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Maximum Number of Alloys

Number: 3095

Difficulty: Medium

Paid? No

Companies: MathWorks, Amazon

Problem Description

You are given n different types of metals, k machines, and a budget. Each machine has a fixed composition requirement for each metal to produce one alloy. You start with a given stock of metals and can buy additional metal units at given costs. All alloys must be created using the same machine, and your goal is to maximize the number of alloys produced while staying within the budget.

Key Insights

  • We must select one machine to produce all alloys.
  • For each machine, determine if it is possible to produce X alloys by calculating the total cost of additional metal required (if needed) beyond the stock.
  • Use binary search on the potential number of alloys since the feasibility check (for producing X alloys) is monotonic: if X alloys can be produced, then producing any number less than X is also possible.
  • Iterate through all machines and compute the maximum number of alloys that can be produced under the budget.

Space and Time Complexity

Time Complexity: O(k * n * log M) where k is the number of machines, n is the number of metal types, and M is the maximum possible number of alloys (derived from budget and composition estimates). Space Complexity: O(1) additional space (excluding input storage).

Solution

For each machine, perform a binary search to determine the maximum number of alloys it is possible to manufacture within the budget. Given a candidate number mid (i.e. mid alloys), for each metal type j, compute the total required units as mid * composition[i][j]. If the available stock is less than this, calculate the units needed to be purchased and multiply by the cost of metal j. Sum these values across all metals; if the sum is within the budget, then producing mid alloys is feasible. Use binary search to find the maximum mid for which the condition holds for each machine, and then select the best answer among all machines.

Code Solutions

# Python solution with detailed comments

def maximumAlloys(n, k, budget, composition, stock, cost):
    # Function to check if it's possible to make 'x' alloys with the given machine configuration (index machine_idx)
    def can_make(x, machine_idx):
        total_cost = 0
        # Iterate over each metal type j
        for j in range(n):
            # Calculate total required units of metal j for x alloys
            required = x * composition[machine_idx][j]
            # Calculate extra units needed if available stock is not enough
            extra = max(0, required - stock[j])
            # Calculate coins needed for this metal
            total_cost += extra * cost[j]
            # Early exit if cost already exceeds budget
            if total_cost > budget:
                return False
        return total_cost <= budget

    max_alloys = 0
    # For each machine, perform binary search to find the maximum alloys possible
    for machine_idx in range(k):
        low, high = 0, budget + max(stock)  # upper bound can be set in a looser manner, budget units might create alloys
        while low <= high:
            mid = (low + high) // 2
            if can_make(mid, machine_idx):
                low = mid + 1
            else:
                high = mid - 1
        # high is the maximum alloys for this machine
        max_alloys = max(max_alloys, high)
    return max_alloys

# Example usage:
n = 3
k = 2
budget = 15
composition = [[1,1,1],[1,1,10]]
stock = [0,0,0]
cost = [1,2,3]
print(maximumAlloys(n, k, budget, composition, stock, cost))  # Expected output: 2
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