Leetcode simple question: the minimum cost of buying candy at a discount

subject

A shop is selling candy at a discount . Every time you buy Two candy , Shop meeting free Send a candy .
The only restriction on free candy is ： Its price needs to be less than or equal to the price of two candies purchased Smaller value .
For example , All in all 4 A candy , The prices are 1 ,2 ,3 and 4 , A customer bought it for 2 and 3 Of candy , Then he can get the price for free 1 Of candy , But you can't get a price of 4 Of candy .
I'll give you a subscript from 0 The starting array of integers cost , among cost[i] It means the first one i The price of a candy , Please return to get all Candy Minimum The total cost .
Example 1：
Input ：cost = [1,2,3]
Output ：5
explain ： Our purchase price is 2 and 3 Of candy , Then get the price for free 1 Of candy .
The total cost is 2 + 3 = 5 . This is the least expensive only programme .
Be careful , We can't buy at 1 and 3 Of candy , And get the price for free 2 Of candy .
This is because the price of free candy must be less than or equal to the purchase 2 The smaller value of the candy price .
Example 2：
Input ：cost = [6,5,7,9,2,2]
Output ：23
explain ： The minimum total cost of buying candy is ：

• The purchase price is 9 and 7 Of candy
• The free price is 6 Of candy
• The purchase price is 5 and 2 Of candy
• The free price is 2 The last candy of
  therefore , The minimum total cost is 9 + 7 + 5 + 2 = 23 .
  Example 3：
  Input ：cost = [5,5]
  Output ：10
  explain ： Because only 2 A candy , We need to buy them all , And there is no free candy .
  So the total minimum cost is 5 + 5 = 10 .
  Tips ：
  1 <= cost.length <= 100
  1 <= cost[i] <= 100
  source ： Power button （LeetCode）

Their thinking

   For a given cost, The maximum value and the second largest value will definitely not be given , Then if the third largest value exists, it will be given away , So you can cost Sort from large to small , Then take out the first three values for operation , The remaining subsequence is another round of operation .

class Solution:
    def minimumCost(self, cost: List[int]) -> int:
        s=0
        cost.sort(reverse=True)
        for i in range(len(cost)):
            if (i+1)%3!=0:  # The third largest value does not need to be included in the total price 
                s+=cost[i]
        return s