Leetcode simple question: the minimum total time required to fill a cup

subject

There is a water dispenser , Cold water can be prepared 、 Warm and hot water . Every second , Can be filled 2 A cup of Different Type of water or 1 Cup any type of water .

I'll give you a subscript from 0 Start 、 The length is 3 Array of integers for amount , among amount[0]、amount[1] and amount[2] Respectively means that it needs to be filled with cold water 、 Number of cups of warm and hot water . Return to fill all cups least Number of seconds .

Example 1：

Input ：amount = [1,4,2]
Output ：4
explain ： Here is a scheme ：
The first 1 second ： Fill a cup of cold water and a cup of warm water .
The first 2 second ： Fill a cup of warm water and a cup of hot water .
The first 3 second ： Fill a cup of warm water and a cup of hot water .
The first 4 second ： Fill a glass of warm water .
It can be proved that at least 4 Seconds to fill all the cups .
Example 2：

Input ：amount = [5,4,4]
Output ：7
explain ： Here is a scheme ：
The first 1 second ： Fill a cup of cold water and a cup of hot water .
The first 2 second ： Fill a cup of cold water and a cup of warm water .
The first 3 second ： Fill a cup of cold water and a cup of warm water .
The first 4 second ： Fill a cup of warm water and a cup of hot water .
The first 5 second ： Fill a cup of cold water and a cup of hot water .
The first 6 second ： Fill a cup of cold water and a cup of warm water .
The first 7 second ： Fill a glass of hot water .
Example 3：

Input ：amount = [5,0,0]
Output ：5
explain ： Fill a glass of cold water every second .

Tips ：

amount.length == 3
0 <= amount[i] <= 100

source ： Power button （LeetCode）

Their thinking

   For three kinds of water , We just need to reduce the two remaining largest water cups by one each time , Until the number of all water cups is less than 1.

class Solution:
    def fillCups(self, amount: List[int]) -> int:
        count=0
        while amount[0]>0 or amount[1]>0 or amount[2]>0 :
            amount.sort()
            amount[1]-=1
            amount[2]-=1
            count+=1
        return count