"""
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

"""

"""
Heap
Time: O(Nlog(N))
Space: O(N)
"""
class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        pq = []
        
        for stone in stones:
            heapq.heappush(pq, -stone)

        while len(pq) > 1:
            a, b = -heapq.heappop(pq), -heapq.heappop(pq)
            heapq.heappush(pq, -abs(a-b))
        
        return -pq[0]

