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1140 - Stone Game II (Medium)

Problem Statement​

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] <= 10^4

Approach 1: Dynamic Programming​

Written by @wingkwong
class Solution {
public:
    // dp[i][j] = max stones you can get from piles[i:] with M = j
    int dp[101][200], n;
    int dfs(vector<int>& piles, int i, int m, int total) {
        // if we calculated the result before, use it directly
        if (dp[i][m] != -1) return dp[i][m];
        // res: used to compare the max number of stones
        // taken: used to record how many stones we've taken
        int res = 0, taken = 0;
        // i is the starting position
        // we can take at most i + 2 * m piles
        // however, it may exceed the size of piles 
        // hence use min to get the max limit
        for (int j = i; j < min(i + 2 * m, n); j++) {
            // take this pile of stones
            taken += piles[j];
            // move to the next position
            // with the new M = max(M, X)
            // where X is how many piles we've taken so far which is j - i + 1
            res = max(res, total - dfs(piles, j + 1, max(m, j - i + 1), total - taken));
        }
        // memoize the result
        return dp[i][m] = res;
    }
    
    int stoneGameII(vector<int>& piles) {
        // init dp with value -1
        memset(dp, -1, sizeof(dp));
        n = piles.size();
        return dfs(piles, 0, 1, accumulate(piles.begin(), piles.end(), 0));
    }
};