Description Link to heading

[2335. Minimum Amount of Time to Fill Cups (Easy)]

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

Solution Link to heading

If max >= mid + min, res = max;

Ifmax <mid + min, res = (max + mid - min) / 2.

Code Link to heading 

class Solution {
public:
    int fillCups(vector<int>& amount) {
        int amax = amount[0], amin = amount[0];
        int asum = amount[0];
        for (int i = 1; i < amount.size(); i++) {
            amax = max(amax, amount[i]);
            amin = min(amin, amount[i]);
            asum += amount[i];
        }
        int amid = asum - amax - amin;
        if (amax >= amin + amid)
            return amax;
        return (asum + 1) / 2;
    }
};