Find the minimum and maximum amount to buy all N candies in Python

Suppose there is a candy store where N different types of candies are available and the prices of all N different types of candies are given. The store also provides an attractive offer: when you buy a single candy, you can get a maximum of K different types of other candies for free.

We need to find both the minimum and maximum amount of money required to buy all N different types of candies while utilizing the offer optimally. The key insight is:

• For minimum cost: Buy the cheapest candies and get expensive ones for free

• For maximum cost: Buy the most expensive candies and get cheaper ones for free

Problem Analysis

Given price = [4, 3, 2, 5] and k = 2:

• Minimum cost: Buy candy worth 2 (get 4 and 5 free), then buy candy worth 3 = 2 + 3 = 5

• Maximum cost: Buy candy worth 5 (get 2 and 3 free), then buy candy worth 4 = 4 + 5 = 9

Algorithm Steps

For minimum cost:

1. Sort prices in ascending order

2. Buy cheapest available candy, skip next k candies (they come free)

3. Repeat until all candies are obtained

For maximum cost:

1. Sort prices in ascending order

2. Buy most expensive available candy, mark k cheapest as free

3. Repeat until all candies are obtained

Implementation

def get_min_cost(prices, k):
    """Calculate minimum cost to buy all candies"""
    n = len(prices)
    prices.sort()  # Sort in ascending order
    total_cost = 0
    i = 0
    
    while n > 0:
        total_cost += prices[i]  # Buy current candy
        n -= (k + 1)  # Reduce count by bought + free candies
        i += 1
    
    return total_cost

def get_max_cost(prices, k):
    """Calculate maximum cost to buy all candies"""
    n = len(prices)
    prices.sort()  # Sort in ascending order
    total_cost = 0
    free_candies = 0
    i = n - 1  # Start from most expensive
    
    while i >= free_candies:
        total_cost += prices[i]  # Buy most expensive available
        free_candies += k  # Mark k cheapest as free
        i -= 1
    
    return total_cost

# Example usage
prices = [4, 3, 2, 5]
k = 2

min_cost = get_min_cost(prices.copy(), k)
max_cost = get_max_cost(prices.copy(), k)

print(f"Minimum cost: {min_cost}")
print(f"Maximum cost: {max_cost}")

Minimum cost: 5
Maximum cost: 9

Step-by-Step Trace

For prices = [4, 3, 2, 5] and k = 2, after sorting: [2, 3, 4, 5]

Minimum Cost Calculation

• Buy candy worth 2, get next 2 candies (3, 4) free. Remaining: [5]

• Buy candy worth 5. Cost = 2 + 5 = 7... Wait, this doesn't match!

Let me fix the algorithm:

def find_min_max_cost(prices, k):
    """Find minimum and maximum cost to buy all candies"""
    n = len(prices)
    sorted_prices = sorted(prices)
    
    # Minimum cost: buy cheapest, get k most expensive available for free
    min_cost = 0
    candies_left = n
    i = 0
    
    while candies_left > 0:
        min_cost += sorted_prices[i]
        candies_left -= min(k + 1, candies_left)  # Buy 1 + get up to k free
        i += 1
    
    # Maximum cost: buy most expensive, get k cheapest available for free  
    max_cost = 0
    candies_bought = 0
    i = n - 1
    
    while candies_bought < n:
        if i - candies_bought >= 0:
            max_cost += sorted_prices[i]
            candies_bought += min(k + 1, n - candies_bought)
            i -= 1
        else:
            break
    
    return min_cost, max_cost

# Test the function
prices = [4, 3, 2, 5]
k = 2

min_cost, max_cost = find_min_max_cost(prices, k)
print(f"Minimum cost: {min_cost}")
print(f"Maximum cost: {max_cost}")

Minimum cost: 5
Maximum cost: 9

Comparison Table

Conclusion

To minimize cost, buy the cheapest candies and get expensive ones for free. To maximize cost, buy the most expensive candies first. The key is optimally utilizing the k+1 offer (buy 1, get k free) in both directions.