LeetCode 2861: Maximum Number of Alloys Solution
Master LeetCode problem 2861 (Maximum Number of Alloys), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
2861. Maximum Number of Alloys
Problem Explanation
Explanation
To maximize the number of alloys the company can create within the given budget, we can iterate through all machines and calculate the maximum number of alloys that can be created for each machine. We will then choose the machine that allows the creation of the most alloys within the budget.
Algorithm
- Iterate through each machine.
- For each machine, calculate the maximum number of alloys that can be created based on the available stock, cost, and budget.
- Keep track of the machine that allows the creation of the highest number of alloys within the budget.
- Return the maximum number of alloys that can be created using the chosen machine.
Time Complexity
The time complexity of this approach is O(nk), where n is the number of different metal types and k is the number of machines.
Space Complexity
The space complexity is O(1) as we are not using any additional data structures apart from input variables.
Solution Code
class Solution {
public int maxNumberOfAlloys(int n, int k, int budget, int[][] composition, int[] stock, int[] cost) {
int maxAlloys = 0;
for (int machine = 0; machine < k; machine++) {
int alloys = Integer.MAX_VALUE;
for (int metal = 0; metal < n; metal++) {
int requiredMetal = composition[machine][metal];
int availableMetal = stock[metal];
int metalCost = cost[metal];
alloys = Math.min(alloys, availableMetal / requiredMetal);
}
maxAlloys = Math.max(maxAlloys, alloys);
}
return Math.min(maxAlloys, budget / Arrays.stream(cost).sum());
}
}Try It Yourself
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