1046-last-stone-weight.py

"""
Problem Link: 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 <= stones.length <= 30
1 <= stones[i] <= 1000
"""
import heapq
class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        q = [-stone for stone in stones]
        heapq.heapify(q)
        while len(q) > 1:
            heapq.heappush(q, heapq.heappop(q) - heapq.heappop(q))
        return -q[0]