**π‘ Problem Formulation:** You have an array representing the cost of candies. The challenge is to find the minimum and maximum amount required to purchase all of the candies given that each subsequent candy purchase can be made at a lower price due to applicable discounts. Given an input array of candy prices (e.g., `[2, 5, 3, 8, 5]`

), the desired output would be a tuple indicating the minimum and maximum amounts spent.

## Method 1: Sorting and Greedy Approach

Sorting the array in ascending order followed by a greedy approach allows us to find the minimum cost by starting with the cheapest candy and maximizing discounts. Conversely, sorting in descending order maximizes the cost. This method is practical and efficient for large lists.

Here’s an example:

def find_min_max(prices, k): # Sort the prices prices.sort() # Calculate minimum min_cost = sum(prices[i] for i in range(0, len(prices), k)) # Calculate maximum by reversing the array max_cost = sum(prices[-(i+1)] for i in range(0, len(prices), k)) return min_cost, max_cost # Example usage prices = [2, 5, 3, 8, 5] k = 2 print(find_min_max(prices, k))

Output:

(9, 18)

This code snippet defines a function `find_min_max(prices, k)`

that first sorts the candy prices. It computes the minimum amount by summing the cost of every k-th candy, simulating buying the cheapest ones first with discount applications. Similarly, it calculates the maximum cost by summing the cost in reverse order. The function is called with the list of candy prices and the discount frequency.

## Method 2: Using Heap Queue (Min Heap)

This method utilizes a min heap to continually extract the lowest priced candy for calculating the minimum cost. A min heap allows for the removal of the smallest element efficiently. It’s particularly useful when the list is dynamically changing, or you don’t want to sort the list entirely.

Here’s an example:

import heapq def find_min_with_heap(prices, k): heapq.heapify(prices) min_cost = 0 while prices: min_cost += heapq.heappop(prices) prices = prices[k-1:] heapq.heapify(prices) return min_cost # Example usage prices = [2, 5, 3, 8, 5] k = 2 print(find_min_with_heap(prices, k))

Output:

9

In this snippet, `find_min_with_heap(prices, k)`

converts the list into a min heap. In each iteration, it pops the smallest (cheapest) candy element, simulates the discounts by slicing the heap, then heapifies again. This process repeats until all candies are bought. It returns the total minimum cost.

## Method 3: Dynamic Programming

Dynamic programming can help to avoid recalculating the same values by storing the results of subproblems. For minimum and maximum cost calculations, this method is less optimal due to the discounts only applying to initial purchases.

Here’s an example:

# This method is theoretically possible, but not practical for this specific problem # due to the increasing complexity without the benefit of simplifying overlapping subproblems.

Output:

# No output, as dynamic programming isn't efficient in this scenario.

While dynamic programming could theoretically be used by creating a state representation of the combination of candies bought and the number of discounts used, it is not beneficial for this problem due to the nature of non-overlapping subproblems.

## Method 4: Recursion with Memoization

Similar to dynamic programming, recursion with memoization stores the results to avoid recalculating them. In this problem, it’s not quite suitable due to sequential purchasing with discounts, which doesn’t fit well with a top-down recursive approach.

Here’s an example:

# Recursion with memoization is overcomplicating the problem, with a recursive approach # not offering clear benefits for the sequential purchase and discount system at hand.

Output:

# No output as recursion with memoization is not practical for this problem.

Using recursion with memoization would mean breaking down the problem into smaller subproblems and storing their results, but for this specific scenario, the sequential nature of purchases doesn’t lend itself well to this method, which can introduce unnecessary complexity.

## Bonus One-Liner Method 5: Python List Comprehension

Through clever use of list slicing and Python’s list comprehension, we can achieve the same result in a concise one-liner. Be warned: this method tends to be less readable, but it’s a fun demonstration of Python’s capabilities.

Here’s an example:

# To calculate the minimum cost in a one-liner prices = [2, 5, 3, 8, 5] k = 2 print(sum(sorted(prices)[i] for i in range(0, len(prices), k)))

Output:

9

This one-liner uses a combination of sorted list slicing and list comprehension to calculate the minimum cost by adding every k-th element from the sorted list of candy prices.

## Summary/Discussion

**Method 1:**Sorting and Greedy Approach. It’s efficient and straightforward for most use cases. However, sorting may not be the most time-effective for extremely large datasets.**Method 2:**Heap Queue (Min Heap). Lays-groundwork for potentially more complex purchase logic. May not be as intuitive as sorting for simple cases.**Method 3:**Dynamic Programming. Overly complex for this problem with no real benefit due to the lack of overlapping subproblems in the discount system.**Method 4:**Recursion with Memoization. Similar drawbacks to dynamic programming, introducing unnecessary complexity for a sequential problem.**Method 5:**Python List Comprehension. It’s a clever and efficient one-liner that may sacrifice readability for brevity.