1.4 KiB
10
First
The first one is really easy. Just sort, make the diff and count. First I take the list and sort it:
[16, 10, 15, 5, 1, 11, 7, 19, 6, 12, 4]
[0, 1, 4, 5, 6, 7, 10, 11, 12, 15, 16, 19, 22] # Sorted and added the wall plug (0) and the phone (biggest + 3)
[1, 3, 1, 1, 1, 3, 1, 1, 3, 1, 3, 3] # The size of each step
Now we can simply count how many 1 and 3 there are with l.count(1).
Second
This is where it gets tricky.
First lets find all the consecutive 1s ad only they can be removed. If we have more than 1 consecutive 1 we can remove one of it. However we need to be careful not to remove to many or the step will be higher than 3 and the chain breaks.
[1, 1, 1, 1] # We can transform this example by adding 2 numbers together and "joining" them.
[1, 2, 1] # Valid
[1, 1, 2] # Valid
[1, 3] # Valid
[4] # Invalid because we can jump a max of 3 steps at a time.
Now we could iterate but I wanted to find a formula. Not sure this is correct but here we go.
Basically we take the length of the consecutive 1 and compute 2**(l-1) to get all possible combinations.
Now we need to subtract the possible 4 which can only be achieved if we have at least 4 numbers -> floor(l/4)
For a grand total of 2**(l-1) - floor(l/4)
Solutions
- 2475
- 442136281481216