Last stone weight

Time: O(NLogN); Space: O(N); easy

We have a collection of stones, each stone has a positive integer weight.

Each turn, we choose the two heaviest stones and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is: * If x == y, both stones are totally destroyed; * If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.

At the end, there is at most 1 stone left. Return the weight of this stone (or 0 if there are no stones left.)

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 last stone.

Notes:

  • 1 <= len(stones) <= 30

  • 1 <= stones[i] <= 1000

Hints:

  1. Simulate the process. We can do it with a heap, or by sorting some list of stones every time we take a turn.

[2]:

import heapq

class Solution1(object):
    """
    Time: O(NLogN)
    Space: O(N)
    """
    def lastStoneWeight(self, stones):
        """
        :type stones: List[int]
        :rtype: int
        """
        max_heap = [-x for x in stones]
        heapq.heapify(max_heap)

        for i in range(len(stones)-1):
            x, y = -heapq.heappop(max_heap), -heapq.heappop(max_heap)
            heapq.heappush(max_heap, -abs(x-y))

        return -max_heap[0]

[3]:

s = Solution1()
stones = [2,7,4,1,8,1]
assert s.lastStoneWeight(stones) == 1