Problem Description

In this problem, we are dealing with fruit shopping with some constraints and aids in order to maximize the total tastiness of the fruits we can buy. We are provided with two arrays – one depicts the price of each fruit, and the other represents the tastiness of each fruit. The length of both arrays is the same, which indicates the number of different fruits available. Additionally, we are given two constraints in the form of integers, maxAmount and maxCoupons. Here’s a summary of the conditions and objectives:

• We aim to choose fruits such that the sum of their tastiness values is maximized.
• The total price of the chosen fruits cannot exceed maxAmount.
• We can use maxCoupons coupons, where each coupon allows buying a fruit at half price (rounded down).
• Each fruit can only be purchased once, and we can apply at most one coupon to each fruit.

The key challenge is to make strategic decisions about which fruits to buy and on which ones to use the coupons, in order to achieve the maximum total tastiness without exceeding the budget constraints.

Intuition

The intuition behind the solution is to explore each possibility of buying a fruit with and without a coupon, and not buying it at all. This kind of problem hints at a recursive approach, likely exploring all options via depth-first search and applying memoization to optimize repeated subproblems — a common dynamic programming strategy.

To arrive at a solution:

1. We need to consider each fruit and decide one of three actions: not buying it, buying it at full price (if we have enough amount), or buying it at half price using a coupon (if we have enough amount and coupons left).
2. This creates a decision tree where each node represents these possibilities, and we traverse these nodes in depth-first order.
3. Since similar subproblems will be solved repeatedly (with overlapping decisions for subsets of fruits, remaining amount, and coupons), we can use memoization to store already computed solutions of subproblems. In Python, this can be achieved using the @cache decorator for our recursive function.
4. The result is the maximum tastiness we can achieve by considering the options for buying, not buying, or coupon-buying each fruit while keeping track of remaining amount and coupons.

The given solution leverages memoization for efficient computation and relies on the recursive call dfs(i, j, k) where i is the current index in the fruit arrays, j is the remaining amount, and k is the remaining coupons. At each step, it makes recursive calls to consider different scenarios and takes the maximum tastiness outcome.

Learn more about Dynamic Programming patterns.

Solution Approach

The problem is solved using a combination of recursive depth-first search (DFS) and dynamic programming (DP) with memoization. Here's a step-by-step explanation of the solution approach based on the provided Python code:

1. A recursive helper function dfs(i, j, k) is defined, which represents the state of the problem after considering the first i fruits, with j being the remaining budget (maxAmount subtracted by the cost of fruits bought so far), and k being the remaining coupons.

2. The base case is when i equals the length of the price array, meaning that we have considered all the fruits. In this case, we cannot increase tastiness anymore, so the function returns 0.

3. For each fruit i, the function computes the maximum tastiness by making a recursive call to itself to represent the scenario of not buying the current fruit: dfs(i + 1, j, k).

4. It then checks if the current fruit can be bought without a coupon (provided the current budget j is at least the price of the fruit). If so, it calculates the tastiness of buying it at full price: dfs(i + 1, j - price[i], k) + tastiness[i] and updates the answer with the better option between not buying and buying at full price.

5. The function also checks if the current fruit can be bought with a coupon (provided there is at least one coupon and the current budget j is sufficient for the half price). In that case, it calculates the tastiness of buying it at half price: dfs(i + 1, j - price[i] // 2, k - 1) + tastiness[i] and updates the answer with the best option among not buying, buying at full price, and buying at half price.

6. Memoization is applied to the recursive function using the @cache decorator which stores the result of each unique state (i, j, k) so that repeated computations for the same state are avoided. This is where dynamic programming comes into play to optimize the solution.

7. Lastly, the dfs(0, maxAmount, maxCoupons) function call is made to kick off the decision process starting with the first fruit, the full budget, and all the coupons available. The returned value from this call will be the maximum possible total tastiness that can be achieved.

