LeetCode in Kotlin
1140. Stone Game II
Medium
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
class Solution {
private val dp = Array(105) { IntArray(105) }
private fun help(i: Int, m: Int, p: IntArray): Int {
if (i >= p.size) {
dp[i][m] = 0
return 0
}
if (dp[i][m] != -1) {
return dp[i][m]
}
var ans = Int.MIN_VALUE
var total = 0
for (j in 0 until 2 * m) {
if (i + j < p.size) {
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
}
fun stoneGameII(piles: IntArray): Int {
var sum = 0
for (arr1 in dp) {
arr1.fill(-1)
}
for (z in piles) {
sum += z
}
return (sum + help(0, 1, piles)) / 2
}
}