Maximum Value of K Coins From Piles

class Solution {
public:
    int maximumValue(vector<vector<int>>& piles, int k) {
        int pileCount = piles.size();
        vector<vector<int>> memo(pileCount + 1, vector<int>(k + 1, -1));
        
        function<int(int, int)> getMaxValue = [&](int index, int remainingCoins) {
            if (index == 0) {
                return 0;
            }
            if (memo[index][remainingCoins] != -1) {
                return memo[index][remainingCoins];
            }
            int sum = 0;
            for (int usedCoins = 0; usedCoins <= min((int)piles[index - 1].size(), remainingCoins); usedCoins++) {
                if (usedCoins > 0) {
                    sum += piles[index - 1][usedCoins - 1];
                }
                memo[index][remainingCoins] = max(memo[index][remainingCoins], getMaxValue(index - 1, remainingCoins - usedCoins) + sum);
            }
            return memo[index][remainingCoins];
        };
        
        return getMaxValue(pileCount, k);
    }
};

This C++ solution tackles the problem of finding the maximum value achievable by selecting up to 'k' coins from multiple piles, where each coin in a pile has a different value. The approach utilizes dynamic programming along with recursion and memoization to store intermediate results and avoid recalculating them, which is essential for optimizing performance in problems involving state.

The way the solution works is as follows:

• Define a 2D vector memo where memo[i][j] stores the maximum value you can obtain from the first i piles with j coins left to choose. If the value is yet to be determined, it is initialized to -1.

• The core of the solution is the recursive function getMaxValue, which calculates the maximum value for a given number of remaining piles (index) and remaining coins (remainingCoins):

  • Base Case: When no piles are left (index is 0), return 0 because no coins can be taken.

  • For each pile, iterate over possible numbers of coins to take (usedCoins), ensuring not to exceed the actual number of coins in the pile or the total number of coins allowed (remainingCoins). For every decision, accumulate the value of coins taken and use the recursive relation to find out the maximum value obtained by taking up to usedCoins coins from this pile and the rest from the previous piles.

  • Use memoization to save and retrieve calculated results of subproblems.

• The function begins its computation from the last pile, trying to make optimal choices for each subproblem, and moves backwards to the first pile.

• Upon exiting the recursive function, getMaxValue(pileCount, k) will provide the maximum value of coins from up to k coins available across all piles.

This solution leverages memoization effectively to reduce the time complexity by avoiding redundant calculations, especially essential when working with considerable data sizes, such as in this problem of optimizing coin collection from multiple piles.

class Solution {
    int memo[][];
    private int calcMaxValue(List<List<Integer>> coinPiles, int index, int availableCoins) {
        if (index == 0) {
            return 0;
        }
        if (memo[index][availableCoins] != -1) {
            return memo[index][availableCoins];
        }
        int sum = 0;
        for (int takenCoins = 0; takenCoins <= Math.min(coinPiles.get(index - 1).size(), availableCoins); takenCoins++) {
            if (takenCoins > 0) {
                sum += coinPiles.get(index - 1).get(takenCoins - 1);
            }
            memo[index][availableCoins] = Math.max(memo[index][availableCoins], calcMaxValue(coinPiles, index - 1, availableCoins - takenCoins) + sum);
        }
        return memo[index][availableCoins];
    }
    public int maxCoinValue(List<List<Integer>> piles, int maxCoins) {
        int pileCount = piles.size();
        memo = new int[pileCount + 1][maxCoins + 1];
        for (int i = 1; i <= pileCount; i++) {
            for (int remainingCoins = 0; remainingCoins <= maxCoins; remainingCoins++) {
                memo[i][remainingCoins] = -1;
            }
        }
        return calcMaxValue(piles, pileCount, maxCoins);
    }
}

The Java solution presented focuses on calculating the maximum value of coins one can collect from several piles with a limitation on the number of coins you can take. The approach utilizes dynamic programming to optimize the solution.

• The core of the solution lies in the Solution class containing a recursive calcMaxValue function, paired with a memoization table, memo, to store intermediate results.
• In calcMaxValue, the function first checks to see if the memoized result for the current subproblem exists. If it does, it returns that stored value to avoid redundant calculations. If not, it calculates the best value by iterating through all possible numbers of coins that can be taken from the current pile.
• The recursion decreases the pile index and the number of available coins accordingly for each recursive call, while accumulating coin values to determine the maximum possible collection.
• In the maxCoinValue method, initialization of the memoization array ensures all values start unset (typically with -1). This method sets up the memo array and kicks off the recursion starting from the last pile with all available coins.
• The maxCoinValue method essentially invokes calcMaxValue with the total number of piles and the total number of available coins, returning the maximum coin value possible under the constraints provided.

This solution effectively manages the exponential nature of exploring each combination of coins via dynamic programming with memoization, ensuring that the solution is computationally feasible even for larger inputs.

class Solution:
    def computeMaxValue(self, piles: List[List[int]], k: int) -> int:
        count_piles = len(piles)
        memo = [[-1] * (k + 1) for _ in range(count_piles + 1)]

        def dfs(idx, remaining):
            if idx == 0:
                return 0
            if memo[idx][remaining] != -1:
                return memo[idx][remaining]
            total_value = 0
            for taken in range(0, min(len(piles[idx - 1]), remaining) + 1):
                if taken > 0:
                    total_value += piles[idx - 1][taken - 1]
                memo[idx][remaining] = max(memo[idx][remaining],
                                           dfs(idx - 1, remaining - taken) + total_value)
            return memo[idx][remaining]

        return dfs(count_piles, k)

This Python code defines a solution to determine the maximum value that can be collected from a collection of piles of coins, where you may choose up to k coins. The implementation uses dynamic programming combined with memoization as an optimization over a simple brute force approach.

• The solution class is named Solution with a function computeMaxValue which accepts a list of lists (piles) where each sublist represents a pile of coins with varying values, and an integer k, the maximum number of coins that can be picked.

• The computeMaxValue function first calculates the number of piles and initializes a 2D list memo to store intermediate values, reducing the redundancy of recomputing results for the same state. Each element in memo is initialized to -1, which represents an uncomputed state.

• This function further includes a nested helper function dfs which uses Depth First Search technique to explore possible combinations of coins from different piles. The base case for dfs checks if the pile index (idx) is zero, where it returns zero since no coins can be taken from zero piles.

• The recursive computation checks previously computed results in memo to avoid re-computation. If not previously computed, it iterates over the number of coins that can be taken from the current pile (up to the minimum of pile size or remaining coin quota k).

• For each possibility, the value collected by taking a certain number of coins from the pile is computed and added to the result of the recursive call with the remaining coin allowance. The maximum of these values is stored in the memo array.

• The recursion terminates when all piles have been considered, and the top-level function call to dfs returns the maximum value obtainable for k coins. This value is then returned by the computeMaxValue function.

Overall, the approach effectively uses memoization to optimize the recursive strategy, allowing the solution to handle larger inputs more efficiently. This is a common technique for problems involving optimal selection from sets (like coins here) under constraints (like the maximum k coins here).