How do you solve Maximum Number of Alloys on LeetCode?

Maximum Number of Alloys (LeetCode #2861) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log m) time complexity and O(1) space complexity. Key insight: Binary search can significantly reduce the number of checks needed.

What is the brute force solution for Maximum Number of Alloys?

The brute force approach for Maximum Number of Alloys is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try to create alloys starting from 0 up to a maximum possible number of alloys and check if we can afford the required metals within the budget. This method is straightforward but inefficient as it checks every possible number of alloys. The optimal Optimal Solution approach improves this to O(n log m).

What algorithm or data structure is used to solve Maximum Number of Alloys?

Maximum Number of Alloys uses the following concepts: Array, Binary Search. The recommended patterns to study are: .

What are the interview tips for Maximum Number of Alloys?

Always clarify the problem and constraints before jumping into coding. Think about edge cases, such as when stocks are already sufficient. Practice explaining your thought process while coding; it helps in interviews.

What are common mistakes when solving Maximum Number of Alloys?

Not considering all machines when calculating costs. Failing to optimize the search space with binary search.

#2861

Maximum Number of Alloys

Medium

ArrayBinary Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log m)
Space	O(1)	O(1)

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Intuition

Time O(n log m)Space O(1)

We can use binary search to efficiently find the maximum number of alloys we can create. By checking the feasibility of creating a certain number of alloys within the budget, we can narrow down our search space.

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Algorithm

3 steps

1Step 1: Define a function to check if a certain number of alloys can be created within the budget for a given machine.
2Step 2: Use binary search on the number of alloys from 0 to a large number (e.g., 10^6).
3Step 3: For each mid-point in the binary search, use the feasibility function to check if that many alloys can be created.

solution.py22 lines

1def canCreateAlloys(alloys, n, k, budget, composition, stock, cost):
2    for machine in range(k):
3        total_cost = 0
4        for metal in range(n):
5            required = composition[machine][metal] * alloys - stock[metal]
6            if required > 0:
7                total_cost += required * cost[metal]
8        if total_cost <= budget:
9            return True
10    return False
11
12def maxAlloys(n, k, budget, composition, stock, cost):
13    left, right = 0, 10**6
14    max_alloys = 0
15    while left <= right:
16        mid = (left + right) // 2
17        if canCreateAlloys(mid, n, k, budget, composition, stock, cost):
18            max_alloys = mid
19            left = mid + 1
20        else:
21            right = mid - 1
22    return max_alloys

ℹ

Complexity note: The time complexity is O(n log m) where n is the number of metals and m is the maximum number of alloys we are checking. The log m comes from the binary search over the number of alloys.

1Binary search can significantly reduce the number of checks needed.
2Feasibility checking is a common pattern in optimization problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.