Leetcode 1648: Sell Diminishing-Valued Colored Balls
You are given an inventory of different colored balls. The customer wants to buy a specific number of balls, and each ball has a value based on how many of that color are still available. Calculate the maximum total value you can obtain after fulfilling the customer’s order. The result should be returned modulo 10^9 + 7.
Problem
Approach
Steps
Complexity
Input: An array of integers representing the inventory of different colored balls and an integer representing the total number of orders the customer wants.
Example: inventory = [4, 7], orders = 5
Constraints:
• 1 <= inventory.length <= 10^5
• 1 <= inventory[i] <= 10^9
• 1 <= orders <= min(sum(inventory[i]), 10^9)
Output: Return the maximum total value that can be attained after selling the orders of colored balls. The result should be modulo 10^9 + 7.
Example: Output: 25
Constraints:
Goal: Maximize the value obtained by selling the balls in the most valuable order, starting with the most abundant colors.
Steps:
• Sort the inventory in descending order to maximize the value for each ball sold.
• Sell the balls from the most abundant colors first, decreasing the value of each ball with each sale.
• Calculate the total value after fulfilling the number of orders, ensuring no more balls are sold than are available.
• Return the total value modulo 10^9 + 7.
Goal: The input constraints are designed to ensure the solution can handle large inventories and orders efficiently.
Steps:
• 1 <= inventory.length <= 10^5
• 1 <= inventory[i] <= 10^9
• 1 <= orders <= min(sum(inventory[i]), 10^9)
Assumptions:
• The input will always contain valid values for inventory and orders, with no need to handle negative values or invalid input.
• Input: inventory = [4, 7], orders = 5
• Explanation: First, sell the highest valued ball (from the largest inventory) to maximize the value. Then, sell balls from the next inventory, continuing the process until all orders are fulfilled.
Approach: The approach involves sorting the inventory, then selling balls in the most valuable order while keeping track of the total value and reducing the value of the balls with each sale.
Observations:
• To maximize the total value, sell balls starting from the color with the most balls in stock.
• We need to consider both the value of the balls and the decreasing stock while fulfilling the order.
Steps:
• Sort the inventory array in descending order to prioritize selling from the most abundant stock.
• Iterate through the sorted inventory, selling balls while calculating their cumulative value.
• Keep track of how many orders are left to process and reduce the stock as balls are sold.
• After fulfilling all orders, return the total value modulo 10^9 + 7.
Empty Inputs:
• Ensure there is always at least one color of balls in the inventory and at least one order.
Large Inputs:
• The solution must efficiently handle large values for inventory sizes and ball counts.
Special Values:
• When the orders are fewer than the number of balls available, we need to ensure we only sell the required number of balls.
Constraints:
• Ensure the solution works efficiently within the given input size limits.
int mod = (int) 1e9 + 7;
int maxProfit(vector<int>& inv, int orders) {
long n = inv.size(), res = 0;
sort(inv.rbegin(), inv.rend());
for(int i = 0, col = 1; i < n && orders > 0; i++, col++) {
long cur = inv[i], prv = i + 1 < n? inv[i + 1]: 0;
long depth = min((long)orders/col, (long)cur - prv);
orders -= depth * col;
res = (res + ((cur * (cur + 1) - (cur - depth) * (cur - depth + 1)) / 2 * col)) % mod;
if(cur - prv > depth) {
res = (res + orders * (cur - depth)) % mod;
break;
}
}
/*
There is a value k, for which all the balls above this value are sold.
*/
return res;
}
1 : Variable Initialization
int mod = (int) 1e9 + 7;
Defines a constant to handle large numbers with modulo arithmetic.
2 : Function Definition
int maxProfit(vector<int>& inv, int orders) {
Begins the function definition for calculating the maximum profit from inventory sales.
3 : Sorting
sort(inv.rbegin(), inv.rend());
Sorts the inventory in descending order to prioritize higher profits.
4 : Loop
for(int i = 0, col = 1; i < n && orders > 0; i++, col++) {
Iterates through inventory items, calculating profit and reducing orders.
5 : Calculation
long cur = inv[i], prv = i + 1 < n? inv[i + 1]: 0;
Determines the current inventory level and the next level for comparison.
6 : Calculation
long depth = min((long)orders/col, (long)cur - prv);
Calculates the depth of items to sell at the current level.
7 : Update
orders -= depth * col;
Decreases the remaining order count by the number of items sold.
8 : Calculation
res = (res + ((cur * (cur + 1) - (cur - depth) * (cur - depth + 1)) / 2 * col)) % mod;
Updates the result with the profit from selling the current level's items.
9 : Condition
if(cur - prv > depth) {
Checks if additional items remain at the current level after depth sales.
10 : Update
res = (res + orders * (cur - depth)) % mod;
Handles remaining orders by selling items at the current level.
11 : Break
break;
Exits the loop once all orders are fulfilled.
12 : Return
return res;
Returns the total profit calculated.
Best Case: O(n log n)
Average Case: O(n log n)
Worst Case: O(n log n)
Description: Sorting the inventory takes O(n log n) time, where n is the number of distinct colors in the inventory.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the storage required for the inventory and the sorted list of inventories.
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