1140. 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

 

Solutions

Solution: Dynamic Programming

• Time complexity: O(n²)
• Space complexity: O(n²)

 

JavaScript

/**
 * @param {number[]} piles
 * @return {number}
 */
const stoneGameII = function (piles) {
  const n = piles.length;
  const memo = new Array(n)
    .fill('')
    .map(_ => new Array(n).fill(0));
  const suffixSum = new Array(n + 1).fill(0);

  for (let index = n - 1; index >= 0; index--) {
    suffixSum[index] = suffixSum[index + 1] + piles[index];
  }

  const takePile = (index, M) => {
    if (index + M * 2 >= n) return suffixSum[index];
    if (memo[index][M]) return memo[index][M];

    let opponent = Number.MAX_SAFE_INTEGER;

    for (let x = 1; x <= 2 * M; x++) {
      opponent = Math.min(opponent, takePile(index + x, Math.max(M, x)));
    }
    return (memo[index][M] = suffixSum[index] - opponent);
  };

  return takePile(0, 1);
};