Sell Diminishing-Valued Colored Balls

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MathWorks

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Topics:

ArraysGreedy AlgorithmsBinary Search

You have an inventory of different colored balls, and there is a customer that wants orders balls of any color.

The customer weirdly values the colored balls. Each colored ball's value is the number of balls of that color you currently have in your inventory. For example, if you own 6 yellow balls, the customer would pay 6 for the first yellow ball. After the transaction, there are only 5 yellow balls left, so the next yellow ball is then valued at 5 (i.e., the value of the balls decreases as you sell more to the customer).

You are given an integer array, inventory, where inventory[i] represents the number of balls of the i^th color that you initially own. You are also given an integer orders, which represents the total number of balls that the customer wants. You can sell the balls in any order.

Return the maximum total value that you can attain after selling orders colored balls. As the answer may be too large, return it modulo 10⁹+ 7.

Example 1:

Input: inventory = [2,5], orders = 4
Output: 14
Explanation: Sell the 1st color 1 time (2) and the 2nd color 3 times (5 + 4 + 3).
The maximum total value is 2 + 5 + 4 + 3 = 14.

Example 2:

Input: inventory = [3,5], orders = 6
Output: 19
Explanation: Sell the 1st color 2 times (3 + 2) and the 2nd color 4 times (5 + 4 + 3 + 2).
The maximum total value is 3 + 2 + 5 + 4 + 3 + 2 = 19.

Constraints:

1 <= inventory.length <= 10⁵
1 <= inventory[i] <= 10⁹
1 <= orders <= min(sum(inventory[i]), 10⁹)

Try coding it in LeetCode

Sell Diminishing-Valued Colored Balls

Medium

MathWorks

1 view

Topics:

ArraysGreedy AlgorithmsBinary Search

You have an inventory of different colored balls, and there is a customer that wants orders balls of any color.

The customer weirdly values the colored balls. Each colored ball's value is the number of balls of that color you currently have in your inventory. For example, if you own 6 yellow balls, the customer would pay 6 for the first yellow ball. After the transaction, there are only 5 yellow balls left, so the next yellow ball is then valued at 5 (i.e., the value of the balls decreases as you sell more to the customer).

You are given an integer array, inventory, where inventory[i] represents the number of balls of the i^th color that you initially own. You are also given an integer orders, which represents the total number of balls that the customer wants. You can sell the balls in any order.

Return the maximum total value that you can attain after selling orders colored balls. As the answer may be too large, return it modulo 10⁹+ 7.

Example 1:

Input: inventory = [2,5], orders = 4
Output: 14
Explanation: Sell the 1st color 1 time (2) and the 2nd color 3 times (5 + 4 + 3).
The maximum total value is 2 + 5 + 4 + 3 = 14.

Example 2:

Input: inventory = [3,5], orders = 6
Output: 19
Explanation: Sell the 1st color 2 times (3 + 2) and the 2nd color 4 times (5 + 4 + 3 + 2).
The maximum total value is 3 + 2 + 5 + 4 + 3 + 2 = 19.

Constraints:

1 <= inventory.length <= 10⁵
1 <= inventory[i] <= 10⁹
1 <= orders <= min(sum(inventory[i]), 10⁹)

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 size of the inventory array and the magnitude of the numbers within it? Also, what is the maximum value for the 'orders' integer?
Can the inventory array contain zero values, and what should be the behavior if orders is zero?
Are the values in the inventory array guaranteed to be non-negative?
If there are multiple ways to maximize the total value after selling 'orders' balls, is any valid solution acceptable?
What should I return if it is impossible to fulfill all 'orders' given the current inventory?

Brute Force Solution

Approach

The brute force approach to selling diminishing-valued balls involves exploring every possible combination of ball sales. We aim to find the combination that yields the highest total value by meticulously considering each sale option. This involves trying every possible quantity for each color, one sale at a time.

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

Start by considering selling only one ball.
For each color, calculate the profit from selling one ball of that color, assuming you sold zero balls of any color previously.
Keep track of the profit for each color, and consider only the color that gives you the most profit to sell.
Then, consider selling two balls.
For each color, calculate the profit from selling a second ball of that color, considering you've already sold one.
Compare the profits for each color and again only consider the color with the highest profit.
Continue this process for three balls, four balls, and so on up to the maximum number of balls you are allowed to sell.
Each time you have to decide which color to sell from, calculate the profit you would get selling the next ball from all available colors, accounting for how many you already sold of that color.
At each step, choose the color that adds the most profit.
Keep track of the total profit you get by selling the balls in the order you chose.
This process gives you the maximum possible profit by checking all possible combinations. Note that this approach doesn't guarantee correctness and is mostly for demonstrative purposes for understanding the problem.

