text: """
  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.
  
  """
  import heapq
  
  class Solution:
      def lastStoneWeight(self, stones: List[int]) -> int:
          stones = [-x for x in stones]
          heapq.heapify(stones)
          while len(stones) > 1:
              a = heapq.heappop(stones)
              b = heapq.heappop(stones)
              if a != b:
                  heapq.heappush(stones,a-b)
          if len(stones) == 0:
              return 0
          return -stones[0]