Problem Description

You are given n piles of coins, each pile containing a stack of coins with varying denominations. The piles are represented by a list piles, where each element piles[i] is itself a list that contains integers representing the coins' values in the ith pile from the top to the bottom of the pile.

The goal is to maximize the total value of coins you can collect by picking coins from the tops of these piles. However, there's a catch: you can only make a total of k moves, and in each move, you can only pick one coin from the top of any pile. Once you take a coin from a pile, you can't skip coins; you can only take another coin from the same pile if it's now the new top coin of that pile.

Knowing this, you need to find the maximum total value of coins you can amass in your wallet assuming you make exactly k moves and do so in the most optimal way possible.

Intuition

To approach this problem effectively, we need a strategy that allows us to make local decisions that contribute to our overall goal of maximizing the value of the coins we collect. This is a classic optimization problem that hints at using dynamic programming (DP) because it involves making a series of decisions that rely on the outcomes of previous decisions.

The intuition behind the solution involves two key insights:

1. We need to track the maximum value we can achieve for any number of moves up to k, as we progress through the piles.
2. We must consider all possible options for each pile, i.e., we can choose 0 coins from the pile, or we can take 1 coin, or we can take 2 coins, and so on, and for each of these choices, we need to update the maximum value possible considering the new total number of moves.

The implementation uses a 2D dynamic programming approach, which is effectively compressed into a 1D DP array dp for space efficiency. The dp array stores the maximum value that can be achieved for each possible number of moves up to k.

The presum array contains the prefix sums for each pile, which allows us to easily compute the total value gained by taking the first i coins from a pile.

The trick is to iterate through all piles and all possible coin counts we might take from each pile, update our dp array to store the best result. The updating is done in reverse to ensure that we don't overwrite a DP state before we're done using it for the current pile.

At the end of the iteration, dp[k] will hold our answer, which is the maximum value we can collect with k coins picked.

Learn more about Dynamic Programming and Prefix Sum patterns.

Solution Approach

The solution involves dynamic programming (DP), a technique that solves problems by breaking them down into simpler subproblems. We use a 1D DP array, dp, where each index j represents the maximum value achievable with j moves. The DP array is initialized with zeros, as initially, no coins are taken.

For each pile s in our presum array, where presum is a list of prefix sums (cumulative sums from the start of the pile to the current index), we iterate through our dp array in reverse. We do this because we want to update our dp values for using a certain number of coins, idx, from the current pile without affecting the other values of dp that we haven't processed yet.

For each index j of our dp array, we loop through all possible coin counts, idx, that we could take from the current pile, represented by the different values in the prefix sum. If taking idx coins is feasible (i.e., j is greater than or equal to idx), then we update our dp[j] by choosing the max between its current value and the sum of dp[j - idx] and the value of the idx-th coin, v. Mathematically, this is represented as follows:

dp[j] = max(dp[j], dp[j - idx] + v)

This line of the code represents the core of the DP transition equation. It updates the dp[j] to the maximum value we could have if we did not take any coins from the current pile or if we took idx coins from the current pile.

The process continues until we have processed all piles and considered all possible moves up to k. The last element of the dp array, dp[-1] or dp[k], will contain the maximum total value achievable by taking k coins across all piles.

Data Structures:

• A list dp for the DP array.
• A list of lists presum for the prefix sums of each pile.

Algorithms and Patterns:

• Dynamic programming for optimal substructure and overlapping subproblems.
• Prefix sums to quickly calculate the sum of taking a certain number of coins from a pile.
• Iterating in reverse to preserve DP states for processing without interference.

By carefully iterating and updating our DP array, we ensure that at the completion of the algorithm, the dp[k] index will hold the maximum value of coins we can collect out of k moves.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Suppose we have the following piles of coins and can make k = 2 moves:

piles = [[1, 2], [2, 1]]

We want to find the maximum value we can collect by making exactly two moves.

Step 1: Create Prefix Sums for Each Pile

First, we calculate the prefix sum for each pile to make summing coins efficient.

