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Suppose there are 3*n number of piles of coins present, and they are of varying size, three players are playing a game like −

In each step, player1 will select any 3 piles of coins.

Of his choice, Player2 will pick the pile with the maximum number of coins.

Player1 will pick the next pile with maximum number of coins.

Player3 will pick the last pile.

Repeat these steps until there are no more piles of coins.

Now if we have an array of integers called piles where piles[i] is the number of coins in the ith pile, then we have to find the maximum number of coins which Player1 can have.

So, if the input is like piles = [2,4,1,2,7,8], then the output will be 9 because at first we can select a triplet (2,7,8), then Player2 selects 8, Player1 selects 7 and the 2 is for Player3. Then again select triplet (1,2,4), then Player2 selects pile with coin 4, then Player1 selects 2 and remaining 1 for Player3. Currently Player1 has 7+2 = 9 coins, this is the maximum.

To solve this, we will follow these steps −

sort the list piles

ans := 0

while size of piles is not same as 0, do

ans := ans + second last element from piles

delete second last element from piles

delete last element from piles

delete first element from piles

return ans

Let us see the following implementation to get better understanding −

def solve(piles): piles.sort() ans = 0 while(len(piles)!=0): ans = ans + piles[-2] del piles[-2] del piles[-1] del piles[0] return ans piles = [2,4,1,2,7,8] print(solve(piles))

[2,4,1,2,7,8]

9

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