# https://leetcode.com/problems/last-stone-weight/
# 1046. Last Stone Weight

# We have a collection of rocks, each rock has a positive integer weight.
#
# Each turn, we choose the two heaviest rocks 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


class Solution(object):
    def lastStoneWeight(self, stones):
        """
        :type stones: List[int]
        :rtype: int
        """
        stones = [-s for s in stones]
        heapq.heapify(stones)

        while stones:
            if len(stones) == 1:
                return -stones[0]

            largest, second_largest = -heapq.heappop(stones), -heapq.heappop(
                stones)

            if largest >= second_largest:
                new_stone = largest - second_largest
                heapq.heappush(stones, -new_stone)

        return 0
