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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 (8)

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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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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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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 •

Did u have a difference of 2 in part1 ?

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philippurwoko profile image
Philip Purwoko •

I don't

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laojala profile image
Laura Ojala (she/they) •

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

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