LeetCode Problem Workspace
Sell Diminishing-Valued Colored Balls
Maximize total value by greedily selling diminishing-valued colored balls based on inventory and customer orders.
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Topics
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Code langs
3
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Practice Focus
Medium · Binary search over the valid answer space
Answer-first summary
Maximize total value by greedily selling diminishing-valued colored balls based on inventory and customer orders.
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This problem requires maximizing total value when selling diminishing-valued colored balls. Use binary search over the valid answer space and greedy strategies to ensure the most valuable balls are sold first. Optimizing the selling order and minimizing wasted value is key to solving this efficiently.
Problem Statement
You have an inventory of colored balls, and a customer wants a specific number of balls. The value of each ball decreases as you sell more of that color, meaning the first ball of a color is worth more than the subsequent ones. The goal is to sell the balls in such a way that maximizes the total value while fulfilling the customer's order.
You are given an array inventory, where inventory[i] represents the number of balls of the ith color. Additionally, you are given an integer orders, which is the total number of balls the customer wants. The task is to determine the maximum total value you can generate from selling exactly orders balls.
Examples
Example 1
Input: inventory = [2,5], orders = 4
Output: 14
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
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)
Solution Approach
Binary Search for Maximum Total Value
To solve this problem, we can use binary search over the valid range of total values. By evaluating the total value generated for different possible values of the highest priced ball, we can narrow down the most optimal number of balls to sell. The search is between 1 (the minimum price of any ball) and the highest value in the inventory.
Greedy Ball Selling Strategy
In each step, we want to greedily sell the most expensive available ball to maximize the total value. After selling a ball of a certain color, the value of the next ball of that color decreases, so it's essential to start selling the most abundant and valuable balls first.
Efficient Calculation of Value
For each potential maximum value (from binary search), calculate the total value by summing the highest possible values across all colors. We can use a heap or simple iteration to calculate how many balls can be sold from each color at the current price point.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity of this solution is dominated by the binary search, which takes O(log(max_inventory_value)), and the process of summing the total value for each search iteration, which takes O(n) where n is the number of colors in the inventory. Hence, the overall time complexity is O(n log(max_inventory_value)). Space complexity is O(n) due to the space required for storing inventory and intermediate calculations.
What Interviewers Usually Probe
- Checks if the candidate uses binary search effectively over the range of possible values.
- Evaluates the candidate's ability to implement a greedy strategy for maximizing value.
- Looks for the candidate’s ability to optimize space and time complexity for large inputs.
Common Pitfalls or Variants
Common pitfalls
- Failing to properly handle the diminishing value of balls as they are sold, which leads to incorrect total value calculations.
- Not using binary search effectively over the answer space, leading to suboptimal performance.
- Improperly calculating the total value from each color, potentially overestimating how many balls can be sold.
Follow-up variants
- What if the number of balls in inventory is fixed, but the number of orders varies?
- What if the customer orders a very small number of balls compared to the inventory?
- What if the inventory contains very large numbers of balls, requiring careful handling of big integer arithmetic?
FAQ
What is the optimal approach for solving the Sell Diminishing-Valued Colored Balls problem?
The optimal approach is to use binary search over the valid range of values and a greedy algorithm to sell the most valuable balls first. This allows us to maximize the total value while fulfilling the customer's orders.
How do I avoid running into performance issues with large inventories in this problem?
To avoid performance issues, ensure that the binary search is implemented efficiently and that the value calculations for each inventory color are done in linear time. The key is optimizing the binary search and minimizing unnecessary calculations.
What are the primary challenges in this problem?
The main challenge is handling the diminishing value of balls as they are sold, and efficiently calculating the total value with respect to the customer’s order.
How does binary search help in solving the Sell Diminishing-Valued Colored Balls problem?
Binary search helps by narrowing down the range of possible maximum values for the balls. This allows us to evaluate different scenarios quickly and find the optimal strategy for selling balls.
What kind of algorithmic pattern is most useful for this problem?
The problem combines binary search over the valid answer space with a greedy algorithm for maximizing the total value, making it essential to understand both patterns for an optimal solution.
Solution
Solution 1
#### Python3
class Solution:
def maxProfit(self, inventory: List[int], orders: int) -> int:
inventory.sort(reverse=True)
mod = 10**9 + 7
ans = i = 0
n = len(inventory)
while orders > 0:
while i < n and inventory[i] >= inventory[0]:
i += 1
nxt = 0
if i < n:
nxt = inventory[i]
cnt = i
x = inventory[0] - nxt
tot = cnt * x
if tot > orders:
decr = orders // cnt
a1, an = inventory[0] - decr + 1, inventory[0]
ans += (a1 + an) * decr // 2 * cnt
ans += (inventory[0] - decr) * (orders % cnt)
else:
a1, an = nxt + 1, inventory[0]
ans += (a1 + an) * x // 2 * cnt
inventory[0] = nxt
orders -= tot
ans %= mod
return ansContinue Topic
array
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