1561. Maximum Number of Coins You Can Get
There are 3n piles of coins of varying size, you and your friends will take piles of coins as follows:
- In each step, you will choose **any **3 piles of coins (not necessarily consecutive).
- Of your choice, Alice will pick the pile with the maximum number of coins.
- You will pick the next pile with maximum number of coins.
- Your friend Bob will pick the last pile.
- Repeat until there are no more piles of coins.
Given an array of integers piles
where piles[i]
is the number of coins in the ith
pile.
Return the maximum number of coins which you can have.
Example 1:
Input: piles = [2,4,1,2,7,8]
Output: 9
Explanation: Choose the triplet (2, 7, 8), Alice Pick the pile with 8 coins, you the pile with **7** coins and Bob the last one.
Choose the triplet (1, 2, 4), Alice Pick the pile with 4 coins, you the pile with **2** coins and Bob the last one.
The maximum number of coins which you can have are: 7 + 2 = 9.
On the other hand if we choose this arrangement (1, **2** , 8), (2, **4** , 7) you only get 2 + 4 = 6 coins which is not optimal.
Example 2:
Input: piles = [2,4,5]
Output: 4
Example 3:
Input: piles = [9,8,7,6,5,1,2,3,4]
Output: 18
Constraints:
3 <= piles.length <= 10^5
piles.length % 3 == 0
1 <= piles[i] <= 10^4
# @lc code=start
using LeetCode
function max_coins(piles::Vector{Int})
len = length(sort!(piles))
return sum(piles[i] for i in len ÷ 3 + 1 : 2 : len - 1)
end
# @lc code=end
max_coins (generic function with 1 method)
This page was generated using DemoCards.jl and Literate.jl.