AdventOfCode2020 (10 Part Series)
1 Advent of code: 2020 Day 01
2 Advent of code: 2020 Day 02
… 6 more parts…
3 Advent of code: 2020 Day 03
4 Advent of Code: 2020 Day 04
5 Advent of Code: 2020 Day 05
6 Advent of Code: 2020 Day 06
7 AoC: 2020 Day 06 AWK
8 Advent of Code: 2020 Day 07
9 Advent of Code: 2020 Day 08
10 Advent of Code: 2020 Day 09 solutions
️ SPOILER ALERT ️
This is a post with my solutions and learning from the puzzle. Don’t continue reading if you haven’t tried the puzzle on your own yet.
If you want to do the puzzle, visit adventofcode.com/2020/day/6.
My programming language of choice is python and all examples below are in python.
TOC
Key learning
- Sets and dictionaries
This puzzle teaches us how to counting unique elements in lists.
Puzzle
The puzzle is about groups of persons answering questions and the goal is to count the results.
The questions are named a-z. Only questions answered “yes” to are present in the input. The questions that one person answered are in one line. The groups of persons are separated by an empty line. Example input:
abc
a
b
c
ab
ac
a
a
a
a
b
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Part 1
For each group, count the number of questions to which anyone answered “yes”.
What is the sum of those counts?
Parse input-data:
If we read the file as a whole without stripping away all the new-line characters (\n) we can quickly group the groups by splitting on double new-line.
f = open('06.input', 'r')
data = f.read()
groups = data.strip().split('\n\n')
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Solution:
Using a set as data structure is the simplest way to solve this puzzle. The built-in set in python will handle duplicates for us.
def part1(groups):
total = 0
for group in groups:
# 1. Remove all new-line characters # 2. Extract unique letters by creating a set # 3. Count size of set total += len(set(group.replace('\n', '')))
return total
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Part 2
For each group, count the number of questions to which everyone answered
“yes”. What is the sum of those counts?
In this puzzle a set won’t do anymore as we need to keep track of occurrences for each letter.
Solution:
Using a dictionary here is more valuable as we can use the key-value store to keep track of occurence per letter.
def part2(groups):
total = 0
for group in groups:
# Use dictionary to count occurrences of letters occurrences = {}
for letter in group:
if letter not in occurrences:
occurrences[letter] = 0
occurrences[letter] +=1
# By splitting on new-line we get each persons answers. person_count = len(group.split('\n'))
# Filter out letters that occur once per persons in group answered_by_all = [x == person_count for x in occurrences.values()]
total += sum(answered_by_all)
return total
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Alternative solution
Python is all about making it easy for the developer. It exists already a library called Counter helping us with counting. Using it would make our code cleaner and would look like this:
from collections import Counter
def part2count(groups):
total = 0
for g in groups:
# Use counter to count occurrences of letters letter_counter = Counter(g)
# By splitting on new-line we get each persons answers. person_count = len(group.split('\n'))
# Filter out letters that occur once per persons in group answered_by_all = [x == person_count for x in occurrences.values()]
total += sum(answered_by_all)
return total
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Read more about Counter here: https://docs.python.org/3/library/collections.html
Set operations
These puzzles are essentially doing unions and intersections for each group. Using set operations it could look like this:
groups = open('06.input', 'r').read().strip().split('\n\n')
def part1set(groups):
sets = [map(set, group.splitlines()) for group in groups]
unions = [set.union(*s) for s in sets]
return sum(map(len, unions))
print part1set(groups)
def part2set(groups):
sets = [map(set, group.splitlines()) for group in groups]
intersections = [set.intersection(*s) for s in sets]
return sum(map(len, intersections))
print part2set(groups)
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Functional style
These function are easy to write in a more functional style. My opinion is that it is less verbose, but a tad more complex to read.
def part1func(groups):
return sum([
len(set(group.replace('\n','')))
for group
in groups
])
def part2func(groups):
return sum([
len([x for x in Counter(group).values() if x == len(group.split('\n'))])
for group
in groups
])
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Python tricks
Instead of filtering on a boolean comparison and check the length, we can just
map the element to the boolean and sum them. The sum will be treat the booleans
as 1 if True or 0 if False. See example below:
answered_by_all = [
x
for x
in occurrences.values()
if x == person_count
]
total += len(answered_by_all)
# Same as above answered_by_all = [
x == person_count
for x
in letter_counter.values()
]
total += sum(answered_by_all)
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Deconstructing parameters
In the set function we do a set.union(*arr). What it does is that it spreads
the arrays elements as parameters. Example:
def my_function(a,b,c):
return a * b * c
arr = [1,2,3]
res = my_function(*arr)
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Thanks for reading!
I hope these solutions were helpful for you.
Complete code can be found at: github.com/cNille/AdventOfCode/blob/master/2020/06.py
AdventOfCode2020 (10 Part Series)
1 Advent of code: 2020 Day 01
2 Advent of code: 2020 Day 02
… 6 more parts…
3 Advent of code: 2020 Day 03
4 Advent of Code: 2020 Day 04
5 Advent of Code: 2020 Day 05
6 Advent of Code: 2020 Day 06
7 AoC: 2020 Day 06 AWK
8 Advent of Code: 2020 Day 07
9 Advent of Code: 2020 Day 08
10 Advent of Code: 2020 Day 09 solutions
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