"~a\""(f(- n 1))))))

Ungolfed

(define (f n)
  (format "One more ~a and I'm out"
          (if (= n 0)
              'LUL
              (format "\"~a\"" (f (- n 1))))))

Unit tests

(module+ test
  (require rackunit)
  (check-equal? (f 0) "One more LUL and I'm out")
  (check-equal? (f 1) "One more \"One more LUL and I'm out\" and I'm out")
  (check-equal? (f 2) "One more \"One more \"One more LUL and I'm out\" and I'm out\" and I'm out")
  (check-equal? (f 7) "One more \"One more \"One more \"One more \"One more \"One more \"One more \"One more LUL and I'm out\" and I'm out\" and I'm out\" and I'm out\" and I'm out\" and I'm out\" and I'm out\" and I'm out"))

Answer 44

Scala, 63 bytes

(n:Int)=>("One more \""*n init)+"LUL"+("\" and I'm out"*n tail)

1-indexed

Answer 45

C++, 183 bytes

Simply input N times you want 'One more LUL' nested.

#include <iostream>
#include <string>
#define s string
using namespace std;main(){int N;cin>>N;s a="LUL";s b="One more \"";s c="\"and I'm out";for(int i=0;i<N;i++){a=b+a+c;}cout<<a;}

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Answer 46

tcl, 88

set g [set f "One more LUL and I'm out"]
time {regsub -all LUL $f "\"$g\"" f} $n
puts $f

demo

Answer 47

C++14, 90 bytes

As unnamed lambda modifying its input. Requires s to be std::string or alike.

[](auto&s,int i){s="LUL",++i;for(char c=0;i--;s="One more "+(c+s+c)+" and I'm out",c=34);}

Ungolfed and usage:

#include<string>
#include<iostream>

auto f=
[](auto&s,int i){
 s="LUL",++i;        //set basic LUL
 for(
  char c=0;          //first string concat without quotes
  i--;               
  s="One more "+(c+s+c)+" and I'm out", //string concat
  c=34                                  //set c to "
 );
}
;

int main(){
 std::string s;
 f(s,0);
 std::cout << s << std::endl;
 f(s,1);
 std::cout << s << std::endl;
 f(s,2);
 std::cout << s << std::endl;
 f(s,3);
 std::cout << s << std::endl;
}

Answer 48

Objective-C, 218

#define p NSString stringWithFormat
-(NSString *)f:(int)a{NSString *o=@"One more ",*k=@" and I'm out",*i=[p:@"%@LUL%@",o,k],*s=@"";for(int j=-a;a-->j+1;s=[p:@"%@\"%@",s,a>0?o:a<0?k:i]);return [s substringFromIndex:1];}

Not the most beautiful language to golf in, but I had fun

Answer 49

QBIC, 72 bytes

:[a|X=X+@One more `+chr$(34)┘Z=Z+chr$(34)+@ and I'm out`]Z=X+A+@LUL`+B+Z

The chr$(34) for the quotes is particularly costly, I need to fix a bug in the interpreter to make them work inside a string literal...

Sample output:

Command line: 0
One more LUL and I'm out

Command line: 3
One more "One more "One more "One more LUL and I'm out" and I'm out" and I'm out" and I'm out

Answer 50

dc, 82 bytes

?dsasw[[LUL ]P]se[34dP]sr[lalw>r[One more ]nlad1-dsar1=e0<y[and I'm out]nP32P]dsyx

Might as well add an answer in dc. This is 1-indexed.

"Calculate" it on the Internet!

Answer 51

Python, 78 bytes

f=lambda n:'One more "'*(n-1)+"One more LUL and I'm out"+'" and I\'m out'*(n-1)

Answer 52

k, 58 bytes

Indexing starts at 0. This is a recursive function.

`0:`c${"One more ",$[x;34,o[x-1],34;"LUL"]," and I'm out"}

Try it online.

Answer 53

GolfScript, 44 bytes

~"LUL"\{'"One more 
 and I\'m out"'n/*}*(;);

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1-indexed.

Explanation:

~"LUL"\{'"One more \n and I\'m out"'n/*}*(;); ("\n" is a newline)
~                                             Eval (input is the only stack element on program start)
 "LUL"                                        Push "LUL"
      \                                       Swap
       {'"One more \n and I\'m out"'n/*}      Push block:
        '"One more \n and I\'m out"'            Push "\"One more \n and I'm out\""
                                    n           Push n (default: newline)
                                     /          Split
                                      *         Join
                                        *     Repeat block
                                         (    Pop first element and push to stack
                                          ;   Pop
                                           )  Pop last element and push to stack
                                            ; Pop

Answer 54

SOGL V0.12, 27 bytes

LUL”.I{" ž-┌πēj░ō|≥π┘⁴‘⁾}jk

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Answer 55

Perl 5, 55 + 1 (-p) = 56 bytes

$_='"One more 'x++$_.LUL.' and I\'m out"'x$_;s/^.|.$//g

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