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Viper
Viper

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Advent of Code 2020: Python Solution Day 10

#adventofcode #algorithms #python #challenge

I nearly forgot about doing the challenge. Part one was easy and part two was bit tricky so I did wrong at first and got help from this repo. Again I tried it first on test input and only then on real one. My solution is at this repository.

Part one 

with open("day10.txt", "r") as fp: lines = [int(line.rstrip()) for line in fp.readlines()] one_jolt = 0 two_jolt = 0 three_jolt = 0 outlet_rating = 0 lines.append(max(lines)+3) # because max jolt is added  while True: #print(1 in lines)  if (outlet_rating + 1) in lines: one_jolt+=1 outlet_rating += 1 elif outlet_rating+2 in lines: two_jolt+=1 elif (outlet_rating + 3) in lines: three_jolt += 1 outlet_rating+=3 else: break print("Part 1 answer is ", one_jolt*three_jolt)
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Part two 

# part 2 sol = {0:1} for line in sorted(lines): sol[line] = 0 if line - 1 in sol: sol[line]+=sol[line-1] if line - 2 in sol: sol[line]+=sol[line-2] if line - 3 in sol: sol[line]+=sol[line-3] print(sol[max(lines)])
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Lets Share Your Solution too.

I write blogs about Computer Vision projects on my GitHub page q-viper.github.io and if you got some time please share yours too.

Top comments (7)

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thetalent profile image
thetalent • • Edited 
data = sorted([int(i) for i in puzzle.input_data.split(lf)]) #heads of paths counted over all paths. resb={0:1} for i in range(len(data)-1): # sum of all paths so far value = sum(resb.values()) # next number differs by3 ? all paths apply n if data[i+1]-data[i]==3: resb={data[i]:value} else: try: # applying is a must for a path with head 3 lower than n # as we know from part 1 - a diff of 2 is not there resb[data[i]-3]=0 except: pass resb[data[i]]=value
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print(sum(resb.values()))

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amanmad1 profile image
Aman Madan •

I really liked the part2 implementation, a subtle dp.

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qviper profile image
Viper •

Yeah it is so dope.

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thetalent profile image
thetalent • • Edited

Part One

data = sorted([int(i) for i in puzzle.input_data.split(lf)]) r={1:0,2:0,3:0} s=0 for i in data: r[i-s]+=1 s=i print(r[1]*(r[3]+1))
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qviper profile image
Viper •

This is very short. And absolutely tricky. Thanks for sharing.

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thetalent profile image
thetalent •

Did u have a difference of 2 in part1 ?

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laojala profile image
Laura Ojala •

Thank you for sharing this! I got "strong inspiration" (read copied 😅) your part 2 solution. My solution in Github.