Problem Description

In this problem, you run a company that specializes in creating alloys from various metals using machines. You have a selection of n different metal types and k machines to use for manufacturing alloys. Each machine has a specific recipe, meaning it requires certain amounts of each metal type to produce one batch of alloy. These recipes are represented by a 2D array composition, where composition[i][j] is the number of units of metal type j required by machine i to create an alloy.

You start with a finite stock of metal units, indicated by a 1-indexed array stock, where stock[i] represents how many units of metal type i you currently have. If you need more of any metal, you can buy additional units, where the cost to purchase one unit of metal type i is cost[i] coins.

Your objective is to maximize the number of alloys you can produce using any one machine, without exceeding your budget of coins. Importantly, all produced alloys must come from the same machine. You're tasked to determine the maximum number of alloys that can be created given your initial resources and financial constraints.

Intuition

To solve this problem, we need to determine, for each machine, how many batches of alloy we can produce given our stock of metals, the ability to buy more metals, and our budget. We repeat this calculation for every machine and then choose the maximum value across all machines.

The intuition behind the solution is to utilize binary search to efficiently find the maximum number of alloys we can produce per machine without surpassing the budget. This is because, for a given machine, if we can produce x alloys, we can certainly produce any number of alloys less than x. Conversely, if we cannot afford to produce x alloys, we cannot afford any number greater than x either. This creates a perfect scenario for using binary search, which is a method for finding an item in a sorted series with an "ordered" relationship by repeatedly halving the search space.

For each machine, we have the following steps:

1. Initialize the search space for the number of alloys (l as the low end and r as the high end), which is determined by the budget and the existing stock of the first metal type.
2. Perform binary search to find the maximum number of alloys:
   • Calculate the midpoint mid between the current low l and high r values.
   • Compute the cost needed to produce mid alloys using the current machine, considering our stock and the prices for additional metals.
   • Check if we can afford this cost with our budget:
     • If yes, we update the low end of our search (l) to mid, as we might be able to produce even more alloys.
     • If no, we update the high end of our search (r) to mid - 1, as we cannot afford mid alloys and need to try a smaller number.
   • Repeat this process until l and r converge to the highest affordable number of alloys for this machine.
3. Keep track of the maximum number of alloys found across all machines.
4. Return the maximum as the final answer.

Learn more about Binary Search patterns.

Solution Approach

The implementation of the solution can be broken down into multiple parts, each aligned with the binary search algorithm used to find the maximum number of alloys that can be produced by a single machine within the budget.

1. Outer Loop - Iterate Over Each Machine: The solution begins with an outer loop for c in composition: that iterates over each machine's alloy composition. The variable c refers to a specific machine's requirement of metal types from the 2D array composition.

2. Binary Search Initialization: For each machine, we initialize the binary search range with l, r = 0, budget + stock[0]. This range is a logical start because the number of alloys that can be possibly created is between 0 and some upper limit proportional to the budget and the initial stock.

3. Binary Search Loop: Within this loop, the algorithm uses binary search to find the optimal number of alloys that can be created with the current machine. The loop while l < r: continues reducing the search space until it converges on the maximum number of alloys.

4. Midpoint and Cost Calculation:

   • The midpoint mid is calculated using (l + r + 1) >> 1, which is equivalent to (l + r + 1) / 2 but faster as it employs bit shifting.
   • The cost required to create mid alloys is computed with s = sum(max(0, mid * x - y) * z for x, y, z in zip(c, stock, cost)), which calculates the costs taking into consideration the stock already available and the additional purchases required for the shortfall.

5. Affordability Check:

   • If the calculated cost s is within the budget (if s <= budget:), the algorithm updates the lower bound of the search space l = mid because it is still possible to create mid alloys or potentially more.
   • If the cost s exceeds the budget (else:), the algorithm adjusts the upper bound r = mid - 1 since mid is too high of a number of alloys to afford.

6. Finding the Maximum:

   • After the binary search converges for a machine, we determine the maximum number of alloys that can be created by this machine without exceeding the budget, which will be the value of l.
   • Then, we update the overall answer ans = max(ans, l), which keeps track of the maximum number of alloys across all machines.

7. Returning Result: After the loop has processed all machines, return ans yields the final result, which is the maximum number of alloys that can be produced by any one machine within the budget.

