Minimum Amount of Time to Fill Cups LeetCode Solution

Problem – Minimum Amount of Time to Fill Cups LeetCode Solution

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

Minimum Amount of Time to Fill Cups LeetCode Solution in Java

    public int fillCups(int[] A) {
        int mx = 0, sum = 0;
        for(int a: A) {
            mx = Math.max(a, mx);
            sum += a;
        }
        return Math.max(mx, (sum + 1) / 2);
    }

Minimum Amount of Time to Fill Cups LeetCode Solution in C++

    int fillCups(vector<int>& A) {
        int mx = 0, sum = 0;
        for(int& a: A) {
            mx = max(a, mx);
            sum += a;
        }
        return max(mx, (sum + 1) / 2);
    }

Minimum Amount of Time to Fill Cups LeetCode Solution in Python

    def fillCups(self, A):
        return max(max(A), (sum(A) + 1) // 2)

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