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/**
* You are given an array of integers stones where stones[i] is the weight of the
* iᵗʰ stone.
* <p>
* We are playing a game with the stones. On each turn, we choose any two stones
* and smash them together. Suppose the stones have weights x and y with x <= y.
* The result of this smash is:
* <p>
* <p>
* 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.
* <p>
* <p>
* At the end of the game, there is at most one stone left.
* <p>
* Return the smallest possible weight of the left stone. If there are no stones
* left, return 0.
* <p>
* <p>
* Example 1:
* <p>
* <p>
* Input: stones = [2,7,4,1,8,1]
* Output: 1
* Explanation:
* We can combine 2 and 4 to get 2, so the array converts to [2,7,1,8,1] then,
* we can combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then,
* we can combine 2 and 1 to get 1, so the array converts to [1,1,1] then,
* we can combine 1 and 1 to get 0, so the array converts to [1], then that's the
* optimal value.
* <p>
* <p>
* Example 2:
* <p>
* <p>
* Input: stones = [31,26,33,21,40]
* Output: 5
* <p>
* <p>
* <p>
* Constraints:
* <p>
* <p>
* 1 <= stones.length <= 30
* 1 <= stones[i] <= 100
* <p>
* <p>
* Related Topics Array Dynamic Programming 👍 3100 👎 118
  */

/*
2024-07-09 20:34:47

Last Stone Weight II
Category	Difficulty	Likes	Dislikes
algorithms	Medium (55.26%)	3100	118
Tags
array | greedy

Companies
Unknown
*/

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    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        //dp: 与416.分割等和子集同类型题目，同属于01背包
        //思路：尽可能分成2组等重量石头，相同石头重量会撞碎，剩下部分即为最后的重量
        public int lastStoneWeightII(int[] stones) {
            int sum = 0;
            for (int i = 0; i < stones.length; i++) {
                sum += stones[i];
            }
            int target = sum / 2;
            //dp[i]容量为i的背包能装下最大价值的物品（石头），物品重量为stone[i],物品价值value[i]也为stone[i]
            int[] dp = new int[target + 1];
            //遍历物品
            for (int i = 0; i < stones.length; i++) {
                //遍历背包
                for (int j = target; j >= stones[i]; j--) {
                    dp[j] = Math.max(dp[j], dp[j - stones[i]] + stones[i]);
                }
            }
            //dp[target]即为容量为target的背包能装下的最大价值的石头
            //将石头分成2堆，一堆石头的总重量是dp[target]，另一堆就是sum - dp[target]
            //因为在计算target时，target = sum / 2向下取整，所以sum - dp[target] 一定大于等于dp[target]
            //那么相撞后，剩下最小石头的重量就是(sum -dp[target]) - dp[target]
            return (sum - dp[target]) - dp[target];
        }
    }
//leetcode submit region end(Prohibit modification and deletion)