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1561. Maximum Number of Coins You Can Get

Description

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

Solutions

Solution 1

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class Solution:
    def maxCoins(self, piles: List[int]) -> int:
        piles.sort()
        return sum(piles[-2 : len(piles) // 3 - 1 : -2])

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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;
    }
}

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class Solution {
public:
    int maxCoins(vector<int>& piles) {
        sort(piles.begin(), piles.end());
        int ans = 0;
        for (int i = piles.size() - 2; i >= (int) piles.size() / 3; i -= 2) ans += piles[i];
        return ans;
    }
};

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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
}

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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;
}

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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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int cmp(const void* a, const void* b) {
    return *(int*) a - *(int*) b;
}

int maxCoins(int* piles, int pilesSize) {
    qsort(piles, pilesSize, sizeof(int), cmp);
    int ans = 0;
    for (int i = 1; i <= pilesSize / 3; i++) {
        ans += piles[pilesSize - 2 * i];
    };
    return ans;
}

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