Problem
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.
Solution
Priority queue, max heap can be used here.
class Solution:
def lastStoneWeight(self, stones):
res = [ -s for s in stones ]
heapq.heapify(res)
while len(res) > 1 :
y = heapq.heappop(res)
x = heapq.heappop(res)
if y != x:
heapq.heappush(res, y-x)
return res[0]*-1 if res else 0