3794. Find The Minimum Amount Of Time To Brew Potions¶

Difficulty: Medium

3794. Find the Minimum Amount of Time to Brew Potions

Medium

You are given two integer arrays, skill and mana, of length n and m, respectively.

In a laboratory, n wizards must brew m potions in order. Each potion has a mana capacity mana[j] and must pass through all the wizards sequentially to be brewed properly. The time taken by the i^th wizard on the j^th potion is time_ij = skill[i] * mana[j].

Since the brewing process is delicate, a potion must be passed to the next wizard immediately after the current wizard completes their work. This means the timing must be synchronized so that each wizard begins working on a potion exactly when it arrives.

Return the minimum amount of time required for the potions to be brewed properly.

Example 1:

Input: skill = [1,5,2,4], mana = [5,1,4,2]

Output: 110

Explanation:

Potion Number	Start time	Wizard 0 done by	Wizard 1 done by	Wizard 2 done by	Wizard 3 done by
0	0	5	30	40	60
1	52	53	58	60	64
2	54	58	78	86	102
3	86	88	98	102	110

As an example for why wizard 0 cannot start working on the 1^st potion before time t = 52, consider the case where the wizards started preparing the 1^st potion at time t = 50. At time t = 58, wizard 2 is done with the 1^st potion, but wizard 3 will still be working on the 0^th potion till time t = 60.

Example 2:

Input: skill = [1,1,1], mana = [1,1,1]

Output: 5

Explanation:

Preparation of the 0^th potion begins at time t = 0, and is completed by time t = 3.
Preparation of the 1^st potion begins at time t = 1, and is completed by time t = 4.
Preparation of the 2^nd potion begins at time t = 2, and is completed by time t = 5.

Example 3:

Input: skill = [1,2,3,4], mana = [1,2]

Output: 21

Constraints:

n == skill.length
m == mana.length
1 <= n, m <= 5000
1 <= mana[i], skill[i] <= 5000

Solution¶

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

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here