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

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 5

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 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, 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 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#, 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 14

Retina, 51 bytes

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

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Answer 15

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 16

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 17

Japt, 39 bytes

Thank you @ETHproductions for helping.

`"O Ú `p°U +"LUL"+` d I'm t"`pU)s1J

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Answer 18

Ruby, 51 bytes

One-indexed. Try it online

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

Answer 19

PowerShell, 72 67 bytes

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

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Answer 20

Go, 104 82 105 bytes

func f(n int)string{if n<0{return"LUL"}
q:=``
if n!=0{q=`"`}
return`One more `+q+f(n-1)+q+` and I'm out`}

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• -22 bytes by @Steffan: string concat instead of Sprintf
• +23 bytes to follow spec (n=0 has no quotes)

Answer 21

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 22

Haskell, 51 bytes

Indexes from 1.

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

Answer 23

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 24

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 25

Python 3, 68 Bytes

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

Answer 26

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 27

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 28

Mathematica, 65 63 Bytes

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

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

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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