LeetCode in Kotlin

3494. 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 ith wizard on the jth potion is timeij = 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:

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

import kotlin.math.max

class Solution {
    fun minTime(skill: IntArray, mana: IntArray): Long {
        val endTime = LongArray(skill.size)
        endTime.fill(0)
        for (k in mana) {
            var t: Long = 0
            var maxDiff: Long = 0
            for (j in skill.indices) {
                maxDiff = max(maxDiff, endTime[j] - t)
                t += skill[j].toLong() * k.toLong()
                endTime[j] = t
            }
            for (j in skill.indices) {
                endTime[j] += maxDiff
            }
        }
        return endTime[endTime.size - 1]
    }
}

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