You are the owner of a company that creates alloys using various types of metals. There are n different types of metals available, and you have access to k machines that can be used to create alloys. Each machine requires a specific amount of each metal type to create an alloy.
For the ith machine to create an alloy, it needs composition[i][j] units of metal of type j. Initially, you have stock[i] units of metal type i, and purchasing one unit of metal type i costs cost[i] coins.
Given integers n, k, budget, a 1-indexed 2D array composition, and 1-indexed arrays stock and cost, your goal is to maximize the number of alloys the company can create while staying within the budget of budget coins.
All alloys must be created with the same machine.
Return the maximum number of alloys that the company can create.
Example 1:
Input: n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,0], cost = [1,2,3] Output: 2 Explanation: It is optimal to use the 1st machine to create alloys. To create 2 alloys we need to buy the: - 2 units of metal of the 1st type. - 2 units of metal of the 2nd type. - 2 units of metal of the 3rd type. In total, we need 2 * 1 + 2 * 2 + 2 * 3 = 12 coins, which is smaller than or equal to budget = 15. Notice that we have 0 units of metal of each type and we have to buy all the required units of metal. It can be proven that we can create at most 2 alloys.
Example 2:
Input: n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,100], cost = [1,2,3] Output: 5 Explanation: It is optimal to use the 2nd machine to create alloys. To create 5 alloys we need to buy: - 5 units of metal of the 1st type. - 5 units of metal of the 2nd type. - 0 units of metal of the 3rd type. In total, we need 5 * 1 + 5 * 2 + 0 * 3 = 15 coins, which is smaller than or equal to budget = 15. It can be proven that we can create at most 5 alloys.
Example 3:
Input: n = 2, k = 3, budget = 10, composition = [[2,1],[1,2],[1,1]], stock = [1,1], cost = [5,5] Output: 2 Explanation: It is optimal to use the 3rd machine to create alloys. To create 2 alloys we need to buy the: - 1 unit of metal of the 1st type. - 1 unit of metal of the 2nd type. In total, we need 1 * 5 + 1 * 5 = 10 coins, which is smaller than or equal to budget = 10. It can be proven that we can create at most 2 alloys.
Constraints:
1 <= n, k <= 1000 <= budget <= 108composition.length == kcomposition[i].length == n1 <= composition[i][j] <= 100stock.length == cost.length == n0 <= stock[i] <= 1081 <= cost[i] <= 100This approach considers maximum potential units of alloys we can produce using binary search over the number of alloys. For each possible number of alloys in the search space, we simulate the verification: we calculate the total cost needed, considering the metal composition and available stock. If the cost is within the budget, we attempt more alloys, otherwise, we reduce the count.
The Python solution uses a binary search approach to maximize the number of alloys produced. We iterate over each machine to evaluate the maximum possible number of alloys that can be produced with the given budget. The binary search checks the feasibility of producing a specific number of alloys and adjusts based on whether the calculated cost is within the budget limits.
JavaScript
Time Complexity: O(k * log(max_possible_alloys) * n) where max_possible_alloys is a heuristic upper bound.
Space Complexity: O(1) since it uses constant extra space.
This approach uses a greedy strategy for each machine. It computes the number of alloys directly by going through each metal type and simulates the cost accumulation based on the composition needs. The aim is to reach the maximum number directly without extensive binary search, ensuring we don't exceed the budget.
This C implementation of a greedy approach iterates over each machine, progressively calculating the total cost to determine the number of alloys while remaining within the budget. It checks each possible increase in the number of alloys one by one, adjusting based on metal requirements.
Time Complexity: O(n * k * budget/cost_min) where cost_min is the minimum cost of any metal per unit. It's less optimal for larger budgets.
Space Complexity: O(1)
| Approach | Complexity |
|---|---|
| Binary Search with Simulation | Time Complexity: O(k * log(max_possible_alloys) * n) where max_possible_alloys is a heuristic upper bound. Space Complexity: O(1) since it uses constant extra space. |
| Greedy Simulation Approach | Time Complexity: O(n * k * budget/cost_min) where cost_min is the minimum cost of any metal per unit. It's less optimal for larger budgets. Space Complexity: O(1) |
Leetcode 149 - Maximum Points on a Line - Python • NeetCodeIO • 25,713 views views
Watch 9 more video solutions →Practice Maximum Number of Alloys with our built-in code editor and test cases.
Practice on FleetCodePractice this problem
Open in Editor