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

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.

Solution

/**
 * @param {number[]} piles
 * @return {number}
 */
var stoneGameII = function (piles) {
  let n = piles.length;
  //dp[i][m] means the maximum number of stones that the current player can get when the current player is playing piles[i:] and the current M is m
  let dp = new Array(n + 1).fill(0).map(() => new Array(n + 1).fill(0));
  //sum[i] means the sum of piles[i:]
  let sum = new Array(n + 1).fill(0);
  for (let i = n - 1; i >= 0; i--) {
    //the sum index is equal to the sum of piles[i:] plus the sum of piles[i + 1:]
    sum[i] = sum[i + 1] + piles[i];
  }
  for (let i = n - 1; i >= 0; i--) {
    for (let m = 1; m <= n; m++) {
      //if i + 2 * m is greater than or equal to n, then the current player can take all the remaining piles
      if (i + 2 * m >= n) {
        dp[i][m] = sum[i];
      } else {
        for (let x = 1; x <= 2 * m; x++) {
          //the current player can take x piles, then the next player can take dp[i + x][Math.max(m, x)] piles
          dp[i][m] = Math.max(dp[i][m], sum[i] - dp[i + x][Math.max(m, x)]);
        }
      }
    }
  }
  return dp[0][1];
};