Stone Game II[Array,Math,Dynamic Programming,Prefix Sum,Game Theory]

https://leetcode.com/problems/stone-game-ii/description/?envType=daily-question&envId=2024-08-20

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

게임 설명을 봐도 당췌 뭔 소린지 모르겠다.

Dynamic Programming을 이용하라는데, 막막하다.

빠른 GG

 

1. C++

#C++ using DP
class Solution {
public:
    int stoneGameII(vector<int>& piles) {
        int n = piles.size();
        
        vector<vector<int>> dp(n, vector<int>(n + 1, 0));
        vector<int> suffixSum(n, 0);
        suffixSum[n - 1] = piles[n - 1];
        
        for (int i = n - 2; i >= 0; i--) {
            suffixSum[i] = suffixSum[i + 1] + piles[i];
        }
        
        for (int i = n - 1; i >= 0; i--) {
            for (int m = 1; m <= n; m++) {
                if (i + 2 * m >= n) {
                    dp[i][m] = suffixSum[i];
                } else {
                    for (int x = 1; x <= 2 * m; x++) {
                        dp[i][m] = max(dp[i][m], suffixSum[i] - dp[i + x][max(m, x)]);
                    }
                }
            }
        }
        
        return dp[0][1];
    }
};

 

2. Python

#Python
class Solution:
    def stoneGameII(self, piles: List[int]) -> int:
        n = len(piles)
        
        dp = [[0] * (n + 1) for _ in range(n)]
        suffix_sum = [0] * n
        suffix_sum[-1] = piles[-1]
        
        for i in range(n - 2, -1, -1):
            suffix_sum[i] = suffix_sum[i + 1] + piles[i]
        
        for i in range(n - 1, -1, -1):
            for m in range(1, n + 1):
                if i + 2 * m >= n:
                    dp[i][m] = suffix_sum[i]
                else:
                    for x in range(1, 2 * m + 1):
                        dp[i][m] = max(dp[i][m], suffix_sum[i] - dp[i + x][max(m, x)])
        
        return dp[0][1]

 

DP는 쉬운 거 부터 차근차근 와야겠다..