LeetCode 1561. The maximum number of coins you can get

1561. The maximum number of coins you can get

Yes 3n A pile of coins in different numbers , You and your friends are going to divide coins in the following way ：

• In every round , You will choose arbitrarily 3 Pile coins （ Not necessarily continuous ）.
• Alice The pile with the largest number of coins will be taken away .
• You will take away the pile with the second largest number of coins .
• Bob Will take the last pile .
• Repeat the process , Until there are no more coins .

Give you an array of integers piles , among piles[i] It's No i The number of coins in the pile .

Returns the maximum number of coins you can get .

Example 1：

 Input ：piles = [2,4,1,2,7,8]
 Output ：9
 explain ： elect  (2, 7, 8) ,Alice  Take away  8  The pile of coins , You take it  7  The pile of coins ,Bob  Take the last pile .
 elect  (1, 2, 4) , Alice  Take away  4  The pile of coins , You take it  2  The pile of coins ,Bob  Take the last pile .
 The maximum number of coins you can get ：7 + 2 = 9.
 Consider the other case , If you choose  (1, 2, 8)  and  (2, 4, 7) , You can only get  2 + 4 = 6  Coin , This is not the optimal solution .

Example 2：

 Input ：piles = [2,4,5]
 Output ：4

Example 3：

 Input ：piles = [9,8,7,6,5,1,2,3,4]
 Output ：18

Tips ：

• 3 <= piles.length <= 10^5
• piles.length % 3 == 0
• 1 <= piles[i] <= 10^4

Two 、 Method 1

greedy , By sorting , Ahead 1/3 All are Bob Of , We reverse the order , Take the second largest pile

class Solution {
    
    public int maxCoins(int[] piles) {
    
        Arrays.sort(piles);
        int round = piles.length / 3;
        int idx = piles.length - 2;
        int res = 0;
        for (int i = 0; i < round; i++) {
    
            res += piles[idx];
            idx -= 2;
        }
        return res;
    }
}

Complexity analysis

• Time complexity ：O(nlogn), among nn Is the length of the array divided by 3（ That is, the length of the array is 3n）. The time complexity of sorting is O(3nlog3n)=O(3n(log3+logn))=O(nlogn), The time complexity of traversing the array to calculate the maximum number of coins is O(n), So the total time complexity is O(nlogn).

• Spatial complexity ：O(logn).