Introduction

In this blog post, we will explore how to solve the LeetCode problem "2931" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 jth item in the ith shop has a value of values[i][j]. Additionally, the items in the ith shop are sorted in non-increasing order of value. That is, values[i][j] >= values[i][j + 1] for all 0 <= j < n - 1. On each day, you would like to buy a single item from one of the shops. Specifically, On the dth day you can: Pick any shop i. Buy the rightmost available item j for the price of values[i][j] d. That is, find the greatest index j such that item j was never bought before, and buy it for the price of values[i][j] d. Note that all items are pairwise different. For example, if you have bought item 0 from shop 1, you can still buy item 0 from any other shop. Return the maximum amount of money that can be spent on buying all m n products. Example 1: Input: values = [[8,5,2],[6,4,1],[9,7,3]] Output: 285 Explanation: On the first day, we buy product 2 from shop 1 for a price of values[1][2] 1 = 1. On the second day, we buy product 2 from shop 0 for a price of values[0][2] 2 = 4. On the third day, we buy product 2 from shop 2 for a price of values[2][2] 3 = 9. On the fourth day, we buy product 1 from shop 1 for a price of values[1][1] 4 = 16. On the fifth day, we buy product 1 from shop 0 for a price of values[0][1] 5 = 25. On the sixth day, we buy product 0 from shop 1 for a price of values[1][0] 6 = 36. On the seventh day, we buy product 1 from shop 2 for a price of values[2][1] 7 = 49. On the eighth day, we buy product 0 from shop 0 for a price of values[0][0] 8 = 64. On the ninth day, we buy product 0 from shop 2 for a price of values[2][0] 9 = 81. Hence, our total spending is equal to 285. It can be shown that 285 is the maximum amount of money that can be spent buying all m n products. Example 2: Input: values = [[10,8,6,4,2],[9,7,5,3,2]] Output: 386 Explanation: On the first day, we buy product 4 from shop 0 for a price of values[0][4] 1 = 2. On the second day, we buy product 4 from shop 1 for a price of values[1][4] 2 = 4. On the third day, we buy product 3 from shop 1 for a price of values[1][3] 3 = 9. On the fourth day, we buy product 3 from shop 0 for a price of values[0][3] 4 = 16. On the fifth day, we buy product 2 from shop 1 for a price of values[1][2] 5 = 25. On the sixth day, we buy product 2 from shop 0 for a price of values[0][2] 6 = 36. On the seventh day, we buy product 1 from shop 1 for a price of values[1][1] 7 = 49. On the eighth day, we buy product 1 from shop 0 for a price of values[0][1] 8 = 64 On the ninth day, we buy product 0 from shop 1 for a price of values[1][0] 9 = 81. On the tenth day, we buy product 0 from shop 0 for a price of values[0][0] 10 = 100. Hence, our total spending is equal to 386. It can be shown that 386 is the maximum amount of money that can be spent buying all m n products. Constraints: 1 <= m == values.length <= 10 1 <= n == values[i].length <= 104 1 <= values[i][j] <= 106 values[i] are sorted in non-increasing order.

Explanation

Here's the solution:

• Key Idea: Use a priority queue (min-heap) to efficiently select the next best item to buy. Store tuples of (value * day, value, shop_index, item_index) in the heap.
• Prioritize: The heap prioritizes items with the highest potential revenue (value * day). We start with all the rightmost items in each shop and iteratively add better items to the heap as we sell them.

• Iterative Selling: Keep track of the current day and process items from the heap, updating the heap with the next available item from the corresponding shop.

• Runtime & Storage Complexity: O(m*n*log(m)), O(m) where m is number of shops and n is number of items in each shop.

Code

    import heapq

def max_spending(values):
    m = len(values)
    n = len(values[0])
    total_spending = 0
    day = 1
    pq = []  # Min-heap

    # Initialize priority queue with the rightmost item from each shop
    for i in range(m):
        heapq.heappush(pq, (-values[i][n - 1] * day, values[i][n - 1], i, n - 1))
        day += 1

    day = m+1

    # Process items until all are bought
    for _ in range(m * n):
        cost, value, shop_index, item_index = heapq.heappop(pq)
        total_spending += -cost
        #total_spending += value * day

        # Add the next item from the same shop to the priority queue
        if item_index > 0:
            heapq.heappush(pq, (-values[shop_index][item_index - 1] * day, values[shop_index][item_index - 1], shop_index, item_index - 1))
        day += 1

    return total_spending