2218. Maximum Value of K Coins From Piles

class Solution:
    def maxValueOfCoins(self, piles: List[List[int]], k: int) -> int:
        presum = [list(accumulate(p, initial=0)) for p in piles]
        n = len(piles)
        dp = [[0] * (k + 1) for _ in range(n + 1)]
        for i, s in enumerate(presum, 1):
            for j in range(k + 1):
                for idx, v in enumerate(s):
                    if j >= idx:
                        dp[i][j] = max(dp[i][j], dp[i - 1][j - idx] + v)
        return dp[-1][-1]

class Solution {
    public int maxValueOfCoins(List<List<Integer>> piles, int k) {
        int n = piles.size();
        List<int[]> presum = new ArrayList<>();
        for (List<Integer> p : piles) {
            int m = p.size();
            int[] s = new int[m + 1];
            for (int i = 0; i < m; ++i) {
                s[i + 1] = s[i] + p.get(i);
            }
            presum.add(s);
        }
        int[] dp = new int[k + 1];
        for (int[] s : presum) {
            for (int j = k; j >= 0; --j) {
                for (int idx = 0; idx < s.length; ++idx) {
                    if (j >= idx) {
                        dp[j] = Math.max(dp[j], dp[j - idx] + s[idx]);
                    }
                }
            }
        }
        return dp[k];
    }
}

class Solution {
public:
    int maxValueOfCoins(vector<vector<int>>& piles, int k) {
        vector<vector<int>> presum;
        for (auto& p : piles) {
            int m = p.size();
            vector<int> s(m + 1);
            for (int i = 0; i < m; ++i) s[i + 1] = s[i] + p[i];
            presum.push_back(s);
        }
        vector<int> dp(k + 1);
        for (auto& s : presum) {
            for (int j = k; ~j; --j) {
                for (int idx = 0; idx < s.size(); ++idx) {
                    if (j >= idx) dp[j] = max(dp[j], dp[j - idx] + s[idx]);
                }
            }
        }
        return dp[k];
    }
};

func maxValueOfCoins(piles [][]int, k int) int {
	var presum [][]int
	for _, p := range piles {
		m := len(p)
		s := make([]int, m+1)
		for i, v := range p {
			s[i+1] = s[i] + v
		}
		presum = append(presum, s)
	}
	dp := make([]int, k+1)
	for _, s := range presum {
		for j := k; j >= 0; j-- {
			for idx, v := range s {
				if j >= idx {
					dp[j] = max(dp[j], dp[j-idx]+v)
				}
			}
		}
	}
	return dp[k]
}

class Solution:
    def maxValueOfCoins(self, piles: List[List[int]], k: int) -> int:
        presum = [list(accumulate(p, initial=0)) for p in piles]
        dp = [0] * (k + 1)
        for s in presum:
            for j in range(k, -1, -1):
                for idx, v in enumerate(s):
                    if j >= idx:
                        dp[j] = max(dp[j], dp[j - idx] + v)
        return dp[-1]