LC 1046 - Last Stone Weight

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.

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

Links

Question here and solution here

Solution

concept

Use a max heap to solve this question

code

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        max_heap = [-1*stone for stone in stones]
        heapq.heapify(max_heap)

        while len(max_heap) > 1:
            stone_1 = heapq.heappop(max_heap)
            stone_2 = heapq.heappop(max_heap)

            if stone_1 != stone_2:
                diff = stone_1 - stone_2
                heapq.heappush(max_heap, diff)

        return -1*max_heap[0] if max_heap else 0

class NeetSolution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        stones = [-s for s in stones]
        heapq.heapify(stones)

        while len(stones) > 1:
            first = heapq.heappop(stones)
            second = heapq.heappop(stones)
            if second > first:
                heapq.heappush(stones, first - second)
        # for edge case where no stone is left after the loop
        stones.append(0)
        return abs(stones[0])

Complexity

time: $O(n \log n)$
space: $O(n)$