Problem Description

You have different colored balls in your inventory, and a customer wants to buy a certain number of balls. The interesting part is how the balls are valued.

Each ball's value equals the current count of balls of that color in your inventory. For example, if you have 6 yellow balls:

• The first yellow ball sells for 6
• After selling it, you have 5 yellow balls left
• The second yellow ball sells for 5
• The third sells for 4, and so on

You're given:

• inventory: an array where inventory[i] represents how many balls of color i you initially have
• orders: the total number of balls the customer wants to buy

You can sell balls of any color in any order you choose. Your goal is to maximize the total value you receive from selling exactly orders balls.

Since the answer might be very large, return it modulo 10^9 + 7.

The key insight is that to maximize profit, you should always sell the most valuable balls first. Since a ball's value equals its current count, you want to sell from the colors with the highest counts, reducing them gradually to keep the values as high as possible throughout the selling process.

For example, if inventory = [2, 5] and orders = 4:

• You could sell balls valued at: 5, 4, 3, 2 (selling from the second color until they're equal, then alternating)
• Total value: 5 + 4 + 3 + 2 = 14

Intuition

Since we want to maximize profit and each ball's value equals its current count, we should always sell the highest-valued balls first. This creates a greedy strategy: keep selling from whichever color currently has the most balls.

Think of it like this - if we have colors with counts [2, 5, 8], we'd want to sell from the color with 8 balls first (getting value 8), then from 7 (getting value 7), and so on. We're essentially "shaving off" the peaks to keep all values as high as possible.

The key observation is that we'll be selling balls in layers. Imagine the inventory counts as bars in a histogram. We want to cut horizontal slices from the top:

• First, we bring down the tallest bar(s) to match the second tallest
• Then we bring those down together to match the third tallest
• And so on

For efficiency, instead of selling balls one by one, we can calculate how many balls we'd sell in each "layer" at once. If we have multiple colors at the same height, we sell from all of them simultaneously.

For example, with inventory = [3, 5]:

• We first sell 2 balls from the second color (values 5 and 4), bringing it down to 3
• Now both colors have 3 balls, so we sell from both alternately

The mathematical insight is that when selling x balls from a color that currently has n balls, we're selling balls valued at n, n-1, n-2, ..., n-x+1. This is an arithmetic sequence with sum (first + last) * count / 2 = (n + (n-x+1)) * x / 2.

By sorting the inventory in descending order and processing these layers efficiently, we can calculate the maximum profit without simulating each individual sale.

Learn more about Greedy, Math, Binary Search, Sorting and Heap (Priority Queue) patterns.

Solution Approach

The implementation follows a greedy approach by processing inventory levels from highest to lowest:

1. Sort the inventory in descending order: This allows us to easily identify and process the highest-valued balls first.

2. Process balls in layers: We iterate through the sorted inventory, selling balls in "layers" - bringing down the tallest bars to match the next level.

3. Key variables:

   • i: tracks how many colors are at the current highest level
   • inventory[0]: always represents the current highest count (since we're modifying it in-place)
   • nxt: the next lower level we need to bring our current colors down to

4. Main loop logic:

   • First, count how many colors (cnt) have the same count as inventory[0]
   • Determine the next level (nxt) - either the next different value in inventory or 0
   • Calculate how many balls would be sold if we bring all cnt colors from inventory[0] down to nxt: tot = cnt * (inventory[0] - nxt)

5. Two cases for selling:

   Case 1: We have more balls to sell than orders remaining (tot > orders):

   • Calculate how many complete levels we can drop: decr = orders // cnt
   • Use arithmetic sequence formula to calculate value: sum from (inventory[0] - decr + 1) to inventory[0], multiplied by cnt colors
   • Formula: (first + last) * count / 2 * cnt = (a1 + an) * decr / 2 * cnt
   • Handle remaining balls: (inventory[0] - decr) * (orders % cnt)

   Case 2: We can sell all balls in this layer (tot <= orders):

   • Sell all balls from inventory[0] down to nxt + 1 for all cnt colors
   • Use arithmetic sequence formula: (nxt + 1 + inventory[0]) * x / 2 * cnt
   • Update inventory[0] = nxt for the next iteration

6. Modulo operation: Apply mod = 10^9 + 7 to keep the answer within bounds.

The algorithm efficiently calculates the maximum profit by:

• Avoiding individual ball sales through batch calculations
• Using arithmetic sequence formulas for sum calculations
• Processing colors at the same level together
• Time complexity: O(n log n) for sorting, where n is the number of colors

Example Walkthrough

Let's walk through a concrete example with inventory = [2, 8, 4, 10, 6] and orders = 20.

Step 1: Sort inventory in descending order

• inventory = [10, 8, 6, 4, 2]
• We'll process from highest to lowest values

Step 2: Process first layer (10 → 8)

• Current highest: inventory[0] = 10
• Count of colors at this level: cnt = 1 (only one color has 10 balls)
• Next level: nxt = 8 (the next highest count)
• Total balls if we bring 10 down to 8: tot = 1 * (10 - 8) = 2
• Since tot = 2 <= orders = 20, we sell all 2 balls
• Value gained: 10 + 9 = 19 (using formula: (8+1 + 10) * 2 / 2 * 1 = 19)
• Update: inventory[0] = 8, orders = 18, result = 19

Step 3: Process second layer (8 → 6)

• Current highest: inventory[0] = 8
• Count of colors at this level: cnt = 2 (two colors have 8 balls: original 10 and 8)
• Next level: nxt = 6
• Total balls if we bring both down to 6: tot = 2 * (8 - 6) = 4
• Since tot = 4 <= orders = 18, we sell all 4 balls
• Value gained: (8 + 7) * 2 = 30 (two colors selling 8 and 7 each)
• Using formula: (6+1 + 8) * 2 / 2 * 2 = 30
• Update: inventory[0] = 6, orders = 14, result = 49