For the first pile [1, 2], the prefix sums are [1, 3] (taking 0 coins gives 0, taking the first coin gives 1, and taking the first two coins gives 3).

For the second pile [2, 1], the prefix sums are [2, 3].

Step 2: Initialize the DP Array

Next, we initialize a dp array with a length of k+1 (for 0 to k moves), starting with all zeros:

dp = [0, 0, 0]

Step 3: Iterate Over Piles and Update DP Array

We iterate over each prefix sum array and update our dp:

• For the first pile's prefix sums [1, 3], we compare:

  • For j = 1: dp[1] = max(dp[1], dp[1 - 0] + 1) = max(0, 0 + 1) = 1 (using 1 coin from the pile)
  • For j = 1: dp[1] = max(dp[1], dp[1 - 1] + 1) = max(1, 0 + 2) = 2 (opting to instead use 2 coins from the pile)
  • For j = 2: dp[2] = max(dp[2], dp[2 - 1] + 1) = max(0, 1 + 2) = 3 (taking 1 coin previously and 1 coin now)
  • For j = 2: dp[2] = max(dp[2], dp[2 - 2] + 3) = max(3, 0 + 3) = 3 (taking both coins from this pile)

In each iteration, idx is considered as the number of coins we pick from the current pile (hence the added values from the prefix sums). We choose the largest possible value to assign to dp[j].

After processing the first pile, our dp array is [0, 2, 3].

• For the second pile's prefix sums [2, 3], we perform similar updates:

  • For j = 1: dp[1] = max(dp[1], dp[1 - 0] + 2) = max(2, 0 + 2) = 2 (no change)
  • For j = 1: dp[1] = max(dp[1], dp[1 - 1] + 2) = max(2, 0 + 2) = 2 (opting to instead use 2 coins from the pile)
  • For j = 2: dp[2] = max(dp[2], dp[2 - 1] + 2) = max(3, 2 + 1) = 3 (taking 1 coin from the first pile and 1 from this one)
  • For j = 2: dp[2] = max(dp[2], dp[2 - 2] + 3) = max(3, 0 + 3) = 3 (no change)

After processing the second pile, our dp array is still [0, 2, 3].

Step 4: Collect the Result

After iterating through all piles, the last entry in the dp array gives us the answer. For k = 2, dp[k] is 3, so the maximum value we can collect is 3.

In this example, the best strategy is to take one coin with a value of 2 from the first pile and one coin with a value of 1 from the second pile, or to take the top coin from each pile, as both strategies yield the total maximum value of 3.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code can be analyzed by breaking down its operations. We need to consider the number of iterations in the nested loops along with the cost of constructing the prefix sum presum.

The first operation is the construction of the prefix sums for each pile, which is O(n) per pile where n is the number of coins in the pile. If there are m piles, this step will be O(m * n).

Next, we look at the nested loops. The outermost loop iterates over each pile's prefix sum list, which will happen m times (the number of piles). The middle loop will always iterate k times since it goes from k to 0. The innermost loop depends on the length of the current prefix sum list, which can be a maximum of k+1, because we will never take more than k coins from a single pile.

Combining the total iterations for the nested loops, we have m (number of piles) multiplied by k (maximum coins we can pick) multiplied by k+1 (coinciding with the prefix sums length). Therefore, the total time complexity of the nested loops is O(m * k^2).

Hence, the overall time complexity of the algorithm is dominated by the nesting loops, resulting in O(m * k^2).

The space complexity of the algorithm can be determined by looking at the extra space used. The presum variable will store up to k+1 elements for each of the m piles, which means that its space complexity is O(m * (k+1)). However, since constants are omitted in Big O notation, it simplifies to O(m * k).

The dp array will have k+1 elements no matter what, leading to a space complexity of O(k).

Since dp does not grow with m, the overall space complexity will be dominated by presum, so the overall space complexity is O(m * k).

Learn more about how to find time and space complexity quickly using problem constraints.