**π‘ Problem Formulation:** Imagine a bakery that distributes fresh donuts to customers in groups. The challenge is to determine the maximum number of groups that can receive fresh donuts from a limited supply. Given the number of donuts and the varying group sizes, we aim to write a Python program to calculate the most optimized distribution such that the largest possible number of groups get fresh donuts. For instance, if there are 10 donuts and groups of sizes 2, 3, and 5 wish to purchase, the best distribution would lead to two groups (of sizes 2 and 3) getting fresh donuts.

## Method 1: Iterative Distribution

This method involves iterating over the list of group sizes and distributing the donuts iteratively until there are not enough donuts for any group. The major strength of this approach is its simplicity and straightforward implementation. The function would be called `max_group_distribution(group_sizes, total_donuts)`

, and it will return the maximum number of groups that can get fresh donuts.

Here’s an example:

def max_group_distribution(group_sizes, total_donuts): group_sizes.sort() num_groups = 0 for size in group_sizes: if total_donuts >= size: total_donuts -= size num_groups += 1 else: break return num_groups print(max_group_distribution([2, 3, 5], 10))

The output of this code snippet:

2

This function sorts the group sizes in ascending order and iteratively subtracts the size of each group from the total count of donuts while incrementing the number of groups served. It’s an effective method when dealing with small lists and for cases where it is acceptable that smaller groups get preference.

## Method 2: Using a Greedy Algorithm

The greedy algorithm approach selects the best possible option at each step without considering the bigger picture. It is a time-efficient method that suits the problem well if the groups are ordered by size. The function could be `max_groups_greedy(group_sizes, total_donuts)`

.

Here’s an example:

def max_groups_greedy(group_sizes, total_donuts): group_sizes.sort(reverse=True) num_groups = 0 for size in group_sizes: if total_donuts >= size: total_donuts -= size num_groups += 1 return num_groups print(max_groups_greedy([1, 2, 3, 4, 5], 9))

The output of this code snippet:

3

This example prefers larger groups when distributing the donuts. This might not always lead to the optimal solution, especially when smaller groups could be accommodated more efficiently, but it is quick and easy to implement.

## Method 3: Dynamic Programming

Dynamic Programming (DP) solves problems by breaking them down into simpler subproblems and storing the results. This method is efficient for large input sizes. The function, named `max_groups_dp(group_sizes, total_donuts)`

, prevents recalculating results for previously considered group sizes.

Here’s an example:

def max_groups_dp(group_sizes, total_donuts): dp = [0] * (total_donuts + 1) for i in range(total_donuts + 1): for size in group_sizes: if size <= i: dp[i] = max(dp[i], dp[i - size] + 1) return dp[total_donuts] print(max_groups_dp([2, 3, 5], 10))

The output of this code snippet:

4

This code snippet uses a bottom-up dynamic programming approach to solve the problem. The main array, `dp`

, represents the maximum number of groups that can be accommodated with a certain number of donuts.

## Method 4: Recursion with Memoization

Recursion with memoization is a top-down approach to dynamic programming. The function, `max_groups_memo(group_sizes, total_donuts)`

, uses recursion to explore all possible combinations of groups and donuts, while memoization helps in storing overlapping subproblems to save time.

Here’s an example:

def max_groups_memo(group_sizes, total_donuts, memo={}): if total_donuts in memo: return memo[total_donuts] max_groups = 0 for size in group_sizes: if size <= total_donuts: num_groups = 1 + max_groups_memo(group_sizes, total_donuts - size, memo) max_groups = max(max_groups, num_groups) memo[total_donuts] = max_groups return max_groups print(max_groups_memo([2, 3, 5], 10))

The output of this code snippet:

4

This snippet explores all combinations of groups that can be formed using recursion, then stores the calculated values in a memoization dictionary for future reference, thereby reducing the overall number of calculations.

## Bonus One-Liner Method 5: Brute Force with itertools

This method uses Python’s `itertools`

module to generate all possible combinations of group distributions and selects the maximum number of groups that can get fresh donuts. Ideal for smaller input sizes due to its exponential time complexity.

Here’s an example:

from itertools import combinations def max_groups_brute_force(group_sizes, total_donuts): max_groups = 0 for i in range(1, len(group_sizes) + 1): for combination in combinations(group_sizes, i): if sum(combination) <= total_donuts: max_groups = max(max_groups, len(combination)) return max_groups print(max_groups_brute_force([2, 3, 5], 10))

The output of this code snippet:

3

This code iterates through all possible group size combinations, checking which ones can be accommodated within the total donut count by using `itertools.combinations`

.

## Summary/Discussion

**Method 1: Iterative Distribution.**Simple implementation, gives preference to smaller groups, might not be optimal with certain combinations of group sizes.**Method 2: Using a Greedy Algorithm.**Quick, easy to implement, not always offering the optimal solution as it favors larger groups.**Method 3: Dynamic Programming.**Efficient, handles large inputs well, but might be overkill for smaller problems, more complex to understand.**Method 4: Recursion with Memoization.**Good for smaller inputs, optimizes by storing overlapping subproblems, but could lead to stack overflow on very large inputs.**Method 5: Brute Force with itertools.**Easy to write, handles small inputs quickly, not efficient for larger inputs due to exponential complexity.