This was straightforward in that all you need to do is implement the algorithm as given. I did optimise a little using the arithmetic progression sum, but this is a speed-up of, at most like, a factor of 9.

I am pretty sure I did an OK job at avoiding edge cases, though I suspect this problem has many of them.

Again, the algorithm is more or less given: Start from the back, look for a hole that’ll work, and put it in if you can. Otherwise, don’t. Then, calculate the contribution to the checksum.

The complex part was the “look for a hole” part. My input size was 19999, meaning an O(n²) solution was probably fast enough, but I decided to optimise and use a min heap for each space size prematurely. I foresaw that you need to split up a space if it is oversized for a particular piece of data, i.e. pop the slot from the heap, reduce the size of the slot, and put it in the heap corresponding to the new size.