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: 10^4
Constraints:
1 <= piles.length <= 1001 <= piles[i] <= 10^4
Solution (Java)
class Solution {
private int[][] dp = new int[10^5][10^5];
private int help(int i, int m, int[] p) {
if (i >= p.length) {
dp[i][m] = 0;
return 0;
}
if (dp[i][m] != -1) {
return dp[i][m];
}
int ans = Integer.MIN_VALUE;
int total = 0;
for (int j = 0; j < 2 * m; j++) {
if (i + j < p.length) {
total += p[i + j];
ans = Math.max(ans, total - help(i + j + 1, Math.max(m, j + 1), p));
}
}
dp[i][m] = ans;
return ans;
}
public int stoneGameII(int[] piles) {
int sum = 0;
for (int[] arr1 : dp) {
Arrays.fill(arr1, -1);
}
for (int z : piles) {
sum += z;
}
return (sum + help(0, 1, piles)) / 2;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).