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: $O(n^2)$
• Space: $O(1)$

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 $O(n)$.
• Time: $O(n \log n)$.
• Space: $O(1)$ 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.