Chocolate Day: Calculate how many chocolates you can buy

Have you ever wondered how to optimize your purchases by taking advantage of tiered discounts? The solution might be more algorithmic than you imagine. 🛍️

🔮 Problem Statement

Lisa, a chocolate enthusiast, finds herself facing a tempting offer at her favorite store. On the occasion of Chocolate Day, the store offers progressive discounts:

• The first chocolate is sold at the defined price (starting price).
• Each subsequent chocolate decreases in value by discount (discount), but never below min_price (minimum price).
• Lisa has a limited budget (budget) to make the most of the promotion.

The challenge is to determine how many chocolates Lisa can purchase with her budget, considering the applied discount strategy.

Parameters:

• price (int): Initial price of the first chocolate.
• discount (int): Discount applied to each chocolate purchased.
• min_price (int): Minimum price the chocolate can reach.
• budget (int): Total budget available for the purchase.

Examples:

>>> day_of_chocolate(20, 3, 6, 80)
6
>>> day_of_chocolate(20, 3, 6, 85)
7
>>> day_of_chocolate(88, 4, 1, 1000)
20

🧩 Step-by-Step Solution

To help Lisa maximize her chocolate purchase, we need to simulate the buying process, updating the price of each chocolate and the remaining budget until she can no longer afford another.

We start by initializing the chocolates variable to zero. This variable will act as our chocolate counter, increasing with each successful purchase.

chocolates = 0

The heart of the solution lies in the while loop, which iteratively simulates the purchase of chocolates while Lisa has enough budget to purchase at least one at the current price.

while budget >= price:

Inside the loop, we deduct the price of the chocolate from Lisa’s budget and update the chocolate counter. 🍫

budget -= price
chocolates += 1

The key piece is updating the price of the chocolate for the next iteration. We reduce the price by the discount value, but ensure it never falls below the established minimum price. The max() function helps us achieve this, guaranteeing that the price stays within the defined limits.

price = max(price - discount, min_price)

Finally, when Lisa can no longer afford another chocolate, the loop ends and we return the total number of chocolates she was able to buy.

Complete solution:

def day_of_chocolate(price, discount, min_price, budget):
	"level: medium; points: 4"
	chocolates = 0
	while budget >= price:
		budget -= price
		chocolates += 1
		price = max(price - discount, min_price)
	return chocolates

🧠 Key Concepts

The solution is based on controlled iteration through a while loop. This type of loop is fundamental when we need to repeat a block of code until a specific condition is met, in this case, that Lisa’s budget is sufficient to buy a chocolate.

Dynamic price updating is another key concept. We use the max() function to ensure that the price of the chocolate never falls below the minimum price. This function takes two arguments and returns the larger of the two. In our case, it compares the discounted price with the minimum price and returns the higher value, ensuring that the price stays within the allowed limits. This is a good example of how real-world constraints can be modeled with simple mathematical functions.

The use of a counter (chocolates) is essential to track the number of chocolates purchased. The counter is incremented by one each time Lisa buys a chocolate, allowing us to determine the total number of chocolates she was able to acquire at the end of the process.

Did you know that the concept of “minimum price” is widely used in game theory and economics to model pricing strategies and competition in the market?

💫 Final Thoughts

The presented solution is efficient and concise, but could be further optimized if we consider scenarios with extremely large budgets and small discounts. In such cases, we could explore the use of mathematical formulas to directly calculate the number of chocolates that can be purchased without the need to iterate through each individual purchase.

This problem illustrates how programming concepts can be applied to solve everyday problems, from optimizing purchases to modeling economic scenarios. The key lies in the ability to break down a complex problem into smaller, more manageable steps, and then implement an algorithmic solution that simulates the decision-making process.

Did you enjoy unraveling this chocolate problem? Don’t keep it to yourself! Explore other programming articles on our blog and discover how algorithmic logic can transform your approach to real-world challenges. Let’s get programming! 🚀