# Last Stone Weight

## Problem statement:

We have a collection of stones, each stone has a positive integer weight.

Each turn, we choose the two heaviest stones and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is:

If x == y, both stones are totally destroyed;
If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.
At the end, there is at most 1 stone left. Return the weight of this stone (or 0 if there are no stones left.)

Example 1:

`Input: [2,7,4,1,8,1]`
`Output: 1`
`Explanation:`
`We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,`
`we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,`
`we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,`
`we combine 1 and 1 to get 0 so the array converts to  then that's the value of last stone.`

Note:

`1 <= stones.length <= 30`
`1 <= stones[i] <= 1000`

## Solution

Happy rainy Omaha Sunday! Today’s challenge wasn’t too bad. I read through the problem a few times and wrote some code and happen to solve it within a few runs.

The time complexity on this is dependent upon the call to `sort.Ints([]int)`. Looking at the source code, we can see that Go implements quicksort for this. We can reference Big-O Cheat Sheet (or google really) to see that we’re on average getting `Θ(n log(n))` for this. I don’t claim to be a master of asymptotic analysis, but I believe having this quicksort inside my while loop promotes this solution to `O(n²)` (please correct me if I’m mistaken).

That being said, our solution’s runtime ended up being 0 milliseconds. Space complexity would be `O(n)`

### Solution Details

Runtime: 0 ms Memory Usage: 2 MB