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2cc595523d
| Author | SHA1 | Date | |
|---|---|---|---|
| 2cc595523d | |||
| ecbbd39af1 |
54
day03/sol.py
54
day03/sol.py
@ -1,34 +1,26 @@
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input = open("input")
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t = 0
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part1 = 0
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part2 = 0
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shift = [0] + [10**i for i in range(12)]
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for line in input:
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cells = list(line.strip())
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cells.reverse()
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b = cells[0]
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max = 0
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for c in cells[1:]:
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j = int(c+b)
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if j > max:
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max = j
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if c > b:
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b = c
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#print(max)
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t += max
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print(t)
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# Part 2
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input = open("input")
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t = 0
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for line in input:
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cells = list(line.strip())
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cells.reverse()
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max = ["0"*i for i in range(12+1)]
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for (k,c) in enumerate(cells):
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digits = list(line.strip())
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# max[n] is the value of the maximal n digit subsequence
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max = [0] * 13
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# iterate over the digits in least-significant to most-significant order.
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# k is the number of digits already processed.
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for (k,c) in enumerate(reversed(digits)):
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c = int(c)
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# the maximal n-digit subsequence we can make that starts with the
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# current digit (c) is equal to the concatenation of c and the
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# maximal (n-1)-digit subsequence.
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# if that's more than the best value we've seen so far, update max[n].
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# we have to update max in reverse order in order to avoid reusing digits.
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for i in reversed(range(1,min(k+2,len(max)))):
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j = c+max[i-1]
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if j > max[i]:
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max[i] = j
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print(max[12])
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t += int(max[12])
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print(t)
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joltage = c*shift[i] + max[i-1]
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if joltage > max[i]:
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max[i] = joltage
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#print(max[12])
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part1 += int(max[2])
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part2 += int(max[12])
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print(part1)
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print(part2)
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