Introduction

In this blog post, we will explore how to solve the LeetCode problem "2861" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 <= 100 0 <= budget <= 108 composition.length == k composition[i].length == n 1 <= composition[i][j] <= 100 stock.length == cost.length == n 0 <= stock[i] <= 108 1 <= cost[i] <= 100

Explanation

Here's the solution to the problem:

• Binary Search: Use binary search to find the maximum number of alloys that can be created for each machine. The search space will be from 0 to a reasonably large number (e.g., 1e9) since we are trying to maximize the number of alloys.
• Cost Calculation: For each potential number of alloys, calculate the total cost required to create them using a given machine. This involves determining how much of each metal needs to be purchased (total required - existing stock) and multiplying by the cost per unit.

• Optimization: Iterate through each machine and perform the binary search to find the maximum number of alloys possible with that machine. Keep track of the overall maximum across all machines.

• Runtime Complexity: O(k n log(10^9)), where k is the number of machines and n is the number of metal types. Storage Complexity: O(1).

Code

    def max_alloys(n: int, k: int, budget: int, composition: list[list[int]], stock: list[int], cost: list[int]) -> int:
    """
    Calculates the maximum number of alloys that can be created within the given budget.

    Args:
        n: The number of different types of metals.
        k: The number of machines.
        budget: The budget.
        composition: A 2D array where composition[i][j] is the amount of metal of type j required for machine i.
        stock: An array where stock[i] is the initial stock of metal type i.
        cost: An array where cost[i] is the cost of purchasing one unit of metal type i.

    Returns:
        The maximum number of alloys that can be created.
    """

    def can_create(machine_index: int, num_alloys: int) -> bool:
        """
        Checks if a given number of alloys can be created with a specific machine within the budget.

        Args:
            machine_index: The index of the machine.
            num_alloys: The number of alloys to create.

        Returns:
            True if the alloys can be created within the budget, False otherwise.
        """
        total_cost = 0
        for i in range(n):
            required = composition[machine_index][i] * num_alloys
            purchase = max(0, required - stock[i])
            total_cost += purchase * cost[i]
        return total_cost <= budget

    max_alloys_possible = 0
    for machine_index in range(k):
        low = 0
        high = 2 * 10**8  # Reasonable upper bound on the number of alloys
        best_count = 0
        while low <= high:
            mid = (low + high) // 2
            if can_create(machine_index, mid):
                best_count = mid
                low = mid + 1
            else:
                high = mid - 1
        max_alloys_possible = max(max_alloys_possible, best_count)

    return max_alloys_possible