Maximum Spending After Buying Items

Hard

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Greedy AlgorithmsArrays

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. 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 d^th 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.

Try coding it in LeetCode

Maximum Spending After Buying Items

Hard

Asked by:

3 views

Topics:

Greedy AlgorithmsArrays

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. 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 d^th 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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the dimensions of the `costs` array (number of shops and items per shop)? What is the maximum value of `k` relative to the number of shops?
Can the costs of the items be negative, zero, or non-integer (e.g., floating-point numbers)?
If it's impossible to buy exactly `k` items (e.g., `k` is greater than the number of shops), what should I return?
Are there any constraints on the number of items available in each shop, and is it guaranteed that each shop contains at least one item?
If there are multiple combinations of items that result in the maximum spending, is any one of those combinations acceptable?

Brute Force Solution

Approach

The brute force method to maximize spending after buying items involves exploring every possible combination of items from different shops. We systematically consider all possible purchases to identify the highest total spending achievable within a given budget. This involves evaluating all combinations to find the optimal solution.

Here's how the algorithm would work step-by-step:

Consider all the possible items we can buy from the first shop, starting with buying nothing.
For each choice from the first shop, explore every possible item (including buying nothing) from the second shop.
Continue this pattern, considering all possible item purchases from each shop, one shop at a time.
For each complete set of purchases across all shops, calculate the total cost.
If the total cost is within our budget, record the total spending (the sum of the prices of items purchased).
Once we have explored every single possible combination of item purchases, compare all recorded spending amounts.
Select the highest recorded spending amount that remained within our budget. This is the maximum spending we can achieve.

Code Implementation

def maximum_spending_brute_force(item_prices, budget): 
    number_of_items = len(item_prices)
    maximum_spent = 0

    # Iterate through all possible combinations of items
    for i in range(1 << number_of_items):
        current_cost = 0
        
        # Calculate the cost of the current combination
        for j in range(number_of_items):
            # Check if the j-th item is included in the current combination
            if (i >> j) & 1:
                current_cost += item_prices[j]

        # If the current cost is within the budget,
        # update the maximum spent if necessary
        if current_cost <= budget:
            maximum_spent = max(maximum_spent, current_cost)

    return maximum_spent

Big(O) Analysis

Time Complexity

O(m^n) – The algorithm explores all possible combinations of items from n shops. For each shop, we can choose any of the m items, including buying nothing. This means each shop presents m options. Since we do this for each of the n shops, the total number of combinations is m multiplied by itself n times, resulting in m^n total combinations. Therefore, the time complexity is O(m^n), where n is the number of shops and m is the maximum number of items in any shop.

Space Complexity

O(1) – The brute force method, as described, primarily iterates through the input shops and items without creating significant auxiliary data structures. It records the maximum spending achieved so far, which requires constant space. It also calculates the total cost for each combination which is also done in constant space. Therefore, the auxiliary space used remains constant regardless of the input size, which makes the space complexity O(1).

Optimal Solution

Approach

The problem asks us to maximize spending given budget constraints on different types of items. The core idea is to prioritize buying the most valuable items first, while respecting the budget limit for each type of item. This approach avoids checking all possible combinations, leading to an efficient solution.

Here's how the algorithm would work step-by-step:

For each type of item, figure out how much satisfaction each dollar spent gives you. We calculate this value by dividing the satisfaction of an item by its cost.
Sort the items from all categories based on this 'satisfaction per dollar' value, starting with the highest value first.
Go through the sorted items one by one.
For each item, check if you still have enough budget left for its category. If you do, buy the item and reduce the budget for that category.
If you can't afford the item, simply skip it and move to the next item in the sorted list.
Continue this process until you have considered all the items, or until you run out of budget in every category.
The total satisfaction from all the items you bought is the maximum spending you can achieve.

Code Implementation

def maximum_spending(item_costs, discounts, budget):
    combined_costs = []
    for i in range(len(item_costs)):
        combined_costs.append(item_costs[i] - discounts[i])

    # Sort items by cost descending to maximize spending.
    combined_costs.sort(reverse=True)

    total_spent = 0

    # Iterate through the sorted costs and purchase if affordable.
    for cost in combined_costs:
        if budget >= cost:

            budget -= cost
            total_spent += cost

    return total_spent

Big(O) Analysis

Time Complexity

O(n log n) – Let n be the total number of items across all categories. Calculating the 'satisfaction per dollar' value for each item takes O(n) time. Sorting the items based on this value using an efficient sorting algorithm (like merge sort or quicksort) dominates the runtime, taking O(n log n) time. The subsequent iteration through the sorted items to determine which ones to buy takes O(n) time. Therefore, the overall time complexity is O(n) + O(n log n) + O(n), which simplifies to O(n log n).

Space Complexity

O(N) – The dominant space complexity comes from sorting all items based on their 'satisfaction per dollar' value into a new list. Where N represents the total number of items across all categories, this sorting process requires an auxiliary list of size N to hold the sorted items. We also store a few variables for iteration and budget tracking, which contribute constant space. Therefore, the auxiliary space is primarily determined by the sorted list of size N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
costs is null or empty	Return 0 if costs is null or empty, as no items can be bought.
k is 0	Return 0 if k is 0, because we can't buy any items.
k is greater than the number of shops	If k exceeds the number of shops, effectively k = number of shops.
costs[i] is empty for some shop i	Treat empty shop as if it contains zero-cost option and continue calculating max cost with remaining shops.
costs contains negative values	Since the problem asks for maximum spending, negative values represent a 'discount', the algorithm should still calculate maximum possible spending by considering the lowest possible (most negative) values.
All costs are zero.	If all costs are zero, return 0, which is the maximum spending.
Integer overflow if costs are large and k is large	Use a data type that can hold larger values like long to prevent integer overflow.
k is equal to the number of shops and each shop contains a single value	The maximum cost will be the sum of the items costs.