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 [1] then that's the value of last stone.Note:
1 <= stones.length <= 30
1 <= stones[i] <= 1000
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)
func lastStoneWeight(stones []int) int {
x, y, err := getHeaviestStones(stones)
for err == nil {
stones = smashStones(x, y, stones)
sort.Ints(stones)
fmt.Println(stones)
x, y, err = getHeaviestStones(stones)
}
if len(stones) > 0 {
return stones[0]
}
return 0
}
func smashStones(x, y int, stones []int) []int {
if x == y {
stones = stones[:len(stones) - 2]
} else {
stoneY := y - x
stones = stones[:len(stones) - 1]
stones[len(stones) - 1] = stoneY
}
return stones
}
func getHeaviestStones(stones []int) (int, int, error) {
if len(stones) < 2 {
return -1, -1, errors.New("Array size must be at least two!")
}
sort.Ints(stones)
return stones[len(stones) - 2], stones[len(stones) - 1], nil
}Runtime: 0 ms Memory Usage: 2 MB