In summary, this solution leverages dynamic programming with memoization to efficiently compute the maximum tastiness achievable, considering the constraints of the budget and coupons. The algorithm tests all permutations of decisions (to buy with or without a coupon, or not to buy) and remembers solutions to subproblems to avoid redundant calculations.

Example Walkthrough

Suppose we have the following small example:

• price: [3, 5, 6]
• tastiness: [5, 6, 5]
• maxAmount: 8
• maxCoupons: 1

Let's walk through the solution approach for this example:

1. We will call our recursive function dfs(i, j, k) where i is the index of the current fruit we are considering, j is the remaining budget, and k is the number of coupons left.

2. Initially, we call dfs(0, 8, 1) because we start with the first fruit (index 0), we have the full budget of 8, and one coupon available.

3. The decision for the first fruit (price: 3, tastiness: 5) is as follows:

   • Not buying it: We move to the next fruit with the same budget and coupons. Recursive call: dfs(1, 8, 1).
   • Buying it at full price: We spend 3 from our budget, so we move to the next fruit with a budget of 5 and the same number of coupons. Recursive call with additional tastiness: dfs(1, 5, 1) + 5.
   • Buying it at half price with a coupon: We spend 3 // 2 = 1 from our budget, keep the tastiness, and reduce the number of coupons by one. Recursive call with additional tastiness: dfs(1, 7, 0) + 5.

4. The helper function will compare these scenarios using the recursive calls and, with the aid of memoization, return the maximum tastiness of these choices.

5. This process continues for the next fruits as well. For example, for fruit with price 5 and tastiness 6 (index 1), we would have these possible calls assuming we didn't buy the first fruit:

   • Not buying it: dfs(2, 8, 1).
   • Buying it at full price, if budget allows: dfs(2, 3, 1) + 6.
   • Buying it at half price with a coupon: dfs(2, 6, 0) + 6.

6. Note that for the second recursive call, the budget would not allow buying the fruit at the full price from the given example, thus that option would be ignored.

7. The memoization would remember outcomes of certain states such as dfs(1, 5, 1) and avoid recalculating them if they come up again in a different branch of the decision tree.

8. The recursion and decision-making continue until all possible combinations of buying/not buying fruits with and without the usage of coupons are exhaustively explored.

Considering our example, an optimal solution would be to:

• Use the coupon to buy the first fruit at a 50% discount, getting 5 tastiness and reducing our budget to 7.
• Buy the second fruit at full price with the remaining budget, getting an additional 6 tastiness.
• We don't have enough budget to buy the third fruit.

The maximum total tastiness can be achieved with these decisions is 11, from the first and second fruit, and this is the value returned by dfs(0, 8, 1).

Solution Implementation

Time and Space Complexity

The given Python code defines a recursive function with memoization to solve a variation of the knapsack problem, where we are trying to maximize the tastiness of items chosen under certain constraints on the price and available coupons.

Time Complexity

The time complexity of the code is dictated by the number of states that need to be computed, which is determined by the number of decisions for each item, the range of maxAmount (denoted as M), and the number of coupons maxCoupons (denoted as C). The function dfs is called with different states represented by a combination of current item index i, remaining amount j, and remaining coupons k.

For each item, there are three choices: skip the item, take the item without a coupon, or take the item with a coupon (if available). Since we only move to the next item in each recursive call, there are N levels of recursion, where N is the total number of items.

There is a unique state for each combination of (i, j, k). Since i can be in the range [0, N], j can take on values from [0, M], and k from [0, C], the number of possible states is roughly N * M * C.

The recursion is memoized to ensure that each state is computed at most once. Therefore, the time complexity is O(N*M*C).

Space Complexity

The space complexity of the code is governed by the storage required for:

1. The memoization cache, which needs to store the result for each unique state (i, j, k), thus requiring a space complexity of O(N*M*C).
2. The call stack for the recursion, which at maximum depth will be O(N), as we have N levels of recursion in the worst case.

Thus, the overall space complexity of the code is also O(N*M*C) due to the cache size dominating the recursive call stack.

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