LeetCode Q 1140 - Stone Game II

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

Solution:

Code:

int[][] memo;
int[] sums;
public int stoneGameII(int[] piles) {
  int len = piles.length;
  memo = new int[len][len];
  sums = new int[len];
  
  sums[len - 1] = piles[len - 1];
  for (int i = len - 2; i >= 0; i--) {
    sums[i] = sums[i + 1] + piles[i];
  }
  
  return backtrack(piles, 0, 1);
}

private int backtrack(int[] piles, int index, int M) {
  if (index >= piles.length) return 0;
  
  // the nex person can get all left stones
  if (2 * M >= piles.length - index) return sums[index]; 
  
  if (memo[index][M] != 0) return memo[index][M];
  
  int min = Integer.MAX_VALUE; //the min value the next player can get
  for (int i = 1; i <= 2 * M; i++) {
      min = Math.min(min, backtrack(piles, index + i, Math.max(i, M)));
  }
  
  //max stones = all the left stones - the min stones next player can get
  memo[index][M] = sums[index] - min;
  
  return memo[index][M];
}

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