LeetCode 2431: Maximize Total Tastiness of Purchased Fruits Solution

Master LeetCode problem 2431 (Maximize Total Tastiness of Purchased Fruits), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2431. Maximize Total Tastiness of Purchased Fruits

Medium

Array Dynamic Programming

Problem Explanation

Explanation:

To maximize the total tastiness of purchased fruits, we can use a dynamic programming approach. We can create a 2D array to store the maximum total tastiness that can be achieved at each store and at each level of tastiness. We iterate through each store and each level of tastiness, calculate the maximum total tastiness that can be achieved at the current store and level of tastiness by either buying or not buying the fruit, and update the corresponding cell in the 2D array. In the end, we return the maximum total tastiness that can be achieved.

Initialize a 2D array dp of size (n+1) x (t+1) where n is the number of stores and t is the maximum level of tastiness.
Initialize maxTastiness to 0.
Iterate through each store from 1 to n:
- For each store, iterate through each level of tastiness from 1 to t:
  - Calculate the maximum total tastiness that can be achieved at the current store and level of tastiness by either buying or not buying the fruit.
  - Update the corresponding cell in the dp array with the maximum total tastiness.
  - If the current total tastiness is greater than maxTastiness, update maxTastiness with the current total tastiness.
Return maxTastiness. :

Solution Code

class Solution {
    public int maximizeTastiness(int n, int t, int[] tastiness, int[] store, int[] price) {
        int[][] dp = new int[n + 1][t + 1];
        int maxTastiness = 0;
        
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= t; j++) {
                dp[i][j] = dp[i - 1][j];
                if (j >= tastiness[i - 1]) {
                    dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - tastiness[i - 1]] + store[i - 1] - price[i - 1]);
                }
                maxTastiness = Math.max(maxTastiness, dp[i][j]);
            }
        }
        
        return maxTastiness;
    }
}

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