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Last Stone Weight

Problem Statement​

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.

Leetcode Link

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

Code​

Python Code

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        stones.sort()
        for _ in range(len(stones) - 1):
            bisect.insort(stones, stones.pop() - stones.pop())
        return stones[0]