1100 - 1199
1140 - Stone Game II (Medium)
Problem Statement
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: 104Constraints:
1 <= piles.length <= 1001 <= piles[i] <= 10^4
Approach 1: Dynamic Programming
class Solution {
public:
// dp[i][j] = max stones you can get from piles[i:] with M = j
int dp[101][200], n;
int dfs(vector<int>& piles, int i, int m, int total) {
// if we calculated the result before, use it directly
if (dp[i][m] != -1) return dp[i][m];
// res: used to compare the max number of stones
// taken: used to record how many stones we've taken
int res = 0, taken = 0;
// i is the starting position
// we can take at most i + 2 * m piles
// however, it may exceed the size of piles
// hence use min to get the max limit
for (int j = i; j < min(i + 2 * m, n); j++) {
// take this pile of stones
taken += piles[j];
// move to the next position
// with the new M = max(M, X)
// where X is how many piles we've taken so far which is j - i + 1
res = max(res, total - dfs(piles, j + 1, max(m, j - i + 1), total - taken));
}
// memoize the result
return dp[i][m] = res;
}
int stoneGameII(vector<int>& piles) {
// init dp with value -1
memset(dp, -1, sizeof(dp));
n = piles.size();
return dfs(piles, 0, 1, accumulate(piles.begin(), piles.end(), 0));
}
};