2218. Maximum Value of K Coins From Piles

Description

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 i^th 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

Solutions

Solution 1: Dynamic Programming (Grouped Knapsack)

We define $f[i][j]$ as the maximum value sum of taking $j$ coins from the first $i$ piles. The answer is $f[n][k]$, where $n$ is the number of piles.

For the $i$-th pile, we can choose to take the first $0$, $1$, $2$, $\cdots$, $k$ coins. We can use a prefix sum array $s$ to quickly calculate the value sum of taking the first $h$ coins.

The state transition equation is:

$$ f[i][j] = \max(f[i][j], f[i - 1][j - h] + s[h]) $$

where $0 \leq h \leq j$, and $s[h]$ represents the value sum of taking the first $h$ coins from the $i$-th pile.

The time complexity is $O(k \times L)$, and the space complexity is $O(n \times k)$. Here, $L$ is the total number of coins, and $n$ is the number of piles.

Python3JavaC++GoTypeScript

class Solution:
    def maxValueOfCoins(self, piles: List[List[int]], k: int) -> int:
        n = len(piles)
        f = [[0] * (k + 1) for _ in range(n + 1)]
        for i, nums in enumerate(piles, 1):
            s = list(accumulate(nums, initial=0))
            for j in range(k + 1):
                for h, w in enumerate(s):
                    if j < h:
                        break
                    f[i][j] = max(f[i][j], f[i - 1][j - h] + w)
        return f[n][k]

class Solution {
    public int maxValueOfCoins(List<List<Integer>> piles, int k) {
        int n = piles.size();
        int[][] f = new int[n + 1][k + 1];
        for (int i = 1; i <= n; i++) {
            List<Integer> nums = piles.get(i - 1);
            int[] s = new int[nums.size() + 1];
            s[0] = 0;
            for (int j = 1; j <= nums.size(); j++) {
                s[j] = s[j - 1] + nums.get(j - 1);
            }
            for (int j = 0; j <= k; j++) {
                for (int h = 0; h < s.length && h <= j; h++) {
                    f[i][j] = Math.max(f[i][j], f[i - 1][j - h] + s[h]);
                }
            }
        }
        return f[n][k];
    }
}

class Solution {
public:
    int maxValueOfCoins(vector<vector<int>>& piles, int k) {
        int n = piles.size();
        vector<vector<int>> f(n + 1, vector<int>(k + 1));
        for (int i = 1; i <= n; i++) {
            vector<int> nums = piles[i - 1];
            vector<int> s(nums.size() + 1);
            for (int j = 1; j <= nums.size(); j++) {
                s[j] = s[j - 1] + nums[j - 1];
            }
            for (int j = 0; j <= k; j++) {
                for (int h = 0; h < s.size() && h <= j; h++) {
                    f[i][j] = max(f[i][j], f[i - 1][j - h] + s[h]);
                }
            }
        }
        return f[n][k];
    }
};

func maxValueOfCoins(piles [][]int, k int) int {
    n := len(piles)
    f := make([][]int, n+1)
    for i := range f {
        f[i] = make([]int, k+1)
    }
    for i := 1; i <= n; i++ {
        nums := piles[i-1]
        s := make([]int, len(nums)+1)
        for j := 1; j <= len(nums); j++ {
            s[j] = s[j-1] + nums[j-1]
        }

        for j := 0; j <= k; j++ {
            for h, w := range s {
                if j < h {
                    break
                }
                f[i][j] = max(f[i][j], f[i-1][j-h]+w)
            }
        }
    }
    return f[n][k]
}

function maxValueOfCoins(piles: number[][], k: number): number {
    const n = piles.length;
    const f: number[][] = Array.from({ length: n + 1 }, () => Array(k + 1).fill(0));
    for (let i = 1; i <= n; i++) {
        const nums = piles[i - 1];
        const s = Array(nums.length + 1).fill(0);
        for (let j = 1; j <= nums.length; j++) {
            s[j] = s[j - 1] + nums[j - 1];
        }
        for (let j = 0; j <= k; j++) {
            for (let h = 0; h < s.length && h <= j; h++) {
                f[i][j] = Math.max(f[i][j], f[i - 1][j - h] + s[h]);
            }
        }
    }
    return f[n][k];
}

Solution 2: Dynamic Programming (Space Optimization)

We can observe that for the $i$-th pile, we only need to use $f[i - 1][j]$ and $f[i][j - h]$, so we can optimize the two-dimensional array to a one-dimensional array.

The time complexity is $O(k \times L)$, and the space complexity is $O(k)$.