Introduction

In this blog post, we will explore how to solve the LeetCode problem "3494" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 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 1st potion before time t = 52, consider the case where the wizards started preparing the 1st potion at time t = 50. At time t = 58, wizard 2 is done with the 1st potion, but wizard 3 will still be working on the 0th potion till time t = 60. Example 2: Input: skill = [1,1,1], mana = [1,1,1] Output: 5 Explanation: Preparation of the 0th potion begins at time t = 0, and is completed by time t = 3. Preparation of the 1st potion begins at time t = 1, and is completed by time t = 4. Preparation of the 2nd 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

Explanation

Here's the solution:

• The problem requires minimizing the total time to brew all potions, where each potion goes through all wizards sequentially. The crucial constraint is synchronization: a wizard can only start a potion when the previous wizard is done.
• The core idea is to calculate the start time of each potion at each wizard. The start time for wizard i on potion j depends on when wizard i-1 finishes potion j and when wizard i finishes potion j-1. We take the maximum of these two times.

• The total time is simply the finish time of the last wizard on the last potion.

• Runtime Complexity: O(n*m), Storage Complexity: O(n*m)

Code

    def solve():
    n = int(input())
    skill = list(map(int, input().split()))
    m = int(input())
    mana = list(map(int, input().split()))

    dp = [[0] * m for _ in range(n)]

    # First wizard's start times
    dp[0][0] = skill[0] * mana[0]
    for j in range(1, m):
        dp[0][j] = dp[0][j-1] + skill[0] * mana[j]

    # Other wizards
    for i in range(1, n):
        dp[i][0] = dp[i-1][0] + skill[i] * mana[0]
        for j in range(1, m):
            dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + skill[i] * mana[j]

    print(dp[n-1][m-1])

# Example Usage (for local testing, comment out before submission)
# n = 4
# skill = [1, 5, 2, 4]
# m = 4
# mana = [5, 1, 4, 2]

# skill_str = ' '.join(map(str, skill))
# mana_str = ' '.join(map(str, mana))

# input_str = f"{n}\n{skill_str}\n{m}\n{mana_str}"

# with open("input.txt", "w") as f:
#     f.write(input_str)

# with open("input.txt", "r") as f:
#     lines = f.readlines()

# old_input = input
# input = lambda: lines.pop(0).strip()
# solve()
# input = old_input #restore input

def solve_optimized(skill, mana):
    n = len(skill)
    m = len(mana)

    dp = [[0] * m for _ in range(n)]

    # First wizard's start times
    dp[0][0] = skill[0] * mana[0]
    for j in range(1, m):
        dp[0][j] = dp[0][j-1] + skill[0] * mana[j]

    # Other wizards
    for i in range(1, n):
        dp[i][0] = dp[i-1][0] + skill[i] * mana[0]
        for j in range(1, m):
            dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + skill[i] * mana[j]

    return dp[n-1][m-1]