Leetcode problem 89 - 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 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

/**
 * @param {number[]} piles
 * @return {number}
 */
var maxCoins = function (piles) {
  // Initialize the variable 'result' to store the final sum of selected piles.
  let result = 0;

  // Sort the 'piles' array in ascending order.
  piles.sort((a, b) => a - b);

  // Iterate over every second element starting from one-third of the length of 'piles'.
  for (let i = piles.length / 3; i < piles.length; i += 2) {
    // Add the value of the current pile to the 'result'.
    result += piles[i];
  }

  // Return the final sum of selected piles.
  return result;
};

The time complexity of this solution is O(n log n) due to the sorting of the ‘piles’ array. The space complexity is O(1) since the sorting is done in-place.