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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1601_1700.s1690_stone_game_vii;

// #Medium #Array #Dynamic_Programming #Math #Game_Theory
// #2022_04_15_Time_18_ms_(96.71%)_Space_42.1_MB_(96.71%)

/**
 * 1690 - Stone Game VII\.
 *
 * Medium
 *
 * Alice and Bob take turns playing a game, with **Alice starting first**.
 *
 * There are `n` stones arranged in a row. On each player's turn, they can **remove** either the leftmost stone or the rightmost stone from the row and receive points equal to the **sum** of the remaining stones' values in the row. The winner is the one with the higher score when there are no stones left to remove.
 *
 * Bob found that he will always lose this game (poor Bob, he always loses), so he decided to **minimize the score's difference**. Alice's goal is to **maximize the difference** in the score.
 *
 * Given an array of integers `stones` where `stones[i]` represents the value of the i^th stone **from the left** , return _the **difference** in Alice and Bob's score if they both play **optimally**._
 *
 * **Example 1:**
 *
 * **Input:** stones = [5,3,1,4,2]
 *
 * **Output:** 6
 *
 * **Explanation:**
 *
 * - Alice removes 2 and gets 5 + 3 + 1 + 4 = 13 points. Alice = 13, Bob = 0, stones = [5,3,1,4].
 *
 * - Bob removes 5 and gets 3 + 1 + 4 = 8 points. Alice = 13, Bob = 8, stones = [3,1,4].
 *
 * - Alice removes 3 and gets 1 + 4 = 5 points. Alice = 18, Bob = 8, stones = [1,4].
 *
 * - Bob removes 1 and gets 4 points. Alice = 18, Bob = 12, stones = [4].
 *
 * - Alice removes 4 and gets 0 points. Alice = 18, Bob = 12, stones = [].
 *
 * The score difference is 18 - 12 = 6.
 *
 * **Example 2:**
 *
 * **Input:** stones = [7,90,5,1,100,10,10,2]
 *
 * **Output:** 122
 *
 * **Constraints:**
 *
 * *   `n == stones.length`
 * *   `2 <= n <= 1000`
 * *   `1 <= stones[i] <= 1000`
**/
public class Solution {
    public int stoneGameVII(int[] stones) {
        int n = stones.length;
        int[] dp = new int[n];
        for (int i = n - 1; i >= 0; i--) {
            int j = i + 1;
            int sum = stones[i];
            while (j < n) {
                sum += stones[j];
                dp[j] = Math.max(sum - stones[i] - dp[j], sum - stones[j] - dp[j - 1]);
                j++;
            }
        }
        return dp[n - 1];
    }
}