Day 6 turned out to be pretty straightforward. Like day 4 you had to deal with two consecutive newlines when parsing the data but assuming you did day 4 that's no problem.

The gist is that a group is formed by consecutive lines and groups are separated by a blank line. Each line in each group is a string of letters representing answers to questions. For instance, for this group:

abc
abd
ab

you have three people. The first answered 'yes' to a, b, and c. the second to a,b, and d and the third to a and b.

Your goal was to figure out how many different questions did each group answer 'yes' to and what was the total across the groups.

It seemed that the easiest thing to do was to take each group, remove the newlines and then count the unique characters.

In Python, something like this (Clojure in the video and on GitHub)

d="abc\nabd\nab"
d = d.replace("\n","")
s = set(d)
ans = len(s)

So, you just have to do that for each group and add them all up.

Part 2 added a small twist. Now you wanted to look at each group and count the number of answers that everyone in that group answered "yes" to. For the above example, it would be 2 - everyone answered a and b.

Assuming your language supports set operations, you can just use union.

Again in Python:

d = "abc\nabd\nab"
d_list = d.split()
result = set(d_list[0])

for item in d_list[1:]:
    result = result & set(item)
print(result)

Again, do it for all the groups and add it up.

This is a great question to go over set operations!!

Check out the Clojure video for that approach where I also talk about easier ways to complete yesterday's problem.

Enjoy!