1140. Stone Game II

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

1140. Stone Game II
struct Solution;
use std::collections::HashMap;

impl Solution {
fn stone_game_ii(piles: Vec<i32>) -> i32 {
let n = piles.len();
let mut memo: HashMap<(usize, usize), (i32, i32)> = HashMap::new();
Self::dp(0, 1, &mut memo, &piles, n).0
}

fn dp(
start: usize,
m: usize,
memo: &mut HashMap<(usize, usize), (i32, i32)>,
piles: &[i32],
n: usize,
) -> (i32, i32) {
if start == n {
(0, 0)
} else {
if let Some(&res) = memo.get(&(start, m)) {
return res;
}
let mut a = 0;
let mut res = (0, 0);
for i in start..(start + 2 * m).min(n) {
a += piles[i];
let x = i - start + 1;
let (b, c) = Self::dp(i + 1, x.max(m), memo, piles, n);
if a + c > res.0 {
res = (a + c, b);
}
}
memo.insert((start, m), res);
res
}
}
}

#[test]
fn test() {
let piles = vec![2, 7, 9, 4, 4];
let res = 10;
assert_eq!(Solution::stone_game_ii(piles), res);
}