Leetcode 2861: Maximum Number of Alloys
You are running a company that manufactures alloys using various types of metals. There are n different metal types available, and you have k machines that can be used to create alloys. Each machine requires a specific amount of each metal type to create an alloy. The i-th machine requires composition[i][j] units of metal type j. You have stock[i] units of metal type i, and purchasing one unit of metal type i costs cost[i] coins. Your task is to maximize the number of alloys you can produce while staying within a budget of budget coins. Each alloy must be made using the same machine.
Problem
Approach
Steps
Complexity
Input: The input consists of: an integer n (the number of different metals), an integer k (the number of machines), and an integer budget (the budget for buying additional metals). A 2D array composition where composition[i][j] gives the units of metal type j needed to produce an alloy using machine i. Arrays stock and cost where stock[i] is the current stock of metal type i and cost[i] is the cost of purchasing one additional unit of metal type i.
Example: n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,0], cost = [1,2,3]
Constraints:
• 1 <= n, k <= 100
• 0 <= budget <= 10^8
• composition.length == k
• composition[i].length == n
• 1 <= composition[i][j] <= 100
• stock.length == cost.length == n
• 0 <= stock[i] <= 10^8
• 1 <= cost[i] <= 100
Output: Return the maximum number of alloys that the company can create within the given budget.
Example: For input n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,0], cost = [1,2,3], the output is 2.
Constraints:
Goal: The goal is to maximize the number of alloys that can be created with a given budget, using any of the k machines.
Steps:
• For each machine, calculate the number of alloys that can be produced while staying within the budget.
• Use binary search to determine the maximum number of alloys that can be made with the budget.
Goal: Constraints are given to limit the size of the input and ensure that the solution runs efficiently.
Steps:
• n and k are at most 100.
• The budget can go up to 10^8, which requires an efficient solution.
Assumptions:
• It is assumed that each machine can be used to produce alloys independently, and the number of alloys produced is limited by the available budget and stock.
• Input: n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,0], cost = [1,2,3]
• Explanation: The total cost for producing 2 alloys using the first machine is within the budget, while any higher number of alloys would exceed the budget.
Approach: A binary search approach is used to determine the maximum number of alloys that can be created under the given budget.
Observations:
• The problem is about maximizing alloys under a given budget, which suggests a binary search solution to check the maximum number of alloys that can be created.
• We can try each machine and check how many alloys can be created based on the current stock and cost of metals.
Steps:
• Use binary search to check the maximum number of alloys that can be created.
• For each machine, calculate the cost to produce mid alloys and compare it with the budget.
Empty Inputs:
• If n or k is 0, return 0 as no alloys can be made.
Large Inputs:
• Ensure that the solution handles the upper limits of the input constraints efficiently.
Special Values:
• If the stock is sufficient to create alloys without any additional purchase, the cost is 0.
Constraints:
• The binary search approach ensures that the solution is efficient even for large inputs.
bool func(ll mid, int budget, vector<vector<int>>& composition, vector<int>& stock, vector<int>& cost)
{
ll minCost = INT_MAX;
for (int i = 0; i < composition.size(); ++i)
{
ll currCost = 0;
for (int j = 0; j < composition[i].size(); ++j)
{
ll curr = mid * composition[i][j];
if (stock[j] >= curr)
continue;
else
{
ll extra = (curr - stock[j]) * cost[j];
currCost += extra;
}
}
minCost = min(currCost, minCost);
}
return (minCost <= budget);
}
int maxNumberOfAlloys(int n, int k, int budget, vector<vector<int>>& composition, vector<int>& stock, vector<int>& cost) {
ll low = 0, high = 1e9, ans = 0;
while (low <= high)
{
ll mid = low + (high - low) / 2;
if (func(mid, budget, composition, stock, cost))
{
ans = mid;
low = mid + 1;
}
else
high = mid - 1;
}
return ans;
}
1 : Function Definition
bool func(ll mid, int budget, vector<vector<int>>& composition, vector<int>& stock, vector<int>& cost)
This function checks whether a given number of alloys (mid) can be produced within the provided budget, taking into account the stock and cost for each material.
2 : Variable Initialization
ll minCost = INT_MAX;
Initialize the minimum cost to the maximum possible value to keep track of the lowest cost encountered.
3 : Loop
for (int i = 0; i < composition.size(); ++i)
Loop over all the alloy compositions to calculate the cost of producing 'mid' number of alloys.
4 : Variable Initialization
ll currCost = 0;
Initialize the current cost for producing the alloys to zero.
5 : Inner Loop
for (int j = 0; j < composition[i].size(); ++j)
Loop over each material in the current composition.
6 : Cost Calculation
ll curr = mid * composition[i][j];
Calculate the amount of material needed for the current alloy composition.
7 : Condition Check
if (stock[j] >= curr)
Check if there is enough stock of the material to produce 'mid' number of alloys.
8 : Continue
continue;
If there is enough stock, skip to the next material.
9 : Else Block
else
If there is not enough stock, calculate the extra cost needed.
10 : Cost Calculation
{
Opening block for calculating the extra cost required.
11 : Cost Calculation
ll extra = (curr - stock[j]) * cost[j];
Calculate the additional cost required to buy the missing materials.
12 : Cost Accumulation
currCost += extra;
Add the extra cost for the material to the current total cost.
13 : Cost Comparison
minCost = min(currCost, minCost);
Update the minimum cost with the lower value between the current cost and the previously stored minimum cost.
14 : Return
return (minCost <= budget);
Return true if the minimum cost is within the provided budget, otherwise return false.
15 : Function Definition
int maxNumberOfAlloys(int n, int k, int budget, vector<vector<int>>& composition, vector<int>& stock, vector<int>& cost) {
This function finds the maximum number of alloys that can be produced within the given budget using binary search.
16 : Variable Initialization
ll low = 0, high = 1e9, ans = 0;
Initialize the binary search range with 'low' as 0 and 'high' as a large number (1e9).
17 : While Loop
while (low <= high)
Binary search loop to find the maximum number of alloys.
18 : Mid Calculation
ll mid = low + (high - low) / 2;
Calculate the midpoint of the current binary search range.
19 : Condition Check
if (func(mid, budget, composition, stock, cost))
Check if it's possible to produce 'mid' number of alloys within the budget using the helper function.
20 : Update Result
ans = mid;
Update the result with the current value of 'mid' which is a valid solution.
21 : Update Low
low = mid + 1;
Move the lower bound up to 'mid + 1' to search for a larger valid solution.
22 : Else Block
else
If 'mid' number of alloys cannot be produced, narrow the search range.
23 : Update High
high = mid - 1;
Move the upper bound down to 'mid - 1' to search for a smaller valid solution.
24 : Return
return ans;
Return the result which is the maximum number of alloys that can be produced within the budget.
Best Case: O(k log(max_possible_alloys))
Average Case: O(k log(max_possible_alloys))
Worst Case: O(k log(max_possible_alloys))
Description: The time complexity depends on the binary search and the number of machines k.
Best Case: O(1)
Worst Case: O(k)
Description: The space complexity is O(k) due to the array used for composition, stock, and cost.
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