LeetCode 3494: Find the Minimum Amount of Time to Brew Potions Solution

Master LeetCode problem 3494 (Find the Minimum Amount of Time to Brew Potions), 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.

3494. Find the Minimum Amount of Time to Brew Potions

Medium

Array Simulation Prefix Sum

Problem Explanation

Explanation:

To solve this problem, we can simulate the process of brewing potions by iterating through each wizard and each potion. We keep track of the time each wizard finishes working on a potion and update the start time for the next wizard accordingly. The key idea is to synchronize the timing so that each wizard starts working on a potion immediately after the previous wizard finishes.

Initialize a 2D array dp to store the minimum time taken for each wizard to brew each potion.
Iterate through each wizard and potion, updating the time taken based on the skill and mana capacities.
Update the start time for each wizard based on the completion time of the previous wizard.
Return the time taken by the last wizard to complete all potions.

Time Complexity: O(n * m)
Space Complexity: O(n * m)

Solution Code

class Solution {
    public int minBrewTime(int[] skill, int[] mana) {
        int n = skill.length;
        int m = mana.length;
        
        int[][] dp = new int[n][m];
        
        for (int j = 0; j < m; j++) {
            dp[0][j] = skill[0] * mana[j];
        }
        
        for (int i = 1; i < n; i++) {
            dp[i][0] = dp[i - 1][0] + skill[i] * mana[0];
        }
        
        for (int i = 1; i < n; i++) {
            for (int j = 1; j < m; j++) {
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]) + skill[i] * mana[j];
            }
        }
        
        return dp[n - 1][m - 1];
    }
}

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