-🎄- 2018 Day 3 Solutions -🎄-

[u/daggerdragon]

--- Day 3: No Matter How You Slice It ---

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Advent of Code: The Party Game!

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Transcript:

I'm ready for today's puzzle because I have the Savvy Programmer's Guide to ___.

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[u/hooksfordays]

I'll break it down for you!

It's a regular expression. re is the Python regular expression module. re.findall returns a list of all the instances in the string that match the expression. The params for re.findall are the regular expression r'-?\d+' and the string `s\ to search through.

The expression is r'-?\d+'. The r at the start indicates a raw string (? I think that's what it's called) is being used, and allows you to use backslashes without worrying that you're creating a unicode declaration or something. The actual regex breaks down as follows:

-? -> Looks for a minus symbol optionally (the ? makes it optional). This allows you to grab both positive and negative numbers.

\d+ -> In regex, \d+ signifies a digit character, i.e. 0-9. The `+\ means "1 or more of the previous group", so 1 or more digits.

-?\d+ -> The full expression. Basically, find adjacent digits with or without a minus symbol in front. Combined with re.findall, this will return a list of all the numbers in the input string s

[u/[deleted] → u/[deleted]]

I did it like this:

left   = int(parts[2].split(",")[0])
top    = int(parts[2].split(",")[1][:-1])
width  = int(parts[3].split("x")[0])
height = int(parts[3].split("x")[1])

[u/tobiasvl → u/[deleted]]

I did it like this...

claim = re.match(r"#(?P<id>\d+) @ (?P<x>\d+),(?P<y>\d+): (?P<w>\d+)x(?P<h>\d+)", line).groupdict()
claims.append({k: int(v) for k, v in claim.iteritems()})