LeetCode [1140] Stone Game II

 1140. Stone Game II

Medium

Alex and Lee 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. 

Alex and Lee take turns, with Alex 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 Alex and Lee play optimally, return the maximum number of stones Alex can get.

 

Example 1:

Input: piles = [2,7,9,4,4]
Output: 10
Explanation:  If Alex takes one pile at the beginning, Lee takes two piles, then Alex takes 2 piles again. Alex can get 2 + 4 + 4 = 10 piles in total. If Alex takes two piles at the beginning, then Lee can take all three piles left. In this case, Alex get 2 + 7 = 9 piles in total. So we return 10 since it's larger. 

 

Constraints:

• 1 <= piles.length <= 100
• 1 <= piles[i] <= 10 ^ 4

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class Solution { int[][] memo; int n; int[] preSum; public int stoneGameII(int[] piles) { n = piles.length; preSum = Arrays.copyOf(piles, n); for(int i=n-2; i>=0; --i){ preSum[i] += preSum[i+1]; } memo = new int[n][n];//memo[i][j]: max stones when M = i in piles[j..n-1] return helper(1, 0); } int helper(int M, int p){ if(p+M*2>=n) return preSum[p];//take all left int res = 0; if(memo[M][p]>0) return memo[M][p]; for(int t=1; t<=M*2; ++t){ int curTake = preSum[p] - preSum[p+t]; res = Math.max(res, curTake+preSum[p+t]-helper(Math.max(M, t), p+t)); } memo[M][p] = res; return res; } }