Anders Englöf Ytterström
4a070e827b
For the first part, I used itertools.combinations to find the highest pairs of batteries. And as expected, that solution did not scale well for pt 2. I figured out that reducing batteries until the top most 12 (and 2) remained was the correct way to go. the _maxj(line, C) function is the hive conclusion from the solution mega thread. I really liked this brilliant use of a while loop to exlude batteries. - The first char just skip the while loop. A char emptying the battery list also does this.
31 lines
640 B
Python
31 lines
640 B
Python
def solve(data):
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p1 = 0
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p2 = 0
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for line in data.splitlines():
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p1 += _maxj(line, 2)
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p2 += _maxj(line, 12)
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return p1, p2
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def _maxj(line, C):
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toexcl = len(line) - C
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batt = []
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for c in line:
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while toexcl and batt and batt[-1] < c:
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toexcl -= 1
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batt.pop()
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batt.append(c)
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return sum(10**x * int(y) for x, y in zip(range(C - 1, -1, -1), batt))
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if __name__ == "__main__":
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with open("./input/03.txt", "r") as f:
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inp = f.read().strip()
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p1, p2 = solve(inp)
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print(p1)
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print(p2)
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assert p1 == 17430
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assert p2 == 171975854269367
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