LeetCode #3494 — MEDIUM

Find the Minimum Amount of Time to Brew Potions

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[1,5,2,4]
[5,1,4,2]

Example 2

[1,1,1]
[1,1,1]

Example 3

[1,2,3,4]
[1,2]

Step 02

Core Insight

What unlocks the optimal approach

Maintain each wizard's earliest free time (for the last potion) as <code>f[i]</code>.
Let <code>x</code> be the current mana value. Starting from <code>now = f[0]</code>, update <code>now = max(now + skill[i - 1] * x, f[i])</code> for <code>i in [1..n]</code>. Then, the final <code>f[n - 1] = now + skill[n - 1] * x</code> for this potion.
Update all other <code>f</code> values by <code>f[i] = f[i + 1] - skill[i + 1] * x</code> for <code>i in [0..n - 2]</code> (in reverse order).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3494: Find the Minimum Amount of Time to Brew Potions
class Solution {
    public long minTime(int[] skill, int[] mana) {
        int n = skill.length;
        long[] f = new long[n];
        for (int x : mana) {
            long tot = 0;
            for (int i = 0; i < n; ++i) {
                tot = Math.max(tot, f[i]) + skill[i] * x;
            }
            f[n - 1] = tot;
            for (int i = n - 2; i >= 0; --i) {
                f[i] = f[i + 1] - skill[i + 1] * x;
            }
        }
        return f[n - 1];
    }
}

// Accepted solution for LeetCode #3494: Find the Minimum Amount of Time to Brew Potions
func minTime(skill []int, mana []int) int64 {
	n := len(skill)
	f := make([]int64, n)
	for _, x := range mana {
		var tot int64
		for i := 0; i < n; i++ {
			tot = max(tot, f[i]) + int64(skill[i])*int64(x)
		}
		f[n-1] = tot
		for i := n - 2; i >= 0; i-- {
			f[i] = f[i+1] - int64(skill[i+1])*int64(x)
		}
	}
	return f[n-1]
}

# Accepted solution for LeetCode #3494: Find the Minimum Amount of Time to Brew Potions
class Solution:
    def minTime(self, skill: List[int], mana: List[int]) -> int:
        max = lambda a, b: a if a > b else b
        n = len(skill)
        f = [0] * n
        for x in mana:
            tot = 0
            for i in range(n):
                tot = max(tot, f[i]) + skill[i] * x
            f[-1] = tot
            for i in range(n - 2, -1, -1):
                f[i] = f[i + 1] - skill[i + 1] * x
        return f[-1]

// Accepted solution for LeetCode #3494: Find the Minimum Amount of Time to Brew Potions
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3494: Find the Minimum Amount of Time to Brew Potions
// class Solution {
//     public long minTime(int[] skill, int[] mana) {
//         int n = skill.length;
//         long[] f = new long[n];
//         for (int x : mana) {
//             long tot = 0;
//             for (int i = 0; i < n; ++i) {
//                 tot = Math.max(tot, f[i]) + skill[i] * x;
//             }
//             f[n - 1] = tot;
//             for (int i = n - 2; i >= 0; --i) {
//                 f[i] = f[i + 1] - skill[i + 1] * x;
//             }
//         }
//         return f[n - 1];
//     }
// }

// Accepted solution for LeetCode #3494: Find the Minimum Amount of Time to Brew Potions
function minTime(skill: number[], mana: number[]): number {
    const n = skill.length;
    const f: number[] = Array(n).fill(0);
    for (const x of mana) {
        let tot = 0;
        for (let i = 0; i < n; ++i) {
            tot = Math.max(tot, f[i]) + skill[i] * x;
        }
        f[n - 1] = tot;
        for (let i = n - 2; i >= 0; --i) {
            f[i] = f[i + 1] - skill[i + 1] * x;
        }
    }
    return f[n - 1];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.