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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1101_1200.s1140_stone_game_ii;

// #Medium #Array #Dynamic_Programming #Math #Game_Theory
// #2023_06_01_Time_9_ms_(43.08%)_Space_43_MB_(39.09%)

import java.util.Arrays;

/**
 * 1140 - Stone Game II\.
 *
 * Medium
 *
 * Alice and Bob continue their games with piles of stones. There are a number of piles **arranged in a row** , and each pile has a positive integer number of stones `piles[i]`. The objective of the game is to end with the most stones.
 *
 * Alice and Bob take turns, with Alice starting first. Initially, `M = 1`.
 *
 * On each player's turn, that player can take **all the stones** in the **first** `X` remaining piles, where `1 <= X <= 2M`. Then, we set `M = max(M, X)`.
 *
 * The game continues until all the stones have been taken.
 *
 * Assuming Alice and Bob play optimally, return the maximum number of stones Alice can get.
 *
 * **Example 1:**
 *
 * **Input:** piles = [2,7,9,4,4]
 *
 * **Output:** 10
 *
 * **Explanation:** If Alice takes one pile at the beginning, Bob takes two piles, then Alice takes 2 piles again. Alice can get 2 + 4 + 4 = 10 piles in total. If Alice takes two piles at the beginning, then Bob can take all three piles left. In this case, Alice get 2 + 7 = 9 piles in total. So we return 10 since it's larger.
 *
 * **Example 2:**
 *
 * **Input:** piles = [1,2,3,4,5,100]
 *
 * **Output:** 104
 *
 * **Constraints:**
 *
 * *   `1 <= piles.length <= 100`
 * *   1 <= piles[i] <= 10⁴
**/
public class Solution {
    private int[][] dp = new int[105][105];

    private int help(int i, int m, int[] p) {
        if (i >= p.length) {
            dp[i][m] = 0;
            return 0;
        }
        if (dp[i][m] != -1) {
            return dp[i][m];
        }
        int ans = Integer.MIN_VALUE;
        int total = 0;
        for (int j = 0; j < 2 * m; j++) {
            if (i + j < p.length) {
                total += p[i + j];
                ans = Math.max(ans, total - help(i + j + 1, Math.max(m, j + 1), p));
            }
        }
        dp[i][m] = ans;
        return ans;
    }

    public int stoneGameII(int[] piles) {
        int sum = 0;
        for (int[] arr1 : dp) {
            Arrays.fill(arr1, -1);
        }
        for (int z : piles) {
            sum += z;
        }
        return (sum + help(0, 1, piles)) / 2;
    }
}