What is the time complexity of Maximum Number of Alloys?

The optimal solution has a time complexity of O(k * n * log(10^9)).

What data structures are used to solve Maximum Number of Alloys?

This problem uses Array, Binary Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Number of Alloys?

Insight 1: Since all alloys must be created with the *same* machine, we can iterate through each machine and find the maximum number of alloys that can be created using that specific machine. Insight 2: For a given machine, we can use binary search to efficiently find the maximum number of alloys we can create. This is because if we *can* create 'x' alloys, we can definitely create any number of alloys less than 'x'. This property makes binary search suitable.

Which companies ask LeetCode 2861. Maximum Number of Alloys in interviews?

This problem is frequently asked at MathWorks.

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Maximum Number of Alloys - LeetCode 2861 Solution

Maximum Number of Alloys - Complete Solution Guide

Maximum Number of Alloys is LeetCode problem 2861, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 i th 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 intege

Detailed Explanation

The problem asks us to find the maximum number of alloys we can create given a budget, a set of machines each requiring different compositions of metals, the current stock of each metal, and the cost of buying each metal. We must create all alloys using the *same* machine. The goal is to choose the machine that maximizes the number of alloys we can produce within the budget constraint.

Solution Approach

The solution iterates through each of the k machines. For each machine, a binary search is performed to find the maximum number of alloys that can be created using that machine within the given budget. A helper function `check` is used within the binary search to determine whether a given number of alloys can be created using the current machine and the given budget. The maximum number of alloys found across all machines is the final result.

Step-by-Step Algorithm

Step 1: Initialize `max_alloys` to 0. This variable will store the maximum number of alloys we can create.
Step 2: Iterate through each machine (composition) in the `composition` list.
Step 3: For each machine, define a helper function `check(num_alloys)` that determines if we can create `num_alloys` using the current machine within the budget.
Step 4: Inside the `check` function, calculate the cost of creating `num_alloys` alloys using the current machine. For each metal, determine how much we need to buy (if the required amount exceeds the stock). Calculate the cost by multiplying the amount we need to buy with the metal's cost, summing the costs across all metals.
Step 5: If the `required_budget` exceeds the given `budget`, return `false`; otherwise, return `true`.
Step 6: Perform binary search to find the maximum number of alloys that can be created using the current machine. The low is 0 and the high is a sufficiently large number (e.g. 3 * 10^8).
Step 7: In each iteration of binary search, call `check(mid)` to determine if `mid` number of alloys can be created.
Step 8: If `check(mid)` returns `true`, update `res_for_machine` to `mid` and move the `low` pointer to `mid + 1` to search for a larger number of alloys.
Step 9: Otherwise, move the `high` pointer to `mid - 1` to search for a smaller number of alloys.
Step 10: After the binary search, update `max_alloys` to the maximum of `max_alloys` and `res_for_machine`.
Step 11: After iterating through all machines, return `max_alloys`.

Key Insights

Insight 1: Since all alloys must be created with the *same* machine, we can iterate through each machine and find the maximum number of alloys that can be created using that specific machine.
Insight 2: For a given machine, we can use binary search to efficiently find the maximum number of alloys we can create. This is because if we *can* create 'x' alloys, we can definitely create any number of alloys less than 'x'. This property makes binary search suitable.
Insight 3: Be careful of integer overflow when calculating the required budget. Use long long (or equivalent) to prevent overflow, especially in languages like Java and C++ where implicit conversions can lead to errors.

Complexity Analysis

Time Complexity: O(k * n * log(10^9))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: MathWorks.

Maximum Number of Alloys | LeetCode 2861

Medium

Array Binary Search

Time: O(k * n * log(10^9))Space: O(1)

Solution Code

Available in:

import java.util.List;

class Solution {
    public int maxNumberOfAlloys(int n, int k, int budget, List<List<Integer>> composition, List<Integer> stock, List<Integer> cost) {
        int max_alloys = 0;

        for (List<Integer> current_composition : composition) {
            
            
            
            
            
            
            
            
            
            
            
            
            
            
            
            
            
            
            
            

            
            java.util.function.Predicate<Integer> check = (num_alloys) -> {
                long required_budget = 0;
                for (int j = 0; j < n; j++) {
                    long metal_needed = (long)num_alloys * current_composition.get(j);
                    if (stock.get(j) < metal_needed) {
                        long to_buy = metal_needed - stock.get(j);
                        required_budget += to_buy * cost.get(j);
                        if (required_budget > budget) {
                            return false;
                        }
                    }
                }
                return true;
            };

            int low = 0;
            int high = (int)3e8;
            int res_for_machine = 0;

            while (low <= high) {
                int mid = low + (high - low) / 2;
                if (check.test(mid)) {
                    res_for_machine = mid;
                    low = mid + 1;
                } else {
                    high = mid - 1;
                }
            }

            max_alloys = Math.max(max_alloys, res_for_machine);
        }

        return max_alloys;
    }
}

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Problem Info

Description

Problem 2861: Maximum Number of Alloys - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

