1046. Last Stone Weight

https://leetcode.com/problems/last-stone-weight/

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. 1 <= stones.length <= 30

2. 1 <= stones[i] <= 1000

由正整数构成的数组，每次选值最大的两个元素让它们减去它们中的最小值后把大于0的插回数组，问最后如果剩下一个元素它的值会是多少。动态找最小元素，并能动态更新 => heap。

from heapq import heappop, heappush, heapify 
class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        # Creating empty heap 
        heap = [] 
        heapify(heap) 
        # Adding items to the heap using heappush 
        # function by multiplying them with -1 
        for s in stones:
            heappush(heap, -1 * s)
        while len(heap) > 1:
            x, y = -1 * heappop(heap), -1 * heappop(heap)
            if x != y:
                heappush(heap, -1 * abs(y - x))
        return 0 if len(heap) == 0 else -1 * heap[0]