Maximum Value of K Coins From Piles Interview Question for Meta

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. For example:

Example 1: piles = [[1,100,3],[7,8,9]], k = 2 Output: 101

Example 2: piles = [[100],[100],[100],[100],[100],[100],[1,1,1,1,1,1,700]], k = 7 Output: 706

What is the most efficient algorithm to solve this problem? What are the time and space complexities?

## Dynamic Programming Solution for Coin Collection

This problem is a classic dynamic programming scenario. We are given a set of piles of coins and asked to find the maximum value of coins we can collect by taking exactly *k* coins.

### 1. Naive Approach (Brute Force)

A brute-force solution would involve exploring all possible combinations of coins from different piles. This quickly becomes computationally infeasible as the number of piles and the value of *k* increase.  The time complexity would be exponential, making it impractical for the given constraints.

### 2. Optimal Solution (Dynamic Programming)

We can use dynamic programming to solve this problem efficiently. Let `dp[i][j]` represent the maximum value we can obtain by considering the first `i` piles and taking exactly `j` coins.

Here's the algorithm:

1.  Initialize a 2D array `dp` of size `(n + 1) x (k + 1)` with all values set to 0.
2.  Iterate through each pile `i` from 1 to `n`.
3.  For each pile, iterate through the number of coins `j` we want to take from 0 to `k`.
4.  For each `j`, iterate through the number of coins `x` we can take from the current pile `i` (from 0 to the minimum of `j` and the number of coins in the pile).
5.  Update `dp[i][j]` with the maximum value we can obtain by either not taking any coins from the current pile or taking `x` coins from the current pile and `j - x` coins from the previous piles.
6.  The final answer will be `dp[n][k]`.

Here's the Python code:

```python
def max_value_of_coins(piles, k):
    n = len(piles)
    dp = [[0] * (k + 1) for _ in range(n + 1)]

    for i in range(1, n + 1):
        for j in range(k + 1):
            dp[i][j] = dp[i-1][j]  # Option 1: Don't take any coins from this pile
            current_pile_sum = 0
            for x in range(1, min(j + 1, len(piles[i-1]) + 1)):
                current_pile_sum += piles[i-1][x-1]
                dp[i][j] = max(dp[i][j], dp[i-1][j-x] + current_pile_sum)

    return dp[n][k]

# Example Usage
piles1 = [[1, 100, 3], [7, 8, 9]]
k1 = 2
print(f"Maximum value for piles {piles1} and k={k1}: {max_value_of_coins(piles1, k1)}") # Output: 101

piles2 = [[100], [100], [100], [100], [100], [100], [1, 1, 1, 1, 1, 1, 700]]
k2 = 7
print(f"Maximum value for piles {piles2} and k={k2}: {max_value_of_coins(piles2, k2)}") # Output: 706

3. Big(O) Time Complexity Analysis

The time complexity of this dynamic programming solution is O(n * k^2), where n is the number of piles and k is the maximum number of coins we can take. The nested loops iterate through each pile, each possible number of coins to take, and each possible number of coins to take from the current pile. Therefore, the overall time complexity is O(nkmin(k, pile_length)), which simplifies to O(n * k^2) because in the worst case, the minimum can be close to k.

4. Big(O) Space Complexity Analysis

The space complexity of this solution is O(n * k), where n is the number of piles and k is the maximum number of coins we can take. This is because we are using a 2D array dp of size (n + 1) x (k + 1) to store the maximum values for each state.

5. Edge Cases

Empty Piles: If any of the piles are empty, the algorithm should still work correctly as the inner loop will simply not execute for that pile.
k = 0: If k is 0, the maximum value will be 0, as we are not allowed to take any coins.
k > Total Coins: If k is greater than the total number of coins in all piles, the algorithm will still work correctly, as the inner loop will be limited by the number of coins in each pile.
Large Coin Values: The coin values can be large (up to 10^5), so we need to make sure that the dp array can store these large values without causing overflow issues.

This dynamic programming approach provides an efficient and accurate solution to the coin collection problem, handling various edge cases and delivering optimal results within the given constraints.