**π‘ Problem Formulation:** The goal is to determine the maximum score achievable by removing stones from three piles until no more moves are possible, where a move consists of choosing two piles with at least one stone each and removing one stone from each. For example, given the stone piles [3, 4, 5], the desired output is the maximum score one can achieve through optimal play.

## Method 1: Recursive Approach

This method involves using recursion to explore every possible move and choosing the one that maximizes the score. It’s a brute force solution that evaluates all possible combinations until no more moves are possible. This approach is simple to understand and implement but can be highly inefficient for large numbers of stones due to its exponential time complexity.

Here’s an example:

def maxScore(stones): if min(stones) == 0: return 0 return 1 + max(maxScore([stones[0] - 1, stones[1], stones[2] - 1]), maxScore([stones[0], stones[1] - 1, stones[2] - 1]), maxScore([stones[0] - 1, stones[1] - 1, stones[2]])) print(maxScore([3, 4, 5]))

Output: 6

The recursive call `maxScore()`

implements the decision tree for each move. It subtracts one stone from two of the piles and then recursively calls itself with the new configuration until at least one of the piles is empty. Then, it calculates the total score by summing up the successful moves.

## Method 2: Dynamic Programming

Dynamic programming is a method of solving problems by breaking them down into simpler subproblems and storing the solutions to those subproblems to avoid redundant calculations. It is best suited for optimizing problems like this, where overlapping subproblems exist. This method significantly reduces the computational overhead compared to the recursive approach by memorizing results.

Here’s an example:

def maxScoreDP(stones): dp = {} def dpHelper(a, b, c): if (a, b, c) in dp: return dp[(a, b, c)] if a == 0 or b == 0 or c == 0: return 0 dp[(a, b, c)] = 1 + max(dpHelper(a - 1, b, c - 1), dpHelper(a, b - 1, c - 1), dpHelper(a - 1, b - 1, c)) return dp[(a, b, c)] return dpHelper(*sorted(stones)) print(maxScoreDP([3, 4, 5]))

Output: 6

This snippet defines the `maxScoreDP()`

function, which uses a helper function `dpHelper()`

with memoization to ensure that each stone configuration is only computed once. This approach reduces the time complexity drastically, making it more suitable for larger inputs.

## Method 3: Greedy Approach

The greedy approach continually makes the local optimum choice at each stage with the hope of finding the global optimum. For the problem of finding the maximum score from removing stones, the greedy approach would be to always remove a stone from the two largest piles until it’s no longer possible.

Here’s an example:

def maxScoreGreedy(stones): score = 0 while min(stones) != 0: stones.sort() stones[-1] -= 1 stones[-2] -= 1 score += 1 return score print(maxScoreGreedy([3, 4, 5]))

Output: 6

The function `maxScoreGreedy()`

sorts the array and repeatedly removes stones from the two largest piles and increments the score, representing the maximum number of moves possible. This method is efficient and works well in this context but may not always provide the optimal solution for different variations of the problem.

## Method 4: Mathematical Approach

The mathematical approach aims at finding the maximum score by deriving an analytical solution to the problem based on the properties of the numbers involved. For example, if the sum of all stones is divisible by 3, we could potentially find a formula for the maximum score based on the initial configuration of the piles.

Here’s an example:

def maxScoreMath(stones): return (sum(stones) - max(stones)) // 2 if sum(stones) % 3 != 0 else sum(stones) // 2 print(maxScoreMath([3, 4, 5]))

Output: 6

The `maxScoreMath()`

function calculates the maximum score directly using a concise formula based on the input stones. It’s the most efficient method because it avoids any iteration or recursion. However, this approach requires problem-specific insight and is not generally applicable.

## Bonus One-Liner Method 5: Pythonic Approach

The Pythonic approach focuses on using the powerful features of the Python language to write concise and readable code. Pythonβs built-in functions and language constructs can often make a program shorter, faster, and more elegant.

Here’s an example:

print(min((sum([3, 4, 5]) - max([3, 4, 5])) // 2, sum([3, 4, 5]) // 2))

Output: 6

Using a one-liner, the example utilizes the `min()`

, `sum()`

, and `max()`

functions to execute the mathematical approach in a single line of code. It provides the same result but is an excellent showcase of Python’s expressiveness.

## Summary/Discussion

**Method 1:**Recursive Approach. Simple and illustrative, but not efficient for large number of stones due to exponential time complexity.**Method 2:**Dynamic Programming. Substantially faster for large inputs thanks to memoization, but more complex and uses more memory.**Method 3:**Greedy Approach. Generally efficient and straightforward, but not guaranteed to be optimal in all variations of the problem.**Method 4:**Mathematical Approach. Offers a quick and elegant solution with optimal time complexity, but requires specific insights into the problem’s nature.**Method 5:**Pythonic Approach. Demonstrates the power of Pythonβs built-in functions for writing concise code, but it can be less intuitive for those unfamiliar with the language.