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Possibly my favourite ever runtime and now favourite number I've generated in Alteryx.
Felt good to solve that one, I knew my logic was accurate yet inefficient so just hoped for the best and let it do its thing.
Keen to look at all these magical <1hr solutions!
Six trillion two hundred thirty-seven billion seventy-five million forty-one thousand four hundred eighty-nine
6,237,075,041,489: what a number!
Part 2 the shift position of 2 caught me out a bit, took a while to handle the updated space.
Solved! I really enjoyed this one. My Part 2 initially ran in 7:20 and then I spent an hour messing around with it and got it to run in 7:10! 😂
Part 1
Spoiler
I split everything into its own cell, threw a Record ID on, and then Joined by Position.
What initially happened was that I wasn't factoring in when a Record ID was already ahead of a space - I threw on a Count of the non-nulls and then only looked at blanks that came before that number (since if it was after, I didn't care about them)
What initially happened was that I wasn't factoring in when a Record ID was already ahead of a space - I threw on a Count of the non-nulls and then only looked at blanks that came before that number (since if it was after, I didn't care about them)
Part 2 v1
Spoiler
For Part 2, I went back to the data before everything was split. I threw it into an Iterative Macro where I checked if I could move it forward. One of the things that I did was if there were multiple blank spots with 3 spaces, I took the one with the lowest ID, since that's where the disk would move. I think that helped with my run time, compared to what other people's runs took. After each 1 was checked and either moved or left alone, I set my field Checked? to 1.
I repeated for 10k iterations without any intelligent stop condition - just find how many disks I have, and keep repeating, starting from the very last disk and working to the very first disk.
Total run time = 7:20
I repeated for 10k iterations without any intelligent stop condition - just find how many disks I have, and keep repeating, starting from the very last disk and working to the very first disk.
Total run time = 7:20
Part 2 v2
Spoiler
As Part 2 was running, I got to thinking about how I knew my disk #2 wasn't going to move because there was no space. I didn't need to wait until check 9,999 to determine that.
I added a Standard macro inside the Iterative macro. After the Iterative macro is mostly done, I take the lowest Disk ("RecordID") that hadn't been checked yet and saw if there was enough room for it to move up. If the answer was no, I would change the "Checked?" field to 1, which would mean that I don't need to run it through iteration 9,999 anymore. Then, the next iteration would check the next lowest Disk.
If the answer was yes (there is space), I left it alone and rechecked after the next iteration.
I wasn't brave enough to check multiple Disks at the same time. That probably would've helped materially reduce my run time.
I also modified the Iterative macro to have an intelligent stop condition which was based on when all the Disks have Checked? = 1
Run time = 7:10 and Iterations = 5,075... 10 whole seconds faster!
I added a Standard macro inside the Iterative macro. After the Iterative macro is mostly done, I take the lowest Disk ("RecordID") that hadn't been checked yet and saw if there was enough room for it to move up. If the answer was no, I would change the "Checked?" field to 1, which would mean that I don't need to run it through iteration 9,999 anymore. Then, the next iteration would check the next lowest Disk.
If the answer was yes (there is space), I left it alone and rechecked after the next iteration.
I wasn't brave enough to check multiple Disks at the same time. That probably would've helped materially reduce my run time.
I also modified the Iterative macro to have an intelligent stop condition which was based on when all the Disks have Checked? = 1
Run time = 7:10 and Iterations = 5,075... 10 whole seconds faster!
