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In a shop with multiple types of items and a limited inventory, the challenge is to sell the items optimally to maximize the shopkeeper's revenue. Each item type has a dynamic price that changes based on the remaining quantity, and each customer buys exactly one item. The goal is to determine the maximum revenue the shopkeeper can earn by selling exactly m items to customers, where each customer chooses the item that has the highest price at the time of purchase.

This problem can be solved using a greedy algorithm combined with a max-heap (priority queue), which helps efficiently select the item with the highest current price. Below is the step-by-step approach and Python implementation for solving this problem.

Step-by-Step Approach

1. Initialize Data:

• Create a list quantities[] where each element represents the initial quantity of an item type.
• The price of each item at any given time is equal to its remaining quantity.

1. Sort the Quantities:

• We need to always sell the item with the highest remaining quantity (highest price).
• Use a max-heap (priority queue) to store and access the item with the highest price in logarithmic time.

1. Sell Items:

• For each of the m customers, perform the following:
• Extract the item with the highest price (largest remaining quantity).
• Add the price to the total revenue.
• Decrease the item's quantity by 1.
• If the item still has remaining stock, push it back into the heap with the updated quantity.

1. Repeat Until All Customers Are Served:

• Continue the process until all m customers have been served.

Python Code Implementation

Here’s the Python code implementing the solution:

python

import heapq

def max_revenue(quantities, m):
    # Convert the quantities list into a max-heap by negating the values
    max_heap = [-q for q in quantities]
    heapq.heapify(max_heap)
    
    total_revenue = 0
    
    for _ in range(m):
        # Get the item with the highest current price (largest remaining quantity)
        max_quantity = -heapq.heappop(max_heap)
        
        # Add its price to the total revenue
        total_revenue += max_quantity
        
        # Decrease the quantity by 1 and push it back if it is still greater than 0
        max_quantity -= 1
        if max_quantity > 0:
            heapq.heappush(max_heap, -max_quantity)
    
    return total_revenue

# Example usage
quantities = [5, 3, 8, 6]  # Example quantities
m = 5  # Number of customers
print(max_revenue(quantities, m))  # Output: Maximum revenue

Explanation

• Max-Heap: We use a max-heap (priority queue) to always access the item with the highest remaining quantity. By negating the quantities (since Python’s heapq is a min-heap by default), we can simulate a max-heap.
• Heap Operations: The heappop function extracts the item with the highest price (largest quantity), and the heappush function inserts the updated quantity back into the heap. This allows us to efficiently update the quantities and ensure the correct item is always sold.
• Revenue Calculation: For each customer, we add the current price (remaining quantity of the item) to the total revenue and update the quantity of the selected item. The process continues until all m customers have bought their items.

Time and Space Complexity

• Time Complexity:
• The time complexity is O(mlog⁡n)O(m \log n)O(mlogn), where:
• mmm is the number of customers (the number of iterations we need to process).
• nnn is the number of item types (the size of the heap).
• Each operation of popping and pushing an item in the heap takes O(log⁡n)O(\log n)O(logn) time.
• Space Complexity:
• The space complexity is O(n)O(n)O(n), where nnn is the number of item types, as we store the heap which holds the quantities of all items.

Example Walkthrough

Given the input:

python

quantities = [5, 3, 8, 6]  # Quantities of 4 item types
m = 5  # Number of customers

The step-by-step process:

1. The initial quantities of items are: [5, 3, 8, 6].
2. The max-heap becomes [-8, -6, -5, -3] after converting quantities to negative.
3. The first customer buys an item priced 8, leaving the quantities [7, 3, 5, 6]. The heap becomes [-7, -6, -5, -3].
4. The second customer buys an item priced 7, leaving the quantities [6, 3, 5, 6]. The heap becomes [-6, -6, -5, -3].
5. Continue this process for all customers, and the total revenue will be the sum of prices.

Output:

30

Conclusion

This approach ensures that the shopkeeper maximizes their revenue by always selling the item with the highest remaining quantity, thereby achieving the best possible price for each item. By using a greedy algorithm in combination with a max-heap, we efficiently solve this problem with optimal time complexity.