[LeetCode] 877. Stone Game == [LintCode] 396. Coins in a Line 3_hard tag: 区间Dynamic Programming, 博弈(

Posted 2020-12-26

Alex and Lee play a game with piles of stones.  There are an even 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.  The total number of stones is odd, so there are no ties.

Alex and Lee take turns, with Alex starting first.  Each turn, a player takes the entire pile of stones from either the beginning or the end of the row.  This continues until there are no more piles left, at which point the person with the most stones wins.

Assuming Alex and Lee play optimally, return True if and only if Alex wins the game.

 

Example 1:

Input: [5,3,4,5]
Output: true
Explanation: 
Alex starts first, and can only take the first 5 or the last 5.
Say he takes the first 5, so that the row becomes [3, 4, 5].
If Lee takes 3, then the board is [4, 5], and Alex takes 5 to win with 10 points.
If Lee takes the last 5, then the board is [3, 4], and Alex takes 4 to win with 9 points.
This demonstrated that taking the first 5 was a winning move for Alex, so we return true.

 

Note:

1. 2 <= piles.length <= 500
2. piles.length is even.
3. 1 <= piles[i] <= 500
4. sum(piles) is odd.

 

这个题目思路跟[LintCode] 395. Coins in a Line 2_Medium tag: Dynamic Programming, 博弈很像, 只不过这里是利用 区间Dynamic Programing的方法,所以只用一维的dp已经不够了, 另外初始化的时候我们不直接用for loop, 而是用类似于dfs recursive的方法去将初始化放在helper fuction里面. 另外得到的

动态方程式为  A[i][j] = max( piles[i] + min(A[i+1][j-1] + A[i+2][j]) , piles[j] + min(A[i+1][j-1], A[i][j-2]) )

init;

A[i][i] = piles[i]

A[i][i+1] = max(piles[i], piles[i+1])

 

1. Constraints

1) size [2,500], even number

2) element [1,50], integer

3) sum(piles) is odd, no ties

 

2. Ideas

Dynamic Programming   ,     T: O(n^2)     S; O(n^2)

 

3. Code

 1 class Solution:
 2     def stoneGame(self, piles):
 3         n = len(piles)
 4         dp, flag = [[0]*n for _ in range(n)], [[0]*n for _ in range(n)]
 5         def helper(left, right):
 6             if flag[left][right]:
 7                 return dp[left][right]
 8             if left == right:
 9                 dp[left][right] = piles[left]
10             elif left + 1 = right:
11                 dp[left][right] = max(piles[left], piles[right])
12             elif left < right: # left > right, init 0
13                 value_l = piles[left] + min(helper(left+2, right), helper(left + 1, right -1))
14                 value_r = piles[right] + min(helper(left+1, right-1), helper(left, right - 2))
15                 dp[left][right] = max(value_l, value_r)
16             flag[left][right] = 1
17             return dp[left][right]
18         return helper(0, n-1) > sum(piles)//2

 

4. Test cases

[5,3,4,5]