1561. Maximum Number of Coins You Can Get.md

1561. Maximum Number of Coins You Can Get

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the second one in Weekly Contest 203.

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Difficulty : Medium

Related Topics : Sort

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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

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Solution

• mine
  • Java
    • Runtime: 26 ms, faster than 83.27%, Memory Usage: 48.9 MB, less than 88.69% of Java online submissions

      // O(N*logN)time
      // O(1)space
      public int maxCoins(int[] piles) {
          Arrays.sort(piles);
          int n = piles.length;
          int count = n / 3;
          int res = 0;
          for(int i = count; i < n; i+=2){
              res += piles[i];
          }
          return res;
      }
      

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