MathWorks

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2861

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize `max_alloys` to 0. This variable will store the maximum number of alloys we can create.
2
Step 2: Iterate through each machine (composition) in the `composition` list.
3
Step 3: For each machine, define a helper function `check(num_alloys)` that determines if we can create `num_alloys` using the current machine within the budget.
4
Step 4: Inside the `check` function, calculate the cost of creating `num_alloys` alloys using the current machine. For each metal, determine how much we need to buy (if the required amount exceeds the stock). Calculate the cost by multiplying the amount we need to buy with the metal's cost, summing the costs across all metals.
5
Step 5: If the `required_budget` exceeds the given `budget`, return `false`; otherwise, return `true`.
6
Step 6: Perform binary search to find the maximum number of alloys that can be created using the current machine. The low is 0 and the high is a sufficiently large number (e.g. 3 * 10^8).
7
Step 7: In each iteration of binary search, call `check(mid)` to determine if `mid` number of alloys can be created.
8
Step 8: If `check(mid)` returns `true`, update `res_for_machine` to `mid` and move the `low` pointer to `mid + 1` to search for a larger number of alloys.
9
Step 9: Otherwise, move the `high` pointer to `mid - 1` to search for a smaller number of alloys.
10
Step 10: After the binary search, update `max_alloys` to the maximum of `max_alloys` and `res_for_machine`.
11
Step 11: After iterating through all machines, return `max_alloys`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(k * n * log(10^9))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The space complexity is O(1) because we are only using a few variables to store intermediate results, and the space used does not depend on the input size. The time complexity arises from iterating through k machines and then doing a binary search (log(10^9) which is the search space for the alloys) and a linear scan of all the material requirements of a machine which has a linear time complexity of n.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Linear Search

Instead of binary search, a linear search could be used to find the maximum number of alloys. We could iterate through possible number of alloys from 0 to some maximum value.

Time: O(k * n * MAX_ALLOYS), where MAX_ALLOYS is the maximum possible number of alloys.Space: O(1)

Trade-offs: Linear search is simpler to implement, but it is much less efficient than binary search when the maximum number of alloys is large. Binary search is always preferable due to its logarithmic time complexity.

✨

Key Insights

1Insight 1: Since all alloys must be created with the *same* machine, we can iterate through each machine and find the maximum number of alloys that can be created using that specific machine.
2Insight 2: For a given machine, we can use binary search to efficiently find the maximum number of alloys we can create. This is because if we *can* create 'x' alloys, we can definitely create any number of alloys less than 'x'. This property makes binary search suitable.
3Insight 3: Be careful of integer overflow when calculating the required budget. Use long long (or equivalent) to prevent overflow, especially in languages like Java and C++ where implicit conversions can lead to errors.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: When `stock[i]` is already greater than or equal to the required amount for a given metal, no additional metal needs to be bought, and the cost for that metal is 0. The code correctly handles this by checking if `stock[j] < metal_needed` before calculating `to_buy`.
Edge case 2: The `budget` can be 0. In this case, the solution should return 0 if any metal needs to be bought to create alloys, or return the number of alloys which can be created from stock directly (though stock is not used in this approach and will return 0). The current code functions correctly in the case of a zero budget.
Edge case 3: Large input values can cause integer overflows. Using `long long` (or equivalent) for calculations involving `num_alloys`, `current_composition[j]`, `stock[j]`, and `cost[j]` is crucial to avoid incorrect results.
Edge case 4: What if all the stocks are very large? The binary search needs a reasonable upper bound. The upper bound chosen in the code (3 * 10^8) is large enough to handle most cases since the cost and composition values are relatively small (<= 100). If the compositions were extremely large, one would have to analyze the stock and budget constraints to find a smaller upper bound to make the binary search faster.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. Failing to use `long long` (or equivalent) can cause incorrect results when calculating the `required_budget`, especially when the number of alloys or the composition values are large. Ensure to use the correct datatype when multiplying/summing variables.

✗Common mistake 2: Incorrect binary search implementation. It is important to correctly update the `low` and `high` pointers in the binary search to ensure that the search space is narrowed down correctly. Pay close attention to when `mid + 1` and `mid - 1` are used.

✗Common mistake 3: Not considering the constraints appropriately to decide on upper bound. While 3*10^8 works in most cases because cost and composition are small, a better analysis of stock values combined with budget could lead to tighter upper bounds.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. This problem is analogous to allocating resources in a manufacturing setting. You have a limited budget and different recipes for producing a product. You need to determine the optimal number of products to produce using a specific recipe to maximize output within the budget.

🏦Real-world scenario 2: Portfolio optimization. In finance, you can think of the metals as different assets, the machines as different investment strategies, the stock as your current holdings, and the cost as the price of acquiring more assets. The goal is to maximize your investment return (number of alloys) while staying within your budget (risk tolerance).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Number of Alloys"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?