Step 4: Process third layer (6 → 4)

• Current highest: inventory[0] = 6
• Count of colors at this level: cnt = 3 (three colors have 6 balls)
• Next level: nxt = 4
• Total balls if we bring all three down to 4: tot = 3 * (6 - 4) = 6
• Since tot = 6 <= orders = 14, we sell all 6 balls
• Value gained: (6 + 5) * 3 = 33 (three colors selling 6 and 5 each)
• Using formula: (4+1 + 6) * 2 / 2 * 3 = 33
• Update: inventory[0] = 4, orders = 8, result = 82

Step 5: Process fourth layer (4 → 2)

• Current highest: inventory[0] = 4
• Count of colors at this level: cnt = 4 (four colors have 4 balls)
• Next level: nxt = 2
• Total balls if we bring all four down to 2: tot = 4 * (4 - 2) = 8
• Since tot = 8 == orders = 8, we sell exactly all remaining orders!
• Value gained: (4 + 3) * 4 = 28 (four colors selling 4 and 3 each)
• Using formula: (2+1 + 4) * 2 / 2 * 4 = 28
• Update: orders = 0, result = 110

Final Answer: 110

The key insight is that we're always selling the highest-valued balls by processing inventory levels from top to bottom, calculating bulk sales using arithmetic sequence formulas rather than selling individual balls.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n log n + n * k) where n is the length of inventory and k is the number of distinct price levels we process.

• Sorting the inventory takes O(n log n)
• In the worst case, we process each distinct value level in the inventory. For each level:
  • The inner while loop finds all items at the current maximum level, which collectively across all iterations examines each element once: O(n) total
  • Each iteration processes a batch of orders, reducing at least one element's value
• The number of distinct price levels k is bounded by the maximum value in inventory, but in practice, we process groups of items together
• Overall: O(n log n + n * k) where k ≤ max(inventory)

Space Complexity: O(1) if we don't count the space used for sorting (or O(log n) for the sorting stack space in Python's Timsort)

• The algorithm uses only a constant amount of extra variables: mod, ans, i, n, nxt, cnt, x, tot, decr, a1, an
• The inventory array is modified in-place
• No additional data structures are created

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow in Arithmetic Calculations

The most critical pitfall in this problem is integer overflow when calculating the sum using the arithmetic sequence formula. Even though Python handles arbitrary precision integers, in languages like Java or C++, or when dealing with very large numbers, the intermediate calculations can overflow.

The Problem: When calculating (first_price + last_price) * count // 2 * items_at_current_price, the multiplication operations can produce values exceeding the integer limit before applying the modulo operation.

Example of Overflow:

# Problematic calculation
total_profit += (first_price + last_price) * full_levels_to_sell // 2 * items_at_current_price

If first_price = 10^9, last_price = 10^9, and items_at_current_price = 10^5, the intermediate result would be astronomical.

Solution: Apply modulo operations at each step and use modular arithmetic properties:

# Safe calculation with modulo at each step
sum_value = ((first_price + last_price) % MOD * full_levels_to_sell % MOD) % MOD
sum_value = (sum_value * pow(2, MOD - 2, MOD)) % MOD  # Modular division by 2
total_profit = (total_profit + sum_value * items_at_current_price % MOD) % MOD

Or alternatively, rearrange the formula to minimize large intermediate values:

# Rearrange to avoid large intermediate products
if full_levels_to_sell % 2 == 0:
    total_profit += ((first_price + last_price) % MOD) * (full_levels_to_sell // 2 % MOD) % MOD * items_at_current_price % MOD
else:
    total_profit += ((first_price + last_price) * full_levels_to_sell % MOD) * pow(2, MOD - 2, MOD) % MOD * items_at_current_price % MOD

2. Incorrect Handling of Edge Cases

The Problem: Not properly handling cases where:

• All inventory counts are the same
• Orders exceed total available balls
• Single color in inventory

Solution: Add boundary checks:

# Add validation at the start
if not inventory or orders == 0:
    return 0

# Ensure we don't process beyond available inventory
total_available = sum(inventory)
orders = min(orders, total_available)

3. Off-by-One Errors in Price Calculations

The Problem: Confusion about inclusive vs exclusive ranges when calculating sums. For example, when bringing prices from inventory[0] down to next_price_level, should we include next_price_level in our sum?

Incorrect:

# This would include next_price_level in the sum
total_profit += (next_price_level + inventory[0]) * (inventory[0] - next_price_level + 1) // 2

Correct:

# We sell from inventory[0] down to (next_price_level + 1)
# So the range is [next_price_level + 1, inventory[0]] inclusive
first_price = next_price_level + 1
last_price = inventory[0]
count = last_price - first_price + 1  # This should equal price_difference
total_profit += (first_price + last_price) * count // 2 * items_at_current_price

4. Not Updating the Inventory Correctly

The Problem: After processing a batch of sales, forgetting to update the inventory state for the next iteration, or updating it incorrectly.

Solution: Ensure inventory[0] is properly updated to reflect the new highest value:

# After selling all units down to next_price_level
inventory[0] = next_price_level  # Critical update for next iteration

5. Inefficient Implementation Using Priority Queue

The Problem: Using a max-heap and popping/pushing individual elements, which seems intuitive but is inefficient:

# Inefficient approach - O(orders * log n)
import heapq
heap = [-x for x in inventory]
heapq.heapify(heap)
total = 0
for _ in range(orders):
    val = -heapq.heappop(heap)
    total = (total + val) % MOD
    heapq.heappush(heap, -(val - 1))

Solution: Use the batch processing approach shown in the main solution, which groups same-valued items and processes them together, achieving O(n log n) complexity.