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Leetcode 2335. Minimum Amount of Time to Fill Cups

Problem Statement

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 water, warm water, and hot water cups you need to fill respectively. You can fill exactly two different types of cups every second.

Return the minimum number of seconds needed to fill all the cups.

Clarifying Questions

  1. Is it always possible to fill exactly two cups of different types each second?
    • Yes, we are given that we can always fill two different types of cups each second.
  2. What if there are cups that can’t be paired with another type of cup?
    • If after pairing, there are cups left over, we have to fill one type of cup in each subsequent second.
  3. Are there specific constraints on the length of the array or the values it can contain?
    • Yes, the length of the array is fixed at 3, and the values can range from 0 to 100.

Strategy

The main idea is to minimize the number of seconds required based on the maximum number of cups for each type:

  1. Sort the Array: This will help in always dealing with the highest values to minimize the time needed.
  2. Check the Maximum Condition: If the maximum value (after sorting) is greater than or equal to the sum of the other two, it will dictate the time required.
  3. General Condition: If the maximum value is not greater than the sum of the other two, then using one second to fill two different types of cups until the cups run out.

Code

Here’s the Java implementation of the above strategy:

import java.util.Arrays;

public class MinimumAmountOfTimeToFillCups {
    public static int fillCups(int[] amount) {
        Arrays.sort(amount);
        
        if (amount[2] >= amount[0] + amount[1]) {
            return amount[2];
        } else {
            return (amount[0] + amount[1] + amount[2] + 1) / 2;
        }
    }

    public static void main(String[] args) {
        int[] amount1 = {1, 4, 2};
        System.out.println(fillCups(amount1));  // Output: 4

        int[] amount2 = {5, 4, 4};
        System.out.println(fillCups(amount2));  // Output: 7
    }
}

Time Complexity

The time complexity of this solution is O(1) since:

Thus, the overall complexity is constant time O(1).

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