# Solution For Maximum Value Of K Coins From Piles

Problem Statement:

You are given an array piles, where piles[i] represents the number of coins in the ith pile. You will be playing a game where you repeatedly perform the following operation until there’s only one pile remaining:

• Choose any 3 piles from the piles available
• Pick one coin from each of the chosen piles and give it to a pile of your choice

Return the maximum number of coins you can get by playing the above game optimally.

Example:

Input: piles = [2,4,1,5,6,7,8,9,3] Output: 25
Explanation: Choose the triplets (2,4,1), (5,6,7), and (8,9,3).
Transfer the last coin from 2 to 5, from 4 to 6, and from 1 to 7.
After this, the piles are [1, 5, 8, 9, 3, 8, 9, 8], and the total number of coins remaining is 51.
You can then repeatedly use the first operation to obtain a maximum number of coins of 25.

Solution:

To solve this problem, we need to find an optimal way to choose the piles for each turn. Intuitively, we want to choose the piles with the maximum number of coins. However, there is also the consideration that we need to transfer one coin from each pile to another pile during each turn. So, choosing the piles with the maximum number of coins may not always result in the maximum number of coins at the end.

To solve this problem optimally, we need to use the following steps:

• Sort the piles in descending order.
• Keep track of the maximum number of coins we can get by choosing a particular number of piles.
• Iterate over all possible numbers of piles we can choose, starting from 3.
• For each number of piles, iterate over all possible combinations of piles.
• For each combination of piles, calculate the maximum number of coins we can get by choosing these piles and transferring the coins optimally.
• Update the maximum number of coins we can get if we have chosen this number of piles.
• Return the maximum number of coins we can get.

To implement the above steps, we can use the following algorithm:

• Sort the piles in descending order.
• Initialize a list dp of length n, where n is the length of piles.
• For i in range(n):
• Initialize a list temp of length i+1.
• For j in range(i-1, -1, -1):
• Iterate over all possible combinations of piles of length j+1 that include piles[i], and calculate the maximum number of coins we can get by choosing these piles and transferring the coins optimally.
• Update temp[j] if the maximum number of coins we can get is greater than the maximum number of coins we have seen so far for a combination of j+1 piles.
• Update dp[j] with the maximum number of coins we can get for a combination of j+1 piles that includes piles[i].
• Return the maximum value in dp.

The time complexity of this algorithm is O(n^4), where n is the length of piles. However, we can optimize this algorithm by using memorization to avoid recomputing the same values multiple times. We can create a dictionary to store the maximum number of coins we can get for a combination of piles. The key for each entry in the dictionary can be a tuple of the indices of the chosen piles. We can update the dictionary with the maximum number of coins we can get for each combination of piles we calculate during the iteration. This optimization reduces the time complexity of the algorithm to O(n^3).

The optimized algorithm is as follows:

• Sort the piles in descending order.
• Initialize a dictionary memo to store the maximum number of coins we can get for a combination of piles.
• For i in range(n):
• Initialize a list temp of length i+1.
• For j in range(i-1, -1, -1):
• Iterate over all possible combinations of piles of length j+1 that include piles[i], and calculate the maximum number of coins we can get by choosing these piles and transferring the coins optimally.
• Update temp[j] if the maximum number of coins we can get is greater than the maximum number of coins we have seen so far for a combination of j+1 piles.
• Update memo with the maximum number of coins we can get for this combination of piles.
• Update dp[j] with the maximum number of coins we can get for a combination of j+1 piles that includes piles[i].
• Return the maximum value in memo.

The implementation of the optimized algorithm in Python is as follows:

def maxCoins(piles: List[int]) -> int:
piles.sort(reverse=True)
n = len(piles)
memo = {}

``````def helper(i, j):
if i == j:
return 0
if (i, j) in memo:
return memo[(i, j)]
res = 0
for k in range(i, j):
res = max(res, piles[k] + helper(i, k-1) + helper(k+1, j))
memo[(i, j)] = res
return res

return helper(0, n-1)
``````

This algorithm takes O(n^3) time and O(n^2) space, where n is the length of piles.

## Step by Step Implementation For Maximum Value Of K Coins From Piles

```class Solution {
public int maxValueAfterKNodes(int[] piles, int k) {
// max value we can get by choosing k nodes from the given piles
int n = piles.length;
int[][] dp = new int[n+1][k+1];

// fill dp table in bottom up manner
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= Math.min(i, k); j++) {
// if there are no piles or no nodes to be selected
if (i == 0 || j == 0) {
dp[i][j] = 0;
}
// if there is only one pile to be selected
else if (i == 1) {
dp[i][j] = piles[0];
}
// after the first move, we can only choose from i - 2 piles
else if (j == 1) {
dp[i][j] = Math.max(piles[i-2], piles[i-1]);
}
// otherwise, we can choose any pile to remove nodes from
else {
dp[i][j] = Math.max(piles[i-2] + dp[i-2][j-1],
piles[i-1] + dp[i-1][j-1]);
}
}
}
// return the max value we can get after removing k nodes from the given piles
return dp[n][k];
}
}```
```def maximum_value_of_k_coins_from_piles(piles, k):
# Base case
if k == 0:
return 0

# If there are only k piles, then the player can only
# pick from those k piles
if len(piles) == k:
return sum(piles)

# If there are more than k piles, then the player can
# pick from any of the piles
return max(piles[0] + maximum_value_of_k_coins_from_piles(piles[1:], k - 1),
maximum_value_of_k_coins_from_piles(piles[1:], k))```
```var kCoins = function(piles, k) {
// create a max heap
let heap = new MaxHeap();

// insert all values from piles array into the heap
for (let i = 0; i < piles.length; i++) {
heap.insert(piles[i]);
}

// remove the top k values from the heap (these will be the largest values)
let maxSum = 0;
for (let i = 0; i < k; i++) {
maxSum += heap.removeMax();
}

return maxSum;
};```
```There are N piles of coins, each containing A[i] coins.

Now, we play a game where we take turns picking up coins from the piles. In each turn, we can pick up 1, 2, or 3 coins from the first pile, and we are not allowed to pick up any more coins from any other pile.

What is the maximum number of coins we can get?

Solution:

#include
#include
#include
#include

using namespace std;

int main()
{
int N;
cin >> N;

vector A(N);
for (int i = 0; i < N; i++) {
cin >> A[i];
}

int ans = 0;

// We can always pick up the coins from the first pile.
ans += A[0];

// We can pick up the coins from the second pile if the number of
// coins in it is greater than or equal to the number of coins in
// the first pile.
if (A[1] >= A[0]) {
ans += A[1];
}

// We can pick up the coins from the third pile if the number of
// coins in it is greater than or equal to the number of coins in
// the second pile.
if (A[2] >= A[1]) {
ans += A[2];
}

cout << ans << endl;

return 0;
}```
```using System;

public class Solution {

// Function to find the maximum value of k coins

// from piles of coins

static int MaximumValue(int k, int[] piles)

{

// Base case

if (k == 0)

return 0;

if (k == 1)

return piles[0];

// Initialize result

int result = 0;

// Consider every coin as the first coin

// and find the maximum value if we

// start picking coins from it

for (int i = 0; i < piles.Length; i++)

{

// Find maximum value if we start

// picking coins from i'th coin

int current = piles[i] + MaximumValue(k - 1, piles);

// Update result if current value

// is greater than result

result = Math.Max(result, current);

}

return result;

}

// Driver Code

public static void Main()

{

int k = 3;

int[] piles = { 2, 4, 5, 3, 2 };

Console.WriteLine(MaximumValue(k, piles));

}

}

// This code is contributed by vt_m```

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