23 lines
1001 B
Python
23 lines
1001 B
Python
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import numpy as np
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# Load input
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with open('input10', 'r') as f:
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data = [0] + sorted([int(l) for l in f.readlines()]) # NB: We add the 0J source ourselves
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n_data = np.array(data)
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# Part 1: Connect all of them together
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diff = n_data[1:] - n_data[:-1]
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# Confirm this by multiplying the 1J differences with the 3J differences.
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# NB: we treat the internal adapter 3J difference as an off-by-one for simplicity.
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print('Part 1: ', (diff==1).sum() * (1 + (diff==3).sum()))
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# Part 2: Find the number of possible chains
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n_possibilities = {data[-1]: 1} # The highest one can only connect to the internal adapter.
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for i in reversed(data[:-1]): # We've already done the highest one as a special case
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delta = n_data - i
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p = 0
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for n in n_data[np.logical_and(delta<=3, delta>0)]: # You could probably plug two of the same together, but our input doesn't require us to handle that, and this simplifies things ;)
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p += n_possibilities[n]
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n_possibilities[i] = p
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print('Part 2: ', n_possibilities[0])
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