Problem
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 <= 1001 <= piles[i] <= 104
Solution
/** * @param {number[]} piles * @return {number} */ var stoneGameII = function(piles) { var dp = Array(piles.length).fill(0).map(() => ({})); var suffixSum = Array(piles.length); for (var i = piles.length - 1; i >= 0; i--) { suffixSum[i] = (suffixSum[i + 1] || 0) + piles[i]; } return helper(piles, 0, 1, dp, suffixSum); }; var helper = function(piles, i, M, dp, suffixSum) { if (dp[i][M]) return dp[i][M]; var res = 0; var sum = 0; for (var j = 0; j < 2 * M && i + j < piles.length; j++) { sum += piles[i + j]; res = Math.max( res, i + j + 1 === piles.length ? sum : sum + suffixSum[i + j + 1] - helper(piles, i + j + 1, Math.max(M, j + 1), dp, suffixSum) ); } dp[i][M] = res; return res; } Explain:
nope.
Complexity:
- Time complexity : O(n ^ 3).
- Space complexity : O(n ^ 2).