Stone Game II

Description

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

Code

Time Complexity: $O(NM)$, Space Complexity: $O(NM)$

關鍵在於定義 play 的回傳值到底是什麼。

定義為自己的 turn 所拿的，以及對手拿完後的 index 之後的 prefix sum，再減掉對手的 play 的回傳。

原本卡住的點是要怎麼紀錄對手拿到多少，然後再開始自己的遞迴（alice 要 call alice 自己的遞迴，bob call bob 的），有點類似 minimax 的味道，但這裡正解是 alice call bob 的，bob 會 call alice 。

class Solution {
public:
    int stoneGameII(vector<int>& piles) {
        vector<int> pre = piles;
        vector<vector<int>> memo(101, vector<int>(202, -1));
        for(int i = pre.size() - 2; i >= 0; i--) {
            pre[i] += pre[i + 1];
        }
        return play(piles, 0, 1, pre, memo);
    }
 
    int play(vector<int>& piles, int idx, int M, vector<int>& pre, vector<vector<int>>& memo) {
        if(idx + 2 * M >= pre.size()) {
            // takes all
            return pre[idx];
        }
 
        if(memo[idx][M] != -1) 
            return memo[idx][M];
 
        int res = 0;
        int curStone = 0;
        for(int i = 1; i <= 2 * M; i++) {
            curStone = pre[idx] - pre[idx + i];
            res = max(res, curStone + pre[idx + i] - play(piles, idx + i, max(M, i), pre, memo));
        }
        
        memo[idx][M] = res;
        return res;
    }
};

Source

• Stone Game II - LeetCode