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#advent of code 2020
#day 10
#very similar (if not identical) to towels puzzle, 2024-d19
#memorize previous paths and just sum up the valid paths for each adapter
#although this one i think is easier due to no duplicates
#and working on numbers instead of characters
bag = [];
PuzzleInput = open("10.in","r");
for num in PuzzleInput: bag.append(int(num));
PuzzleInput.close();
#first and last items are from puzzle description
bag = [0] + [x for x in sorted(bag)] + [max(bag) +3];
connections = dict();
connections[1] = 0;
connections[2] = 0;
connections[3] = 0;
#since there are no duplicates and all adapters must be connected
#we can just check against the previous value for part 1;
for index in range(1,len(bag)):
adapter1 = bag[index];
adapter2 = bag[index-1];
diff = adapter1 - adapter2;
connections[diff] += 1;
p1 = connections[1] * connections[3];
p2 = 0;
options = dict();
options[0] = 1;
for index in range(1,len(bag)):
previous = bag[:index];
possible = [x for x in previous if bag[index] - x <=3];
options[bag[index]] = 0;
for poss in possible:
options[bag[index]] += options[poss];
p2 = options[max(bag)];
print("part 1 =",p1);
print("part 2 =",p2);
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