Advent of Code – Day 4

So since we’ve had a little bit of free time lately, I’ve been doing some python scripting. The other day I decided to go back to Advent of Code and work on it a little bit more.

— Day 4: Secure Container —

You arrive at the Venus fuel depot only to discover it’s protected by a password. The Elves had written the password on a sticky note, but someone threw it out.

However, they do remember a few key facts about the password:

  • It is a six-digit number.
  • The value is within the range given in your puzzle input.
  • Two adjacent digits are the same (like 22 in 122345).
  • Going from left to right, the digits never decrease; they only ever increase or stay the same (like 111123 or 135679).

Other than the range rule, the following are true:

  • 111111 meets these criteria (double 11, never decreases).
  • 223450 does not meet these criteria (decreasing pair of digits 50).
  • 123789 does not meet these criteria (no double).

Part 1

How many different passwords within the range given in your puzzle input meet these criteria?

So for me, my range was 124075 – 580769. This was a pretty straight forward problem. First, iterate through all numbers within the range. As you go through each number, check to see if it meets a few conditions.

total = 0
for n in range (124075, 580769):
	num = [int(i) for i in str(n)]
	adjacent = 0
	decrease = 1
	#Check for adjacent values
	for i in range (0, 5):
		if num[i] == num[i+1]:
			adjacent = 1
			break
	#Check for decrease
	for i in range (0, 5):
		if num[i+1] < num[i]:
			decrease = 0
			break
	if adjacent and decrease:
		total += 1
		print(n)	
print(total)

Part 2

An Elf just remembered one more important detail: the two adjacent matching digits are not part of a larger group of matching digits.

Given this additional criterion, but still ignoring the range rule, the following are now true:

  • 112233 meets these criteria because the digits never decrease and all repeated digits are exactly two digits long.
  • 123444 no longer meets the criteria (the repeated 44 is part of a larger group of 444).
  • 111122 meets the criteria (even though 1 is repeated more than twice, it still contains a double 22).

How many different passwords within the range given in your puzzle input meet all of the criteria?

Again, pretty straight forward. Really only had to make a minor change to my code to see if the adjacent numbers were exactly two, or not.

total = 0
for n in range (124075, 580769):
	num = [int(i) for i in str(n)]
	adjacent = 0
	decrease = 1
	#Check for adjacent values
	for i in range (0, 5):
		if num[i] == num[i+1]:
			if num.count(num[i]) == 2:
				adjacent = 1
	#Check for decrease
	for i in range (0, 5):
		if num[i+1] < num[i]:
			decrease = 0
			break
	if adjacent and decrease:
		total += 1
		print(n)	
print(total)

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