Maximum Number of Alloys
Medium
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Topics:
Binary Search
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] <= 100
Solution
Clarifying Questions
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
- What are the upper bounds for `n`, `k`, `budget`, and the values in the `composition` and `stock` arrays? What are their data types (integers, floats)?
- Can the values in the `composition` and `stock` arrays be zero?
- Is it possible that no alloys can be produced with the given budget and resources? If so, what should I return in that case?
- Is there a minimum amount of each metal type required for a valid alloy, or can the composition array contain zeros and still be considered a valid alloy?
- Is it possible to produce a fractional number of alloys, or must the output be a whole number (integer)? If it must be a whole number, should I return the floor or ceiling of the maximum possible number of alloys?
Brute Force Solution
Approach
The brute force strategy for maximizing alloys involves systematically trying out every possible quantity of alloys we can make. For each quantity, we check if we have enough metal to make it using any of the available machines.
Here's how the algorithm would work step-by-step:
- Start by assuming we can make zero alloys.
- Then, check if we can make one alloy using any of the machines.
- Next, check if we can make two alloys, again using any of the machines.
- Continue increasing the number of alloys we attempt to make, one at a time.
- For each number of alloys, go through each machine and determine if it's possible to make that many alloys using that specific machine, given the available metal.
- If a machine can make that many alloys, that number of alloys is a possible solution.
- Stop when you find a number of alloys that can't be made by any machine because we've run out of available metal or resources.
- The largest number of alloys we successfully made using any machine is the answer.
Code Implementation
def maximum_number_of_alloys_brute_force(number_of_machines, available_metal, cost_of_alloys):
maximum_alloys = 0
for number_of_alloys in range(1, 10000):
possible_to_make = False
for machine_index in range(number_of_machines):
metal_needed = 0
# Calculate the total metal needed for this machine
for metal_type_cost in cost_of_alloys[machine_index]:
metal_needed += metal_type_cost * number_of_alloys
# Check if the machine can produce the alloys with available resources
if metal_needed <= available_metal:
possible_to_make = True
break
#If we can make alloys increase max count
if possible_to_make:
maximum_alloys = number_of_alloys
else:
# If can't make any more alloys, return
break
return maximum_alloys
Big(O) Analysis
Time Complexity
O(m * n * k) – The described brute-force approach iterates through possible numbers of alloys, let's say up to 'm'. For each quantity of alloys, it iterates through 'n' machines. For each machine, it needs to check if it can produce the target number of alloys given available metal, which involves iterating through 'k' types of metal. Therefore, the time complexity is O(m * n * k), where m is the maximum number of alloys to check, n is the number of machines, and k is the number of metal types. In the worst case, m could be significantly large depending on the constraints.
Space Complexity
O(1) – The brute force algorithm described checks each possible alloy quantity incrementally without persistently storing intermediate results. It iterates through machines and calculates if the current alloy quantity is feasible. Since it only needs to store a few variables like the number of alloys to try and potentially a boolean flag to indicate feasibility for a particular machine, the auxiliary space required remains constant, independent of the input size (number of machines, available metal, metal composition of alloys).
Optimal Solution
Approach
The key to this problem is recognizing we can efficiently search for the maximum number of alloys. Instead of trying every possible number of alloys, we can use a technique that quickly narrows down the search using a binary-search like approach.
Here's how the algorithm would work step-by-step:
- Realize that the number of alloys we can make has a minimum (probably zero) and a maximum (based on the available machines and metal).
- Guess a number of alloys between the minimum and maximum. Let's call it the 'target number'.
- Check if we can actually make the 'target number' of alloys with the available resources.
- To check, see if we have enough metal for each machine to produce the 'target number' of alloys.
- If we have enough metal for all machines, then this 'target number' might be too low. We can try a higher 'target number' next time.
- If we don't have enough metal for all machines, then this 'target number' is too high. We should try a lower 'target number' next time.
- Keep guessing and checking, each time halving the range of possible numbers, until we find the largest number of alloys we can make.
Code Implementation
def max_number_of_alloys(number_of_machines,
budget,
cost_per_alloy,
metal_needed,
metal_available):
low_alloys = 0
high_alloys = 10**9 # A large upper bound
max_alloys = 0
while low_alloys <= high_alloys:
target_alloys = (low_alloys + high_alloys) // 2
can_make_alloys = True
# Check each machine to see if it can produce target_alloys
for machine_index in range(number_of_machines):
cost_for_machine = (target_alloys *
metal_needed[machine_index] -
metal_available[machine_index])
if cost_for_machine > 0:
cost_for_machine *= cost_per_alloy[machine_index]
else:
cost_for_machine = 0
budget -= cost_for_machine
# If the budget goes negative, we cannot produce target_alloys
if budget < 0:
can_make_alloys = False
# Reset budget for the next iteration
budget = budget
for machine_index in range(number_of_machines):
cost_for_machine = (target_alloys *
metal_needed[machine_index] -
metal_available[machine_index])
if cost_for_machine > 0:
cost_for_machine *= cost_per_alloy[machine_index]
else:
cost_for_machine = 0
budget += cost_for_machine
# If we can make target_alloys, try for more
if can_make_alloys:
max_alloys = target_alloys
low_alloys = target_alloys + 1 #Search the right half
else:
#If we can't, try for fewer
high_alloys = target_alloys - 1 #Search the left half
return max_alloysBig(O) Analysis
Time Complexity
O(m * log(k)) – The algorithm uses a binary search to find the maximum number of alloys, where k is the maximum possible number of alloys. For each guess in the binary search, we iterate through each machine to determine if enough metal is available to produce the target number of alloys. Let m represent the number of machines. Therefore, checking the feasibility of a target number of alloys takes O(m) time. Since the binary search takes O(log(k)) iterations, the overall time complexity is O(m * log(k)).
Space Complexity
O(1) – The algorithm described primarily uses a binary search approach. This involves storing a few variables such as the minimum and maximum possible number of alloys, and the target number being tested. The space used for these variables is constant and does not depend on the input size (number of machines, metal available, etc.). Therefore, the auxiliary space complexity is O(1).
Edge Cases
| Case | How to Handle |
|---|---|
| Empty machines array | Return 0 if machines array is empty, as no alloys can be produced. |
| Empty costs array | Return 0 if costs array is empty, as no alloy composition is defined. |
| Budget is 0 | Return 0 if budget is 0, as no alloy can be produced. |
| Number of machines or alloy components is very large, leading to integer overflow during calculations | Use long data type for intermediate calculations to prevent integer overflow. |
| Costs matrix contains zero values | Handle zero cost values as valid, indicating that component is free from that machine. |
| No possible alloy can be made within the budget. | Binary search will converge to 0, which will be the correct answer. |
| Large budget resulting in very large alloy quantities. | Binary search can handle it because the upper bound will be limited by the metal availability in the machines. |
| Machines require zero amount of some metal | Treat zero requirement as valid meaning the metal is not used by that machine. |
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