1046. Last Stone Weight

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

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#https://leetcode.com/problems/last-stone-weight/description/

'''
1. 아이디어 :
    % Prioirty queue를 구현하는게 heap이다.
    1) priority queue(heapq) 를 이용하여, 하나씩 뺴면서(heappop) x!=y일때 다시 삽입(heappush)을 한다,
2. 시간복잡도 :
    1) O(logn)
3. 자료구조 :
    1) Prioirty Queue
'''

import heapq
class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        print(stones)
        heapq.heapify(stones)
        hstones = []
        for stone in stones:
            heapq.heappush(hstones,(-stone, stone))

        while len(hstones)>=2:
            y = heapq.heappop(hstones)[1]
            x = heapq.heappop(hstones)[1]
            if y!=x:
                heapq.heappush(hstones, (-(y-x),(y-x)))
        return heapq.heappop(hstones)[1] if len(hstones)!=0 else 0