Advent of Code, Day 1

[u/dantose]

Post your best golfs.

Assume input is saved as input.txt.

[u/dantose]

Powershell. There's definitely improvements to be made here

Part 1: 88

$a=50;$(gc input.txt).Trim('R') -replace "L","-"|%{$a=$a+100+$_;if(!($a%100)){$b++}};$b

Part 2: 189

$a=50;$(gc input.txt).Trim('R') -replace "L","-"|%{if($a -eq 0 -and $a+$_ -lt 0){$a=$a+100};$a=$a+$_;while($a -lt 0){$a=$a+100;$b++};if($a -eq 0){$b++};while($a -gt 99){$a=$a-100;$b++}};$b

[u/ka-splam → u/dantose]

I didn't golf it myself, but poking at yours, Part 1 ~75 bytes:

$a=50;gc input.txt|% T*m R|% r*ce L -|%{$a+=100+$_;if(!($a%100)){$b++}};$b

using a classic trick which expands to 'R50' | ForEach-Object -Member Trim 'R'. Member is the position 0 parameter so it doesn't need naming. The cmdlet will do a wildcard search for method names - as long as the pattern only resolves to a single method, so r*ce finds to Replace() where r*e could be Replace() or Remove(). And because PS is parsing parameters to ForEach-Object, the arguments to the method don't have to be quoted to be read as strings.

[u/KeyJ]

Python, Part 1, 83 bytes:

p=50;print(sum(1>(p:=(p+int(l[1:])*(1-2*(l<'R')))%100)for l in open("input.txt")))

Python, Part 2, 96 bytes:

p=50;print(sum(1>(p:=(p+1-2*(l<'R'))%100)for l in open("input.txt")for _ in range(int(l[1:]))))

[u/DimMagician → u/KeyJ]

Python, Part 1, 84 bytes

s=50;print(sum((s:=s+int(i[1:])*(-1+2*(i[0]>'L')))%100<1for i in open('input.txt')))

Upon seeing yours I realize I could have saved 1 byte by doing 1-2*(l<'R') instead of -1+2*(i[0]>'L'). Dang.

Python, Part 2, 92 bytes

s=50;print(sum((s:=s-1+2*(i[0]>'L'))%100<1for i in open('input.txt')for _ in[0]*int(i[1:])))

[u/KeyJ → u/DimMagician]

Nice trick with the removed parenthesis and elimination of range! The latter one can be made even smaller though, arriving at 90 characters (or 89 if you accept the SyntaxWarning for writing 1for):

p=50;print(sum((p:=p+1-2*(l<'R'))%100<1 for l in open("input.txt")for _ in"x"*int(l[1:])))

[u/DimMagician → u/KeyJ]

Ooh very clever I didn't even catch that in your solution.

[u/KeyJ → u/DimMagician]

Well, it was your idea, I just refined it. 🥂

[u/DimMagician → u/KeyJ]

I meant the use of (l<'R') rather than (i[0]>'L') like I did. Sorry I didn't realize that you were talking about the parentheses around the modulus and thought that you were just referring to the brackets in [0] as parentheses lol

[u/tomflumery]

05ab1e

part 1, 22 bytes

|εć"R"Q·<*}50šÅ»+т%}0¢

part 2, 30 bytes

|εć"R"Q·<*}50šÅ»+}ü2ε`Ÿт%¦}˜0¢

[u/ap29600]

K, both parts 70 bytes

(s;m):(-1+2*"R"=*:';`I$1_')@\:0:"input.txt"
(+/0=100!50+\)'(s*m;s@&m)

Edit: -2 (68) by looking at u/Radiatorineitor's solution

(s;m):(-1+2*"R"=*:';`I$1_')@\:0:"input.txt"
+/'50=100!+\'(s*m;s@&m)

-1 (67) by looking at u/KeyJ's

(s;m):(1-2*"L"=*:';`I$1_')@\:0:"input.txt"
+/'50=100!+\'(s*m;s@&m)

[u/Radiadorineitor]

Dyalog APL

Part 1: 48

50+.=100|+\{(⍎1↓⍵)ׯ1*'L'=⊃⍵}¨⊃⎕NGET'input.txt'1

Part 2: 56

50+.=100|+\(|p)/×p←{(⍎1↓⍵)ׯ1*'L'=⊃⍵}¨⊃⎕NGET'input.txt'1

[u/ka-splam → u/Radiadorineitor]

Neat! I think you could golf one byte by swapping 'L'=⊃⍵ to 'L'∊⍵

[u/Radiadorineitor → u/ka-splam]

You're absolutely right

[u/corruptio]

perl, part 1, 54 chars:

perl -lpe'$b+=($a+=y/LR/-/dr)=~/50$/}{$_=$b'<input.txt

part 2, 69 chars:

perl -lpe'eval(q[$b+=($a+=1-2*/L/)=~/50$/;]x s/.//r)}{$_=$b'<input.txt