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Fairly straight forward logic with this one, just a lot of tools to do it.
I found the work I did on the blog about trying to find the earliest you could complete a pack of Disney collectable cards I wrote influenced my approach with this one, as I straight away knew I could do it without an iterative macro.
Also nice that the solution to part two only needed a sample and sort added to solve the second part without needing to re-work the logic.
Chris Check out my collaboration with fellow ACE Joshua Burkhow at AlterTricks.com
I ended up with a brute force (rather than efficient) iterative macro. It found the solution, but it's not pretty. I just changed the filter logic to get the first bingo vs all bingos. Interface tools added later.
Tool golf, this was certainly not. But I am super happy with how I ended up where I did - I had a vision at the start while reading through the problem about how I would need to build it, and more or less was able to build it out as I went along without too much going backwards to redo logic. ALSO I got to use my favorite dark horse tool - MAKE COLUMNS!! What a glorious end to the week.
After I made it without macros, I realized that it was very simple. We can calculate when the card win very easily. Because it is that we join the order to each number in the bingo and then take max order at each row and column. After specifying the win card and the order, we can calculate the required result easily. On the star2, we know the result easily by changing the sort way.
I should know the way before going to macro route.
1. I parsed out all the possible winning sequences (columns and rows) 2. Joined the values from the picked number string with their spot in the selection 3. For challenge 1 grouped by table and sequence to find the min step id per for a sequence 4. For challenge 2 used the same group by as in (3), and added another group by to get the table with the largest step id for it's first sequence.