LeetCode-in-Java

1561. Maximum Number of Coins You Can Get

Medium

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 the 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 that 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 <= 105
• piles.length % 3 == 0
• 1 <= piles[i] <= 104

Solution

import java.util.Arrays;

public class Solution {
    public int maxCoins(int[] piles) {
        Arrays.sort(piles);
        int j = 0;
        int coins = 0;
        for (int i = piles.length - 2; i > 0; i -= 2) {
            coins += piles[i];
            if (++j == piles.length / 3) {
                return coins;
            }
        }
        return coins;
    }
}

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