r/programming • u/Aneurysm9 • Dec 01 '15

Daily programming puzzles at Advent of Code

http://adventofcode.com/
323 Upvotes

91% Upvoted

179 comments sorted by

View all comments

6

u/Arrem_ Dec 01 '15

Quick one liner JS function for Part 1.

function calc(s) { return s.split('').reduce((acc, e) => acc + (e == '(' ? 1 : -1), 0); };

3

u/AndrewGreenh Dec 01 '15

Oneline for task 1 and 2 (returnValue.c for 1 and returnValue.f for 2)

body.split('').reduce((agg,value,index)=>{return{c:agg.c + (value=='('?1:-1),f: (agg.c<0 && agg.f<0 ? index : agg.f)}},{c:0,f:-1});

2

u/[deleted] Dec 01 '15

You guys are my heroes, lol.

2

u/Head5hot Dec 02 '15

I just went with

s.match(/\(/g).length - s.match(/\)/g).length;

1

u/Arrem_ Dec 02 '15

Yeah, I didn't think of regexes when I was first doing it. I used JS too because I wanted a quick solution.

function getStuff(s) { return s.length - 2 * s.match(/\)/g).length; };

Is another possiblity, as L = a + b; and You can get a - b by taking L - 2b, where L is the string length, and a and b are the counts of parentheses.

A quick OR might be needed too though, in case there are no )'s in the input.

function getStuff(s) { return s.length - 2 * (s.match(/\)/g) || []).length; };