Maximum Number of Coins You Can Get

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

This C++ solution addresses the problem by calculating the maximum number of coins one can obtain, given a set of piles where each pile contains a number of coins. The algorithm works as follows:

• First, sort the given vector of coins in ascending order to ensure that the selected piles are maximizing the coins count.
• Initialize a counter for the total number of coins collected.
• Start iterating through the sorted piles from a third of the way through the list (since two other players pick coins before your turn), and select every second pile thereafter to maximize your share of the coins.
• For each selected pile in this subset, add the number of coins to your total.
• Finally, return the cumulative total coins collected.

This method ensures you always take the most valuable coin piles by skipping over the least valuable third (picked by others) and then picking two out of every three piles. The sorted array allows for picking maximum values consistently.

class Solution {
    public int maximumCoins(int[] pile) {
        Arrays.sort(pile);
        int total = 0;
        
        for (int j = pile.length / 3; j < pile.length; j += 2) {
            total += pile[j];
        }
        
        return total;
    }
}

This Java solution addresses the problem of maximizing the number of coins you can get. The strategy involves sorting the coin piles and then iterating through them to maximize gains. Here's how the solution works:

• First, sort the array of coins. This allows you to directly access the largest piles efficiently.
• Initialize a total counter to accumulate the number of coins.
• Use a loop starting from one-third the length of the array, continuing to the end, incrementing by two. This specific starting point and increment are chosen to maximize your coin count based on the problem's rules (picking coins from piles strategically).
• In each iteration, add the coin count from the selected pile to your total.
• Finally, return the total count of coins collected.

The solution leverages sorting for easy access to the largest elements and arithmetic operations to select the optimal piles as per the conditions described in the problem, ensuring an efficient and straightforward approach to maximizing the coin count.

class Solution:
    def maximumCoins(self, piles: List[int]) -> int:
        piles.sort()
        total_coins = 0
        
        for index in range(len(piles) // 3, len(piles), 2):
            total_coins += piles[index]
        
        return total_coins

The Python3 solution provided focuses on maximizing the number of coins one can get from a list of coin piles. The algorithm follows a simple yet effective logic outlined below:

• First, sort the list of coin piles in increasing order. This arrangement helps in managing the distribution strategy efficiently.
• Initialize a variable to store the total number of coins collected.
• Iterate through the sorted list starting from one-third of its length till the end. By selecting every second pile for this segment, it ensures that you're picking up piles while leaving the smallest ones, hence maximizing the coins collected.
• During each iteration, add the value of the current pile to the total coins collected.
• Finally, the function returns the total coins collected.

This approach ensures that by avoiding the smallest piles and focusing on the larger ones in the sorted list, one maximizes the total number of coins collected.