Day 1
Part 1
I solved part 1 with Excel. Despite all its shortcomings, Excel remains an excellent tool to visualise and manipulate data. The logic was as follows: if any 2 unique numbers add up to 2020, then I cannot have any numbers with a sum greater than 2020. First step was to sort the data and find the smallest number, which for me was 60. Now for any number k in the list, if a different number from the list, say b, sums to 2020, then 2020-k-b would be equal to 0. The check is therefore to add an additional column where the values are given by 2020-k-b. When k+b is greater than 2020, the check returns a negative number. By deleting all entries with negative values it is possible to reduce the list and I found that 60 did not give any solutions. The next number in the list was 107. 107 was sadly also not a part of the solution, but did reduce the amount of testing data significantly. Continuing with this method until eventually the check column provided a 0 and I had the 2 numbers required.
Part 2
Part 2 was also solvable using Excel, but not in 5 minutes as was the case for part 1. Part 2 required finding 3 numbers in the list whose sum was 2020. First instinct was to have a symmetrical matrix, setting the values within to be 2020-ColumnHeader-RowHeader and then using conditional formatting such that if a number was a part of the original date it would get a green background.
Instead I went with Python for the solution. After sorting the data, I was able to use the below to get the solution.
for i in range(len(data)-3):
n1 = data[i]
for j in range(i+1, len(data)-2):
n2 = data[j]
if (n1 + n2 < 2020):
for k in range(len(data)-1, j, -1):
n3 = data[k]
if n1 + n2 + n3 == 2020:
print(n1*n2*n3)
exit()
This takes advantage of the fact that at any point, i < j < k slightly reducing the calculation time although not the time complexity, which remains at O(n3). A colleague managed to reduce it to O(n2) by making a smart use of sets.
Day 2
Day 2 was all about data parsing. Nothing better for data parsing than headache inducing regular expressions. The best regex engine is from perl. Most other languages invoke ‘perl-equivalent’ regex, but the syntax is never quite as simple. After looking at the problem again, I could have solved it in a much simpler regex than I did, but then, Advent of Code is about getting a solution, rather than the solution.
$data =~ /(\d)-(\d)*([a-z])\:\s([az]+)/;
my $min = $1;
my $max = $2;
my $check = $3;
my $pw = $4;
Not sure if that could be combined into a single line by making the initialisation a part of the regex check. But then, that’s why people say perl is unreadable so back to good, human readable code. Instead I did several substitution expressions at each line to transform it into a CSV line, split it into an array and then used variables such that I wouldn’t need to be referencing array indices in the rest of the code. In order to solve the first part, I removed all characters which were not a part of the password policy and then verified that the remaining string had a length between the min & max character occurences.
The second part involved checking a different policy, verifying if the password had the given letter at one of the 2 positions provided, but not both. I wasn’t able to get perl to treat the passwords as an array of characters as I remember doing in C eons ago. This meant splitting the pw into an array. Because the elves don’t recognise that arrays start at 0 there is a large risk of OBOE errors => reduce both min & max variables by 1, and then through the use of xor verify if only 1 of the 2 conditions works. Could it have been done with regex? Yes. Was using the array index much easier to implement, and later on, read? I certainly hope so.