The Final Stone
You have a list of stones with different weights. The result of hitting two stones will be a new stone with a weight difference between the two stones.
Your goal is to find the weight of the final stone. If no stones left, the result is 0.
The algorith is very simple:
- get the two biggest stones in the batch
- hit and get the resulting stone
- put the resulting stone back in the batch.
For the Speedy category, you can think about your solution for a million stones
Input: List of stones as list of ints
Output: Int.
Example:
assert final_stone([3, 5, 1, 1, 9]) == 1 assert final_stone([1, 2, 3]) == 0 assert final_stone([1, 2, 3, 4]) == 0 assert final_stone([1, 2, 3, 4, 5]) == 1
Precondition: only positive, non-zero values. The list of stones can be empty
