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1046 Last Stone Weight

Solved at: 2022-08-30

Question​

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 weight x is destroyed, and the stone of weight y has new weight y - 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.

Solution​

class Solution:
def lastStoneWeight(self, stones: List[int]) -> int:
while(len(stones) > 1):
greatest = max(stones)
stones.remove(greatest)
second = max(stones)
stones.remove(second)
if greatest - second > 0:
stones.append(greatest - second)
if stones:
return stones[0]
return 0
  • Time: O(n2)O(n^2)
  • Space: O(1)O(1)

Improved​

class Solution:
def lastStoneWeight(self, stones: List[int]) -> int:
for idx, val in enumerate(stones):
stones[idx] = -1 * val
heapq.heapify(stones)
while(len(stones) > 1):
first = heapq.heappop(stones)
second = heapq.heappop(stones)
if first - second < 0:
heapq.heappush(stones, first - second)
return -1 * heapq.heappop(stones) if len(stones) else 0
  • heapq.heapify() is O(n)O(n).
  • Time: O(nlog⁡n)O(n \log n).
  • Space: O(1)O(1) in Python

Results​

Runtime​

75 ms, faster than 5.53% of Python3 online submissions for Last Stone Weight.

Memory Usage​

14 MB, less than 14.37% of Python3 online submissions for Last Stone Weight.