Maximize Total Tastiness of Purchased Fruits

  1. Problem Statement: You have an array of fruits where each fruit has a price and tastiness value. You want to buy some fruits to maximize the total tastiness without exceeding a given budget (maxAmount). You also have coupons that allow you to buy fruits at half price, but you can use only a limited number of coupons (maxCoupons).

  2. Traverse the Array of Fruits: You go through the array of fruits one by one, considering what to do with each fruit. You have three possible actions for each fruit.

  3. Three Possible Operations for Any Fruit:

    • Buy the Fruit Without Coupon: You can purchase the fruit at its full price, and the tastiness value of that fruit is added to the total tastiness.
    • Buy the Fruit With Coupon: If you have coupons left, you can buy the fruit at half its price (rounded down), and again, its tastiness is added to the total.
    • Do Not Buy the Fruit: You can also decide not to buy the fruit and move on to the next one.

  4. Using Recursion: Since you have three choices for each fruit, and you need to explore all possible combinations to find the best solution, recursion is a natural choice. A recursive function goes through each fruit and explores all three possibilities, calling itself for the next fruit until all options are explored.

  5. Cache (Memoization) to Avoid Recomputation: As the recursive function explores many possible paths, it may calculate the same result multiple times. To save time, you store the results of calculations in a “memo” (a cache), so if the function is called again with the same parameters, it simply returns the stored result instead of recalculating. This prevents Time Limit Exceeded (TLE) errors in coding challenges and makes the code more efficient.

  6. Final Result: The recursion explores all possible combinations of buying/not buying each fruit with and without coupons, and the solution with the maximum total tastiness within the budget is found.

  7. Example to Illustrate: Consider you have two fruits with prices [10, 15] and tastiness [5, 8], maxAmount = 10, and maxCoupons = 1. You would start with the first fruit and try all three possibilities, then do the same for the second fruit. The recursion would explore all paths, and using memoization, it would find the best solution, which in this case would be buying the second fruit with a coupon for a total tastiness of 8.

By using recursion to explore all possibilities and memoization to avoid unnecessary recomputation, this approach efficiently finds the maximum tastiness within the constraints.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37

class Solution:
    def maxTastiness(self, price: List[int], tastiness: List[int], maxAmount: int, maxCoupons: int) -> int:
        n = len(price)
        # Create a memoization table to store the results of subproblems
        memo = {}

        def dfs(idx, remainingAmount, remainingCoupons):
            # Base case: if we reach the end of the array, return 0
            if idx == n:
                return 0

            # Check if the result is already computed
            if (idx, remainingAmount, remainingCoupons) in memo:
                return memo[(idx, remainingAmount, remainingCoupons)]

            # Option 1: Do not buy the fruit
            no_buy = dfs(idx + 1, remainingAmount, remainingCoupons)

            # Option 2: Buy the fruit without coupon
            buy_without_coupon = 0
            if remainingAmount >= price[idx]:
                buy_without_coupon = tastiness[idx] + dfs(idx + 1, remainingAmount - price[idx], remainingCoupons)

            # Option 3: Buy the fruit with coupon
            buy_with_coupon = 0
            if remainingCoupons > 0 and remainingAmount >= price[idx] // 2:
                buy_with_coupon = tastiness[idx] + dfs(idx + 1, remainingAmount - price[idx] // 2, remainingCoupons - 1)

            # Compute the maximum tastiness among the three options
            result = max(no_buy, buy_without_coupon, buy_with_coupon)

            # Store the result in the memoization table
            memo[(idx, remainingAmount, remainingCoupons)] = result
            return result

        # Start the recursive function from the first index with the given maxAmount and maxCoupons
        return dfs(0, maxAmount, maxCoupons)

This code correctly implements the recursive approach with memoization and considers the three possible operations for each fruit, finding the maximum total tastiness that can be purchased.