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Formatted question description: https://leetcode.ca/all/1561.html

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 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

Related Topics:
Sort

Solution 1.

  • class Solution {
        public int maxCoins(int[] piles) {
            Arrays.sort(piles);
            int length = piles.length;
            int rounds = length / 3;
            int maxCoins = 0;
            int index = length - 2;
            for (int i = 0; i < rounds; i++) {
                maxCoins += piles[index];
                index -= 2;
            }
            return maxCoins;
        }
    }
    
    ############
    
    class Solution {
        public int maxCoins(int[] piles) {
            Arrays.sort(piles);
            int ans = 0;
            for (int i = piles.length - 2; i >= piles.length / 3; i -= 2) {
                ans += piles[i];
            }
            return ans;
        }
    }
    
  • // OJ: https://leetcode.com/problems/maximum-number-of-coins-you-can-get/
    // Time: O(NlogN)
    // Space: O(1)
    class Solution {
    public:
        int maxCoins(vector<int>& A) {
            sort(begin(A), end(A));
            int i = 0, j = A.size() - 1, ans = 0;
            while (i < j) {
                ++i;
                --j;
                ans += A[j--];
            }
            return ans;
        }
    };
    
  • class Solution:
        def maxCoins(self, piles: List[int]) -> int:
            piles.sort()
            return sum(piles[-2 : len(piles) // 3 - 1 : -2])
    
    
    
  • func maxCoins(piles []int) int {
    	sort.Ints(piles)
    	ans, n := 0, len(piles)
    	for i := n - 2; i >= n/3; i -= 2 {
    		ans += piles[i]
    	}
    	return ans
    }
    
  • function maxCoins(piles: number[]): number {
        piles.sort((a, b) => a - b);
        const n = piles.length;
        let ans = 0;
        for (let i = 1; i <= Math.floor(n / 3); i++) {
            ans += piles[n - 2 * i];
        }
        return ans;
    }
    
    
  • impl Solution {
        pub fn max_coins(mut piles: Vec<i32>) -> i32 {
            piles.sort();
            let n = piles.len();
            let mut ans = 0;
            for i in 1..=n / 3 {
                ans += piles[n - 2 * i];
            }
            ans
        }
    }
    
    

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