# 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.)
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
# Solution
Approach 1: Priority queue.
Approach 2: Bucketing (as in bucket sort).
# Code (Python)
Approach 1:
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[0]
Approach 2:
# Code (C++)
Approach 1:
Approach 2: