LeetCode 1046. Last Stone Weight



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: [2,7,4,1,8,1]
Output: 1
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.


  1. 1 <= stones.length <= 30
  2. 1 <= stones[i] <= 1000


just simulate the process to find the last stone weight

Python Solution

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        while (len(stones) > 1):
            max_stone = max(stones)
            second_max_stone = max(stones)
            new_stone = max_stone - second_max_stone
            if new_stone != 0:
        if (len(stones) == 1):            
            return stones[0]
        return 0
  • Time complexity: O(N).
  • Space complexity: O(1).

One Thought to “LeetCode 1046. Last Stone Weight”

Leave a Reply

Your email address will not be published.