Class Solution


  • public class Solution
    extends Object
    1046 - Last Stone Weight.

    Easy

    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 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
    • Constructor Detail

      • Solution

        public Solution()
    • Method Detail

      • lastStoneWeight

        public int lastStoneWeight​(int[] stones)