**π‘ Problem Formulation:** Suppose you are given an array `coins[]`

, which represents piles of coins. In each turn, you can take coins from the pile but only the pile with the maximum number of coins. The task is to find the maximum number of coins you can get if you can take any number of turns. Given `coins = [2, 4, 5, 8, 3]`

, the desired output is the maximum coins that can be obtained, which would be `19`

(8+5+4+2).

## Method 1: Sorting and Greedy Approach

This method involves sorting the array in descending order and then iteratively picking the largest pile until all piles are exhausted. The function signature would typically be `def max_coins(coins: List[int]) -> int`

.

Here’s an example:

def max_coins(coins): coins.sort(reverse=True) total_coins = 0 for pile in coins: total_coins += pile return total_coins # Example usage: print(max_coins([2, 4, 5, 8, 3]))

Output: 22

The example sorts the list of coins in descending order and then iterates through the piles, adding the values to a total. The final result is the sum of all the piles, which is the maximum number of coins one could get.

## Method 2: Max Heap

This method uses a Max Heap to always have the largest pile at the top. By popping from the heap and accumulating the values, we can efficiently find the maximum number of coins. The Python `heapq`

module can be utilized here in a reverse manner to create a Max Heap.

Here’s an example:

import heapq def max_coins(coins): max_heap = [-x for x in coins] # Creating a Max Heap by inserting negated values heapq.heapify(max_heap) total_coins = 0 while max_heap: total_coins -= heapq.heappop(max_heap) # Since we negated the values, subtract to add the positive value return total_coins # Example usage: print(max_coins([2, 4, 5, 8, 3]))

Output: 22

The code snippet uses the `heapq`

module to create a Max Heap structure, then pops the largest element (in negated form) and adds it to the total count of coins until the heap is empty.

## Method 3: Dynamic Programming

Dynamic Programming can solve this problem by finding the optimal solution at each step and building up to the final solution. However, considering the problem does not have overlapping subproblems or optimal substructure typically needed for DP, this might not be the best fit.

Here’s an example:

def max_coins(coins): # In this specific case, DP does not provide a more efficient solution # Hence, this function would be similar to Method 1 return sum(coins) # Example usage: print(max_coins([2, 4, 5, 8, 3]))

Output: 22

This example simplifies the dynamic programming approach since the problem itself does not benefit from DP and adopts a simple summing strategy similar to Method 1.

## Method 4: Recursion with Memoization

If the problem involved additional constraints, like limitations on the number of turns or choosing among non-adjacent piles, recursion with memoization could be used to explore different combinations and keep track of the maximum coins obtained.

Here’s an example:

def max_coins(coins): # Again, for this problem, recursion does not provide additional advantages # This function mirrors the Method 1's logic with unnecessary complexity return sum(coins) # Example usage: print(max_coins([2, 4, 5, 8, 3]))

Output: 22

This snippet demonstrates that recursion with memoization is an overcomplication for this problem as the sought value is simply the total sum of the coins.

## Bonus One-Liner Method 5: Using Python’s Built-In Functions

A much simpler approach making use of Python’s built-in functions might work best here; `max_coins`

can simply return the sum of the array using the `sum()`

function.

Here’s an example:

max_coins = lambda coins: sum(coins) # Example usage: print(max_coins([2, 4, 5, 8, 3]))

Output: 22

The one-liner uses a lambda function to directly apply the `sum()`

function to the array of coins, providing the total coin count in the most straightforward manner.

## Summary/Discussion

**Method 1: Sorting and Greedy Approach.**Simple and intuitive. Can be inefficient for large arrays due to sorting.**Method 2: Max Heap.**Efficient for repeated queries. Overkill for a single total calculation and introduces complexity.**Method 3: Dynamic Programming.**Generally powerful for many problems. Not useful in this specific case since there are no overlapping subproblems.**Method 4: Recursion with Memoization.**Good for problems with constraints and branching choices. Unnecessary for straightforward summation.**Bonus One-Liner Method 5: Using Python’s Built-In Functions.**Extremely simple and efficient for this problem. May lack flexibility for more complex scenarios.