Maximum Number of Alloys
MedPrompt
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
Approaches
2 approaches with complexity analysis and trade-offs.
This approach simulates the process for each machine one by one. For each of the k machines, we try to find the maximum number of alloys it can produce. We do this by iteratively checking the cost for producing 1 alloy, 2 alloys, 3 alloys, and so on, until the cost exceeds the given budget. The maximum number of alloys found across all machines is the answer.
Algorithm
- Initialize a variable
max_alloysto 0. - Loop through each machine
ifrom 0 tok-1.- Initialize
num_alloys_for_machine_i = 0. - Start a loop, checking the number of alloys
x = 1, 2, 3, .... - For each
x, calculate the cost to producexalloys using machinei. - The cost is calculated as
cost(x) = sum(max(0, x * composition[i][j] - stock[j]) * cost[j])for all metalsjfrom0ton-1. - If
cost(x)is less than or equal to thebudget, it means we can afford to makexalloys. We updatenum_alloys_for_machine_i = xand proceed to checkx+1. - If
cost(x)exceeds thebudget, we cannot affordxalloys, and since the cost is always increasing, we cannot afford any number greater thanxeither. We break the inner loop. - After the inner loop, update
max_alloys = max(max_alloys, num_alloys_for_machine_i).
- Initialize
- After checking all
kmachines, returnmax_alloys.
Walkthrough
The main idea is to exhaustively check every possibility. Since we must use only one type of machine, we can iterate through each of the k machines.
For a fixed machine i, we need to find the maximum number of alloys, let's call it x, that we can create. We can find this x by checking x = 1, 2, 3, ... sequentially. For each value of x, we calculate the total cost.
The cost to produce x alloys with machine i is calculated as follows:
- For each metal
j(from 0 ton-1), the total units needed arex * composition[i][j]. - We have
stock[j]units available. - The units of metal
jto purchase aremax(0, (long)x * composition[i][j] - stock[j]). - The cost for metal
jis(units to purchase) * cost[j]. - The total cost is the sum of costs for all
nmetals. Note that intermediate calculations can exceed 32-bit integer limits, so using 64-bit integers (long) is necessary.
We continue incrementing x as long as the total cost is within the budget. The last value of x for which the cost was within budget is the maximum for machine i. We repeat this for all k machines and keep track of the overall maximum number of alloys possible.
class Solution { public int maxNumberOfAlloys(int n, int k, int budget, List<List<Integer>> composition, List<Integer> stock, List<Integer> cost) { int maxAlloys = 0; for (int i = 0; i < k; i++) { int numAlloys = 0; while (true) { long alloysToTest = numAlloys + 1; long currentCost = 0; for (int j = 0; j < n; j++) { long required = alloysToTest * composition.get(i).get(j); if (required > stock.get(j)) { currentCost += (required - stock.get(j)) * cost.get(j); } if (currentCost > budget) { break; } } if (currentCost <= budget) { numAlloys++; } else { break; // Cannot afford alloysToTest, so numAlloys is the max for this machine } } maxAlloys = Math.max(maxAlloys, numAlloys); } return maxAlloys; }}Complexity
Time
O(k * M * n), where `M` is the maximum possible number of alloys. Given the constraints, `M` can be up to `budget + max(stock)`, which is around `2 * 10^8`. This makes the complexity `O(100 * 2*10^8 * 100)`, which is too high for the given time limits.
Space
O(1), as we only use a few variables for calculations, not counting the input storage.
Trade-offs
Pros
Simple to understand and implement.
Directly translates the problem statement into code without complex algorithms.
Cons
Extremely inefficient for the given constraints. The maximum number of alloys can be very large (up to
2 * 10^8), causing the inner loop to run too many times.This approach will almost certainly lead to a 'Time Limit Exceeded' (TLE) error on competitive programming platforms.
This approach improves upon the brute-force method by recognizing a key property: the cost to produce alloys is a monotonically increasing function of the number of alloys. If we can afford to make x alloys, we can also afford any number of alloys less than x. This property allows us to use binary search to efficiently find the maximum number of alloys for each machine.
Algorithm
- Initialize
max_alloys = 0. - For each machine
ifrom0tok-1:- Perform a binary search to find the maximum number of alloys
x_ithis machine can produce. - Set a search range
low = 0,high = 2 * 10^8 + 7(a safe upper bound). - Initialize
ans_i = 0. - While
low <= high:mid = low + (high - low) / 2.- Check if it's possible to create
midalloys. This check involves calculating the total cost formidalloys and comparing it with thebudget. - If it is possible (cost <= budget):
midis a potential answer. Store it (ans_i = mid) and try for more by searching in the upper half (low = mid + 1). - If it is not possible (cost > budget):
midis too many. Try for fewer by searching in the lower half (high = mid - 1).
- After the binary search, update
max_alloys = max(max_alloys, ans_i).
- Perform a binary search to find the maximum number of alloys
- Return
max_alloys.
Walkthrough
The overall structure is to iterate through each of the k machines, find the maximum alloys possible with that machine, and then take the overall maximum. The key improvement is using binary search to find the maximum alloys for a single machine.
