2020-12-09 11:03:34 +00:00
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Day 9 Notes
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+--------+
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| Part 1 |
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+--------+
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$ elixir day9part1.exs
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18272118
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Thoughts:
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Searching inside a sliding window.
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Tricky to write efficiently in an immutable language.
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I knew what I wanted to do, but working out how to write that took a lot of thought.
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The solution completes in just over 1ms, so I'm pretty happy with it.
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Every time the window slides, I decided to sort it, which means I could take the following shortcuts:
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* When selecting the first number in the pair, stop early if added to the first number in the preamble it's
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greater than the target, because we know the rest of the numbers are also too large.
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* Similarly can do the same when checking for the second number in the pair.
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2020-12-09 15:36:55 +00:00
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* Avoids comparing pairs we already compared, by only searching from the location of the first value onwards.
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2020-12-09 11:03:34 +00:00
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+--------+
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| Part 2 |
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+--------+
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$ elixir day9part2.exs
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2186361
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Thoughts:
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Initially hard-coded the answer from part 1, but then decided to re-use part one and pass it as an
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input to keep the solution fully dynamic.
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Similar to part 1, but now we have to check an unknown number of adjacent items. Again stop checking each
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group as soon as the sum is too big, and move to the next group.
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This answer also completes in about a millisecond (not including part 1).
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P.S. I accidentally copied part 1 over part 2 after solving this, and had to write the solution again 🤦🏼♂️
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+------------------+
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| Overall Thoughts |
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+------------------+
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The most challenging thing here was working out how to write it in an immutable, recursive way cleanly.
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I was considering using the :array module, but I think I managed to get everything working nicely with
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lists without the need to do lots of expensive list-indexing. Pretty sure both answers are O(nlogn).
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Probably will tidy up this more later, especailly Part 1, but done for now.
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