Day 2 solutions in Python

02a.py

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+""" Advent of Code 2018
+
+December 02, puzzle 1
+"""
+
+def main(lines):
+	""" For each line, count characters.
+
+		If any character appears exactly twice, increment x2.
+		If any character appears exactly three times, increment x3.
+		Ignore any doubles or triples past the first set.
+		At the end, multiply x2 by x3 to get the checksum.
+	"""
+	x2, x3 = 0, 0
+	for line in lines:
+		chars = {}
+		gota2, gota3 = False, False
+		for char in line:
+			if char in chars.keys():
+				chars[char] += 1
+			else:
+				chars[char] = 1
+		for k,v in chars.items():
+			if v == 2 and not gota2:
+				x2 += 1
+				gota2 = True
+			if v == 3 and not gota3:
+				x3 += 1
+				gota3 = True
+			if gota2 and gota3:
+				break
+	print("Totals: x2 {}, x3 {}".format(x2, x3))
+	print("Checksum: {}".format(x2 * x3))
+
+
+if __name__ == "__main__":
+	lines = []
+	with open("02in.txt","r") as f:
+		for line in f:
+			lines.append(line.strip())
+	main(lines)

02b.py

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+""" Advent of Code 2018
+
+December 02, puzzle 2
+"""
+
+def main(lines):
+	""" For each line, check to see if another line varies by only one character.
+
+		There's an easy but expensive way to do this. I'll start with that
+		and see if I can refine it later.
+	"""
+	
+	for i, line in enumerate(lines):
+		if i + 1 != len(lines):
+			for word in lines[(i+1):]:
+				diffs = 0
+				idx = 0
+				for j in range(len(line)):
+					if line[j] != word[j]:
+						diffs += 1
+						idx = j
+				if diffs == 1:
+					print("Found a match: {} and {}".format(line, word))
+					print("Matched value: {}".format("".join([word[:idx],word[(idx+1):]])))
+					return
+
+if __name__ == "__main__":
+	lines = []
+	with open("02in.txt","r") as f:
+		for line in f:
+			lines.append(line.strip())
+	main(lines)