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

import heapq
from typing import List


class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:

        max_heap = []
        for stone in stones:
            heapq.heappush(max_heap, stone*(-1))

        while len(max_heap) > 1:
            stone1 = heapq.heappop(max_heap)*(-1)
            stone2 = heapq.heappop(max_heap)*(-1)

            diff = abs(stone1 - stone2)
            if diff > 0:
                heapq.heappush(max_heap, diff*(-1))

        return (-1)*max_heap[-1] if max_heap else 0
