You are given an array of integers stones
where stones[i]
is the weight of the ith
stone.
We are playing a game with the stones. On each turn, we choose the heaviest two stones and smash them together. Suppose the heaviest two stones have weights x
and y
with x <= y
. The result of this smash is:
- If
x == y
, both stones are destroyed, and - If
x != y
, the stone of weightx
is destroyed, and the stone of weighty
has new weighty - x
.
At the end of the game, there is at most one stone left.
Return the weight of the last remaining stone. If there are no stones left, return 0
.
Example 1:
Input: stones = [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 the last stone.
Example 2:
Input: stones = [1]
Output: 1
Constraints:
-
1 <= stones.length <= 30
-
1 <= stones[i] <= 1000
SOLUTION:
import heapq
class Solution:
def lastStoneWeight(self, stones: List[int]) -> int:
heap = []
for s in stones:
heapq.heappush(heap, -s)
while len(heap) > 1:
y = -heapq.heappop(heap)
x = -heapq.heappop(heap)
if x != y:
heapq.heappush(heap, x - y)
if len(heap) == 1:
return -heap[0]
return 0
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