g1001_1100.s1049_last_stone_weight_ii.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1001_1100.s1049_last_stone_weight_ii;

// #Medium #Array #Dynamic_Programming #2022_02_28_Time_2_ms_(95.98%)_Space_41.8_MB_(31.51%)

/**
 * 1049 - Last Stone Weight II\.
 *
 * Medium
 *
 * You are given an array of integers `stones` where `stones[i]` is the weight of the i^th 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`
**/
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;
    }
}