Introduction

In this blog post, we will explore how to solve the LeetCode problem "2335" 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 have a water dispenser that can dispense cold, warm, and hot water. Every second, you can either fill up 2 cups with different types of water, or 1 cup of any type of water. You are given a 0-indexed integer array amount of length 3 where amount[0], amount[1], and amount[2] denote the number of cold, warm, and hot water cups you need to fill respectively. Return the minimum number of seconds needed to fill up all the cups. Example 1: Input: amount = [1,4,2] Output: 4 Explanation: One way to fill up the cups is: Second 1: Fill up a cold cup and a warm cup. Second 2: Fill up a warm cup and a hot cup. Second 3: Fill up a warm cup and a hot cup. Second 4: Fill up a warm cup. It can be proven that 4 is the minimum number of seconds needed. Example 2: Input: amount = [5,4,4] Output: 7 Explanation: One way to fill up the cups is: Second 1: Fill up a cold cup, and a hot cup. Second 2: Fill up a cold cup, and a warm cup. Second 3: Fill up a cold cup, and a warm cup. Second 4: Fill up a warm cup, and a hot cup. Second 5: Fill up a cold cup, and a hot cup. Second 6: Fill up a cold cup, and a warm cup. Second 7: Fill up a hot cup. Example 3: Input: amount = [5,0,0] Output: 5 Explanation: Every second, we fill up a cold cup. Constraints: amount.length == 3 0 <= amount[i] <= 100

Explanation

Here's the solution:

• Key Idea: The minimum number of seconds is determined by the largest value in amount array, or the sum of the two largest value. If the largest value is greater than the sum of other two values, we need at least the number of seconds equivalent to the largest value. However, if we can always fill two cups at the same time, the sum of all cups divided by 2 gives us the lower bound. Therefore, we take the maximum of the largest value and the average time.

• Optimization: We sort the amount array and apply a simple condition to calculate the answer in O(1) time.

• Complexity: Time: O(1), Space: O(1)

Code

    class Solution:
    def fillCups(self, amount: list[int]) -> int:
        amount.sort()
        a, b, c = amount[0], amount[1], amount[2]
        return max(c, (a + b + c + 1) // 2)