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).