5259_max_value_of_k_coins.py

'''
5269. Maximum Value of K Coins From Piles
There are n piles of coins on a table. Each pile consists of a positive number of coins of assorted denominations.
In one move, you can choose any coin on top of any pile, remove it, and add it to your wallet.
Given a list piles, where piles[i] is a list of integers denoting the composition of the ith pile from top to bottom, and a positive integer k, return the maximum total value of coins you can have in your wallet if you choose exactly k coins optimally.

Example 1:
Input: piles = [[1,100,3],[7,8,9]], k = 2
Output: 101
Explanation:
The above diagram shows the different ways we can choose k coins.
The maximum total we can obtain is 101.

Example 2:
Input: piles = [[100],[100],[100],[100],[100],[100],[1,1,1,1,1,1,700]], k = 7
Output: 706
Explanation:
The maximum total can be obtained if we choose all coins from the last pile.

Constraints:
n == piles.length
1 <= n <= 1000
1 <= piles[i][j] <= 105
1 <= k <= sum(piles[i].length) <= 2000
'''
class Solution:
    def maxValueOfCoins(self, piles, k: int) -> int:
        N = len(piles)
        def dp(i, to_choose):
            if to_choose <= 0 or i == N:
                return 0
            res = dp(i+1, to_choose)
            cur = 0
            for j in range(min(to_choose, len(piles[i]))):
                cur += piles[i][j]
                res = max(res, cur+dp(i+1, to_choose-j-1))
            # print(i, to_choose, res)
            return res
        return dp(0, k)

s = Solution()
piles = [[100],[100],[100],[100],[100],[100],[1,1,1,1,1,1,700]]
k = 7
res = s.maxValueOfCoins(piles, k)
print(res)