🔒 Need Help Finding A Good Approach

[hollow hornet]

Given an interger N, delete 1 or continuous digits of N to make a new int such that it's divisable by 3 (%3==0)
Find All possible ways to do that
Note: The N number itself if %3==0 also consider 1 solution

[lunar cosmos]

@hollow hornet

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[hollow hornet]

sample input/output

i just need help with the approach to solve this

if anyone respond pls ping me

[frozen spoke]

one way to check divisibility by three is to see if the sum of all digits is divisible by three

so if the sum of digits isn't divisible by 3, you just have to find what digits can be deleted to make it so

[sage hinge]

wow never knew this trick

interesting

[frozen spoke]

for example in 2009, you can delete consecutive digits (2), (2, 0), and (2, 0, 0) => three ways

[sage hinge]

why is the first one output 4

[frozen spoke]

in 1005, it's already divisible by three, so that counts as one way
then you can delete the first zero, the second zero, and both zeros together (since they're consecutive), so 4 ways in total

[sage hinge]

I see

[hollow hornet]

but how do i do that tho? i can only delete 1 digits at a time with that thinking

the problems says we can delete the digits but has to be consecutive

[sage hinge]

I imagine the simplest way is to try every combination possible and check the outcome of all of them. The fun would be how to optimize that

[hollow hornet]

i can also think of brute force approach like you but i do believe there would be a better apporach

[frozen spoke]

(I haven't tested this, just some ideas)
first find what the remainder is
then iterate through the digits:
if you can find a digit that is that remainder OR gives that remainder when divided by 3, that's one way
if that digit is followed by a number of zeros, each of those counts as a way
else consider the next digit plus the previous one, is that equal to or gives the remainder, and so on

[hollow hornet]

ok i can imagine it now

!close

[lunar cosmos]

Python help channel closed

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