This algorithm leverages the binary search pattern due to the sorted nature of the problem space — there's a clear ordering in the problem setup that if x number of alloys can be produced, then any number less than x can also be produced (and vice versa). The binary search pattern effectively narrows down the possibilities, leading to an efficient solution.

Example Walkthrough

Let's walk through a small example to illustrate how the solution approach would be applied. Imagine we have the following input:

• Metal types: n = 2 (two types of metal)
• Machines: k = 1 (one machine)
• Composition for the machine: composition = [[3, 2]] (requires 3 units of metal type 1 and 2 units of metal type 2 for one batch)
• Initial stock: stock = [5, 4] (5 units of metal 1 and 4 units of metal 2)
• Costs to purchase metals: cost = [2, 1] (2 coins per unit of metal 1, 1 coin per unit of metal 2)
• Budget: budget = 10 coins

We want to calculate the maximum number of alloys we can produce using this machine without exceeding our budget.

Outer Loop - Iterate Over Each Machine

We only have one machine, so our loop will run a single iteration. For our machine, the composition is c = [3, 2].

Binary Search Initialization

We initialize our binary search range with l, r = 0, budget + stock[0]. Since the budget is 10 and we have 5 units of metal 1, our range is l, r = 0, 15.

Binary Search Loop

We now enter the binary search loop:

1. First Iteration:

   • Calculate midpoint mid = (l + r + 1) >> 1 which is (0 + 15 + 1) >> 1 = 8.
   • Calculate cost s: we would need 8 * 3 = 24 units of metal 1 and 8 * 2 = 16 units of metal 2. We have 5 and 4 units respectively, so we need to purchase 19 more units of metal 1 and 12 units of metal 2, totaling a cost of 19 * 2 + 12 * 1 = 38 + 12 = 50 coins.
   • Since 50 coins exceeds our budget of 10 coins, we set r = mid - 1, so r = 7.

2. Second Iteration:

   • Midpoint mid = (l + r + 1) >> 1 which is (0 + 7 + 1) >> 1 = 4.
   • Calculate cost s: we would need 4 * 3 = 12 units of metal 1 and 4 * 2 = 8 units of metal 2. This would require buying 7 units of metal 1 and 4 units of metal 2 for 7 * 2 + 4 * 1 = 14 + 4 = 18 coins.
   • Since 18 coins still exceed our budget, we update r = mid - 1, so r = 3.

3. Third Iteration:

   • Midpoint mid = (l + r + 1) >> 1 which is (0 + 3 + 1) >> 1 = 2.
   • Calculate cost s: Total units needed are 2 * 3 = 6 for metal 1 and 2 * 2 = 4 for metal 2. We need to buy 1 additional unit of metal 1 and no additional units of metal 2, totaling a cost of 1 * 2 + 0 * 1 = 2 coins.
   • This is affordable, so we can update l = mid, which will be l = 2.

The lower and upper bounds of our search l and r have now converged, indicating that with our budget of 10 coins, we can afford to produce up to 2 batches of alloys with the given machine and available stock.

Maximum Number of Alloys

Since we only have a single machine in this example, the maximum number of alloys is the same as what we've determined for the machine: 2 batches.

The number of alloys we can produce by this one machine within our budget of 10 coins is 2, which would be the final output of the algorithm for this example.

Solution Implementation

Time and Space Complexity

The time complexity of the given code can be analyzed by breaking it down into its fundamental operations. The outer loop iterates over each composition:

for c in composition:

Since there are n compositions, the outer loop runs n times. Inside this loop, there is a binary search:

while l < r:
    mid = (l + r + 1) >> 1
    s = sum(max(0, mid * x - y) * z for x, y, z in zip(c, stock, cost))
    if s <= budget:
        l = mid
    else:
        r = mid - 1

Binary search runs in O(log(budget + stock[0])) time because it repeatedly divides the search interval in half until it becomes too small to divide (in this case, the initial range is from 0 to budget + stock[0]). Inside the binary search, there is a sum operation processed in linear time with respect to k, which is the length of the composition, stock, and cost lists.

Combined, the time complexity for the above operation is O(k), because it iterates over the k elements to calculate s. Since the binary search is within a loop over n compositions, we multiply the above to get the total time complexity:

O(n * k * log(budget + stock[0]))

For space complexity, the code uses a constant amount of extra space, excluding the input data structures. There are no dynamically sized data structures that grow with the size of the input. Therefore, the space complexity is:

O(1)

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