LeetCode-in-Java

1049. Last Stone Weight II

Medium

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 any two 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 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 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.

Example 2:

Input: stones = [31,26,33,21,40]

Output: 5

Constraints:

• 1 <= stones.length <= 30
• 1 <= stones[i] <= 100

Solution

public class Solution {
    public int lastStoneWeightII(int[] stones) {
        // dp[i][j] i is the index of stones, j is the current weight
        // goal is to find max closest to half and use it to get the diff
        // 0-1 knapsack problem
        int sum = 0;
        for (int stone : stones) {
            sum += stone;
        }
        int half = sum / 2;
        int[] dp = new int[half + 1];
        for (int cur : stones) {
            for (int j = half; j >= cur; j--) {
                dp[j] = Math.max(dp[j], dp[j - cur] + cur);
            }
        }
        return sum - dp[half] * 2;
    }
}

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