For a given machine i, we want to find the maximum x such that cost(x) <= budget. Since cost(x) is monotonic, we can binary search for x.
We define a search space for x. The lower bound is 0. The upper bound can be estimated. The number of alloys is limited by the budget and stock. A safe upper bound is budget + max(stock), which is roughly 2 * 10^8. We can use a slightly larger constant like 200000007.
The binary search for a machine i works as follows:
- Set
low = 0,high = 200000007,ans = 0. - While
low <= high: a. Calculatemid = low + (high - low) / 2. b. Check if it's possible to createmidalloys within the budget using a helper functioncanCreate(mid). c. ThecanCreate(mid)function calculates the cost formidalloys:cost(mid) = sum(max(0, mid * composition[i][j] - stock[j]) * cost[j]). d. Ifcost(mid) <= budget, it meansmidis a possible number of alloys. We might be able to do even better, so we storemidas a potential answer and search in the upper half:ans = mid,low = mid + 1. e. Ifcost(mid) > budget,midis too many alloys. We must try a smaller number. We search in the lower half:high = mid - 1.
After the binary search loop finishes, ans will hold the maximum number of alloys for machine i. We do this for all k machines and return the maximum ans found.
class Solution { public int maxNumberOfAlloys(int n, int k, int budget, List<List<Integer>> composition, List<Integer> stock, List<Integer> cost) { int maxAlloys = 0; for (int i = 0; i < k; i++) { // Binary search for the max alloys for machine i int low = 0; // A safe upper bound. Max budget is 10^8, min cost is 1, min composition is 1. // If we have 0 stock, we can make at most budget alloys. // If we have max stock 10^8, we can make 10^8 alloys for free. // So, a safe upper bound is budget + max_stock, roughly 2*10^8. int high = 200000007; int ansForMachine = 0; while (low <= high) { int mid = low + (high - low) / 2; if (canCreate(mid, i, n, budget, composition, stock, cost)) { ansForMachine = mid; low = mid + 1; // Try for more } else { high = mid - 1; // Too expensive, try for less } } maxAlloys = Math.max(maxAlloys, ansForMachine); } return maxAlloys; } private boolean canCreate(long numAlloys, int machineIndex, int n, long budget, List<List<Integer>> composition, List<Integer> stock, List<Integer> cost) { if (numAlloys == 0) return true; long totalCost = 0; List<Integer> machineComp = composition.get(machineIndex); for (int j = 0; j < n; j++) { long requiredMetal = numAlloys * machineComp.get(j); long stockMetal = stock.get(j); if (requiredMetal > stockMetal) { long metalToBuy = requiredMetal - stockMetal; totalCost += metalToBuy * cost.get(j); } if (totalCost > budget) { return false; } } return true; }}Complexity
Time
O(k * n * log(M)), where `M` is the size of the search space for the number of alloys (e.g., `2 * 10^8`). With the given constraints, this is approximately `100 * 100 * log(2*10^8)`, which is very efficient.
Space
O(1), aside from the input storage. The binary search is done in-place and does not require additional data structures proportional to the input size.
Trade-offs
Pros
Highly efficient due to the logarithmic time complexity of the search for each machine.
Guaranteed to find the optimal solution within the time limits for the given constraints.
Cons
Slightly more complex to conceptualize and implement than the brute-force approach.
Requires careful handling of potential integer overflows by using 64-bit integers (
long) for cost calculations.
Solutions
Solution
class Solution { int n; int budget; List<List<Integer>> composition; List<Integer> stock; List<Integer> cost; boolean isValid(long target) { for (List<Integer> currMachine : composition) { long remain = budget; for (int j = 0; j < n && remain >= 0; j++) { long need = Math.max(0, currMachine.get(j) * target - stock.get(j)); remain -= need * cost.get(j); } if (remain >= 0) { return true; } } return false; }public int maxNumberOfAlloys(int n, int k, int budget, List<List<Integer>> composition, List<Integer> stock, List<Integer> cost) { this.n = n; this.budget = budget; this.composition = composition; this.stock = stock; this.cost = cost; int l = -1; int r = budget / cost.get(0) + stock.get(0); while (l < r) { int mid = (l + r + 1) >> 1; if (isValid(mid)) { l = mid; } else { r = mid - 1; } } return l; }}class Solution {public: int maxNumberOfAlloys(int n, int k, int budget, vector<vector<int>> &composition, vector<int> &stock, vector<int> &cost) { auto isValid = [&](long long target) { for (int i = 0; i < k; i++) { long long remain = budget; auto currMachine = composition[i]; for (int j = 0; j < n && remain >= 0; j++) { long long need = max(0LL, target * currMachine[j] - stock[j]); remain -= need * cost[j]; } if (remain >= 0) { return true; } } return false; }; long long l = 0, r = budget + stock[0]; while (l < r) { long long mid = (l + r + 1) >> 1; if (isValid(mid)) { l = mid; } else { r = mid - 1; } } return l; }};class Solution: def maxNumberOfAlloys(self, n: int, k: int, budget: int, composition: List[List[int]], stock: List[int], cost: List[int], ) -> int: ans = 0 for c in composition: l, r = 0, budget + stock[0] 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 ans = max(ans, l) return ansVideo walkthrough
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