Code Implementation

def sell_diminishing_valued_colored_balls_brute_force(inventory, orders):
    total_profit = 0

    # Iterate up to the number of orders
    for _ in range(orders):
        max_profit_increase = 0
        color_to_sell = -1

        # Find the color with the highest profit
        for color_index in range(len(inventory)):
            current_profit = inventory[color_index]
            if current_profit > max_profit_increase:
                max_profit_increase = current_profit
                color_to_sell = color_index

        # Ensure a color can be sold.
        if color_to_sell != -1:
            total_profit += inventory[color_to_sell]

            # Reduce inventory of chosen color
            inventory[color_to_sell] -= 1

    return total_profit

Big(O) Analysis

Time Complexity

O(orders * colors) – The algorithm iterates up to 'orders' times (the maximum number of balls to sell). In each iteration, it calculates the profit for each available color, which takes 'colors' operations, to determine which color to sell next. Therefore, the time complexity is proportional to the product of 'orders' and 'colors', representing the total number of profit calculations performed. Thus, the overall time complexity is O(orders * colors).

Space Complexity

O(1) – The described brute force algorithm primarily uses a fixed number of variables to track the current profit, the number of balls sold for each color, and the total profit. The space needed to store these variables remains constant irrespective of the number of colors or the number of balls to sell. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The key to maximizing profit is to sell the most valuable balls first. We use a strategy to quickly figure out how many balls of different values we should sell to meet the sales goal, taking into account the diminishing value as we sell more of each color.

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

Figure out the range of possible ball values we could sell, from the lowest to the highest available.
Use a process like binary search to efficiently find the best minimum ball value to sell. This helps us quickly narrow down the possibilities.
For a given minimum ball value, calculate how many balls of each color we can sell at or above that value.
Sum up the number of balls we can sell across all colors at or above that minimum value.
Compare the total number of balls we can sell with our target sales number.
If we can sell enough balls, try a higher minimum value. If we can't sell enough balls, try a lower minimum value. Continue this search until you find the optimal minimum value.
Once you have the optimal minimum value, calculate the profit from selling balls above that minimum value and the profit from selling balls at that minimum value to reach the target sales quantity exactly.
Add these profits together to find the total maximum profit.

Code Implementation

def max_profit(inventory, orders):
    left_value = 1
    right_value = max(inventory)
    maximum_profit = 0
    modulo_value = 10**9 + 7

    while left_value <= right_value:
        mid_value = (left_value + right_value) // 2
        number_of_balls_we_can_sell = 0

        for ball_value in inventory:
            if ball_value >= mid_value:
                number_of_balls_we_can_sell += ball_value - mid_value + 1

        # Narrow the search to find the optimal minimum ball value.
        if number_of_balls_we_can_sell >= orders:
            left_value = mid_value + 1
        else:
            right_value = mid_value - 1

    minimum_value = right_value
    for ball_value in inventory:
        if ball_value > minimum_value:
            maximum_profit += (ball_value * (ball_value + 1)) // 2 - (minimum_value * (minimum_value + 1)) // 2
            maximum_profit %= modulo_value
            orders -= ball_value - minimum_value

    # Calculate profit from selling remaining balls at the minimum value.
    maximum_profit += minimum_value * orders
    maximum_profit %= modulo_value

    return maximum_profit

Big(O) Analysis

Time Complexity

O(n log m) – The algorithm uses binary search to find the optimal minimum ball value, where m is the range of possible ball values (from the lowest to the highest value in the inventory). Within each step of the binary search, we iterate through the inventory of balls (n colors) to calculate the number of balls we can sell above the current minimum value and the profit we would make. Therefore, each binary search step takes O(n) time. Since binary search takes O(log m) steps, the overall time complexity is O(n log m), where n is the number of colors and m is the range of ball values.

Space Complexity

O(1) – The algorithm described uses binary search which generally involves a few variables to track the low, high, and mid points of the search range. It also calculates sums and counts based on the input array but these are stored in constant space variables. Therefore, the auxiliary space used remains constant regardless of the input array size, making it O(1).

Edge Cases

Case	How to Handle
inventory is null or empty	Return 0, as no balls can be sold.
orders is 0	Return 0, as no balls need to be sold.
inventory contains only one color	Handle by calculating the sum of an arithmetic series from inventory[0] down to inventory[0] - orders + 1 (or 1, whichever is greater).
inventory contains very large numbers and orders is also very large, potentially leading to integer overflow during calculation	Use long data type for intermediate calculations and the final result to prevent integer overflow.
orders is larger than the total number of balls in the inventory	Calculate the revenue by selling all the balls until each color has 0 balls.
All colors in inventory have the same number of balls	Calculate revenue by selling balls equally from each color until orders is met.
inventory contains extreme boundary values (e.g., Integer.MAX_VALUE)	Ensure calculations avoid exceeding the maximum integer value, possibly by using long data types and careful ordering of operations.
Inventory is very large, impacting memory usage and potentially causing performance issues	Binary search approach ensures performance is logarithmic relative to ball counts, and space complexity is constant/linear depending on whether in-place sorting is used.