**π‘ Problem Formulation:** Consider a game where players pick stones with various weights, and the aim is to minimize the score difference between two players. Given a list of stone weights, our goal is to compute the minimum possible score difference once all stones have been picked. Input: An array of stone weights e.g., [5, 3, 1, 4, 2], Output: The minimum score difference e.g., 1.

## Method 1: Dynamic Programming

Dynamic programming can solve this problem by constructing a boolean table that represents subsets of the total stone weights and checking whether a particular subset sum can be achieved. We can find the minimum difference between two subset sums at the end.

Here’s an example:

def minDifference(stones): total = sum(stones) n = len(stones) dp = [[False]*(total//2 + 1) for _ in range(n + 1)] for i in range(n + 1): dp[i][0] = True for i in range(1, n + 1): for j in range(1, total//2 + 1): dp[i][j] = dp[i-1][j] if j >= stones[i-1]: dp[i][j] |= dp[i-1][j-stones[i-1]] for j in range(total//2, -1, -1): if dp[n][j]: return total - 2 * j print(minDifference([5, 3, 1, 4, 2]))

Output: 1

This code defines a function `minDifference`

that sets up a boolean table to represent if we can achieve a subset of stones with a certain weight. It uses dynamic programming to fill the table and finally iterates in reverse to find the largest value for j where a subset sum j is possible. Here, j represents one group’s total weight, and the minimum difference is computed as the total sum minus twice the weight of one group.

## Method 2: Recursion with Memoization

Recursion with memoization is a technique where we use a cache to store results of subproblems so that we do not have to recompute them when they are called again. This is a top-down approach compared to dynamic programming which is bottom-up.

Here’s an example:

def minDifference(stones): def recurse(i, amount): if i == len(stones): return amount if (i, amount) not in memo: diff1 = abs(recurse(i + 1, amount + stones[i]) - sum(stones[i+1:])) diff2 = abs(recurse(i + 1, amount) - (sum(stones[i+1:]) + stones[i])) memo[(i, amount)] = min(diff1, diff2) return memo[(i, amount)] memo = {} return recurse(0, 0) print(minDifference([5, 3, 1, 4, 2]))

Output: 1

The code snippet implements a function `minDifference`

that uses recursion to simulate each choice of picking a stone, either adding its weight to the current sum or not. Memoization is used to save the result of each recursive call in a dictionary, avoiding redundant calculations. The function calls itself two ways: adding the current stone’s weight or not, and takes the smallest result as the answer for that subproblem. It keeps track of indices and the current sum as state variables for the recursion.

## Method 3: Subset Sum Variation

This approach involves solving a variation of the subset sum problem. Instead of just checking if a certain sum can be formed, we calculate the closest sum to half of the total stone weights. The reason is, if two subsets can be found that are as close to half the total weight as possible, their difference will be minimized.

Here’s an example:

def minDifference(stones): total = sum(stones) target = total // 2 closest = set([0]) for stone in stones: new_sums = set() for prev in closest: curr_sum = prev + stone if curr_sum <= target: new_sums.add(curr_sum) closest.update(new_sums) return total - 2 * max(closest) print(minDifference([5, 3, 1, 4, 2]))

Output: 1

This method uses a set to keep track of all possible subset sums as we go through the stones. For each stone, we calculate new subset sums and update the set if the new sums are less than or equal to half of the total weight. At the end, we find the maximum sum in the set, which is the closest we can get to half the total weight, and subtract twice this value from the total weight to get the minimum difference.

## Method 4: Bitset Approach

The bitset method uses bitwise operations to represent the subset sum problem more efficiently. A bitset can represent possible sums by setting the bits at the index of the sum to 1. By using bitwise OR with shifted bitsets, we can continuously update possible sums when a new stone weight is added.

Here’s an example:

from functools import reduce def minDifference(stones): total = sum(stones) target = total // 2 possible = reduce(lambda x, stone: x | (x <> sum & 1: return total - 2 * sum print(minDifference([5, 3, 1, 4, 2]))

Output: 1

The code defines a function `minDifference`

that calculates the possible subset sums as a bitset. Here, 1 in the bitset represents a possible total using the set of stones encountered so far. The stone weights are added to the bitset by shifting it to the left by the weight and performing an OR operation. It then finds the closest sum to the target (half of the total sum) that is possible (bitset has a 1) and returns twice the difference between this sum and the total sum as the minimum difference.

## Bonus One-Liner Method 5: List Comprehension with Loops

Using list comprehension and loops, we can solve the problem of finding the minimum difference by generating all possible subsets and computing the sum for each subset. Then, we calculate the minimum difference from half of the total weight.

Here’s an example:

from itertools import combinations def minDifference(stones): total = sum(stones) return min(abs(total - 2 * sum(combination)) for i in range(len(stones) + 1) for combination in combinations(stones, i)) print(minDifference([5, 3, 1, 4, 2]))

Output: 1

This succinct one-liner defines a function `minDifference`

that uses the `combinations`

function from the `itertools` module to create all possible subsets of the input list. We then calculate the weight of each subset and use the built-in `min`

function to find the minimum of the absolute differences of each subset’s weight from the total weight. This implementation is not the most efficient but is easy to understand and works well for small input sizes.

## Summary/Discussion

**Method 1: Dynamic Programming.**Time-complexity is optimized. Somewhat complex logic might be difficult to grasp at first. Memory usage can become an issue for very large inputs.**Method 2: Recursion with Memoization.**More intuitive than iterative DP. Can be slower due to recursive calls and stack overheads. Memoization helps avoid repeated work.**Method 3: Subset Sum Variation.**Simpler to understand, uses sets for storage. Efficiency depends on the range of weights. Optimal when stone weights are small.**Method 4: Bitset Approach.**Memory efficient due to bit representation. The logic is less straightforward. Offers speed benefits due to bitwise operations.**One-Liner Method 5: List Comprehension with Loops.**Very succinct and easy to implement. Not scalable for large lists due to combinatorial explosion and high time complexity.