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Maximum Spending After Buying Items

Difficulty: Hard

Problem Description

You are given a 0-indexed m * n integer matrix values, representing the values of m * n different items in m different shops. Each shop has n items where the j-th item in the i-th shop has a value of values[i][j]. Additionally, the items in the i-th shop are sorted in non-increasing order of value. On each day, you can buy a single item from one of the shops at a price of values[i][j] * d, where d is the day number. Your goal is to determine the maximum amount of money that can be spent on buying all m * n products.

Key Insights

  • Each shop's items are sorted in non-increasing order, which means we can always consider the rightmost available item for maximum spending on each day.
  • The maximum spending can be achieved by strategically selecting the last available items from different shops over the days.
  • Using a max-heap (priority queue) can help efficiently manage the selection of items across multiple shops on each day.

Space and Time Complexity

Time Complexity: O(m * n log(m)), where m is the number of shops and n is the number of items per shop. The log(m) factor comes from using a max-heap to manage the items. Space Complexity: O(m), for storing the max-heap.

Solution

To solve the problem, we can utilize a max-heap to keep track of the rightmost available items from each shop. The algorithm works as follows:

  1. Initialize a max-heap and populate it with the rightmost items from each shop along with their respective shop indices.
  2. For each day from 1 to m * n, extract the maximum item from the heap, calculate its price based on the day, and add this price to the total spending.
  3. After purchasing an item, push the next rightmost item from the same shop onto the heap.
  4. Repeat until all items have been purchased.

This approach guarantees that we always get the maximum possible spending by selecting the most valuable available items each day.

Code Solutions

import heapq

def maxSpending(values):
    m = len(values)
    n = len(values[0])
    
    # Create a max-heap
    max_heap = []
    
    # Populate the heap with the rightmost items from each shop
    for i in range(m):
        heapq.heappush(max_heap, (-values[i][n-1], i, n-1))  # Store negative for max-heap behavior

    total_spending = 0
    
    # Buy items over m * n days
    for day in range(1, m * n + 1):
        # Get the item with the maximum value
        value, shop_index, item_index = heapq.heappop(max_heap)
        total_spending += -value * day  # Convert back to positive
        
        # Push the next item from the same shop if it exists
        if item_index > 0:
            next_value = values[shop_index][item_index - 1]
            heapq.heappush(max_heap, (-next_value, shop_index, item_index - 1))
    
    return total_spending
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