1046-last-stone-weight¶

Description¶

You are given an array of integers stones where stones[i] is the weight of the i^th 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 smallest possible weight of the left 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(Python)¶

class Solution:
    # Time Complexity: O(nlogn)
    # Space Complexity: O(n)
    def lastStoneWeight(self, stones: List[int]) -> int:
        heap = [-stone for stone in stones]
        heapq.heapify(heap)

        while len(heap) > 1:
            diff = heapq.heappop(heap) - heapq.heappop(heap)

            if diff != 0:
                heapq.heappush(heap, diff)

        return -heap[0] if len(heap) == 1 else 0