# 1046. 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.)
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  then that's the value of last stone.
1 <= stones.length <= 30 1 <= stones[i] <= 1000
Approach 1: Priority queue.
Approach 2: Bucketing (as in bucket sort).
# Code (Python)
import heapq class Solution: def lastStoneWeight(self, stones: List[int]) -> int: # Priority queue. Could also use bucketing (as in bucket sort) since the range of weights are bounded. heap = [-item for item in stones] heapq.heapify(heap) while len(heap) > 1: val1, val2 = -heapq.heappop(heap), -heapq.heappop(heap) if val1 > val2: heapq.heappush(heap, -(val1 - val2)) return 0 if not heap else -heap
# Code (C++)