LeetCode 2335. Minimum Amount of Time to Fill Cups

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.

可以同时接2个不同类型的咖啡，也可以只接一种咖啡；

那么我们如何保证接的次数最少，就要同时接2种咖啡的次数最多即可；

所以每次都要去看剩下的是哪2种咖啡剩的最多，所以每接一次就要排一次序，

于是就想到了优先队列，只是排序的规则是从大到小，而不是从小到大，

所以需要重写一下compare的方法就行了；

下面是代码：

Runtime: 3 ms, faster than 35.27% of Java online submissions for Minimum Amount of Time to Fill Cups.

Memory Usage: 40.4 MB, less than 35.50% of Java online submissions for Minimum Amount of Time to Fill Cups.