What is the time complexity of Minimize the Maximum of Two Arrays?

The optimal solution has a time complexity of O(log(n)).

What data structures are used to solve Minimize the Maximum of Two Arrays?

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

What are the key insights for solving Minimize the Maximum of Two Arrays?

Insight 1: Binary search is applicable because we are looking for the minimum possible maximum value from a range. We can check if a given maximum value is feasible or not. Insight 2: The core idea is to determine how many numbers are available for each array up to a given `mid` value (the potential maximum). This involves calculating the number of integers not divisible by `divisor1`, `divisor2`, or both.

Which companies ask LeetCode 2513. Minimize the Maximum of Two Arrays in interviews?

This problem is frequently asked at Goldman Sachs.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Minimize the Maximum of Two Arrays - LeetCode 2513 Solution

Minimize the Maximum of Two Arrays - Complete Solution Guide

Minimize the Maximum of Two Arrays is LeetCode problem 2513, 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

We have two arrays arr1 and arr2 which are initially empty. You need to add positive integers to them such that they satisfy all the following conditions: arr1 contains uniqueCnt1 distinct positive integers, each of which is not divisible by divisor1 . arr2 contains uniqueCnt2 distinct positive integers, each of which is not divisible by divisor2 . No integer is present in both arr1 and arr2 . Given divisor1 , divisor2 , uniqueCnt1 , and uniqueCnt2 , return the minimum possible maximum integer t

Detailed Explanation

The problem requires us to find the smallest possible maximum integer that can be present in either of two arrays, `arr1` and `arr2`. These arrays must satisfy the following conditions: `arr1` must contain `uniqueCnt1` distinct positive integers not divisible by `divisor1`, `arr2` must contain `uniqueCnt2` distinct positive integers not divisible by `divisor2`, and no integer can be present in both `arr1` and `arr2`. Essentially, we need to distribute integers such that we minimize the largest number used, while respecting the divisibility rules and uniqueness requirements.

Solution Approach

The solution uses binary search to find the minimum possible maximum integer. For a given mid value (representing a potential maximum integer), we determine the number of available slots for `arr1` and `arr2` by calculating the count of numbers up to `mid` that are not divisible by `divisor1` and `divisor2` respectively. We use the inclusion-exclusion principle to handle numbers divisible by both divisors using their least common multiple (LCM). Based on the available slots and the required counts (`uniqueCnt1` and `uniqueCnt2`), we check if the `mid` value is feasible. If it is, we try a smaller `mid` value; otherwise, we increase `mid`.

Step-by-Step Algorithm

Step 1: Initialize `low` to 1 and `high` to a large value (e.g., 4 * 10^9) to define the search space for the minimum maximum integer.
Step 2: Calculate the Least Common Multiple (LCM) of `divisor1` and `divisor2` using the formula: `LCM(a, b) = (a * b) / GCD(a, b)`. The GCD (Greatest Common Divisor) can be found using the Euclidean algorithm.
Step 3: Perform binary search within the range `[low, high]`. In each iteration, calculate the `mid` value: `mid = low + (high - low) // 2`.
Step 4: Calculate the number of slots available for `arr1` only: `arr1_only_slots = mid // divisor2 - mid // lcm`. This represents numbers divisible by `divisor2` but not `divisor1`.
Step 5: Calculate the number of slots available for `arr2` only: `arr2_only_slots = mid // divisor1 - mid // lcm`. This represents numbers divisible by `divisor1` but not `divisor2`.
Step 6: Calculate the number of common slots (numbers that can be placed in either `arr1` or `arr2`): `common_slots = mid - (mid // divisor1 + mid // divisor2 - mid // lcm)`. This represents numbers not divisible by either `divisor1` or `divisor2`.
Step 7: Calculate the remaining counts `needed1` and `needed2`. `needed1` is the count left for arr1 after utilizing arr1_only_slots, and the same for `needed2` and arr2.
Step 8: Check if the current `mid` is a feasible solution. If `common_slots >= needed1 + needed2`, it means we can accommodate all the required numbers within the given `mid`. Update `ans = mid` and narrow the search space by setting `high = mid - 1`.
Step 9: If the current `mid` is not feasible, it means we need a larger `mid`. Update `low = mid + 1` to search in the upper half.
Step 10: Repeat steps 3-9 until `low > high`. The final `ans` value will be the minimum possible maximum integer.
Step 11: Cast ans to an integer and return it.

Key Insights

Insight 1: Binary search is applicable because we are looking for the minimum possible maximum value from a range. We can check if a given maximum value is feasible or not.
Insight 2: The core idea is to determine how many numbers are available for each array up to a given `mid` value (the potential maximum). This involves calculating the number of integers not divisible by `divisor1`, `divisor2`, or both.
Insight 3: We need to handle the overlap (numbers divisible by both `divisor1` and `divisor2`) correctly using the Least Common Multiple (LCM) to avoid double-counting.

Complexity Analysis

Time Complexity: O(log(n))

Space Complexity: O(1)

Topics

This problem involves: Math, Binary Search, Number Theory.

Companies

Asked at: Goldman Sachs.

Minimize the Maximum of Two Arrays | LeetCode 2513

Medium

Math Binary Search Number Theory

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

Solution Code

Available in:

class Solution {
    public int minimizeSet(int divisor1, int divisor2, int uniqueCnt1, int uniqueCnt2) {
        long low = 1;
        long high = (long) 4e9;
        long ans = high;

        long lcm = (divisor1 * divisor2) / gcd(divisor1, divisor2);

        while (low <= high) {
            long mid = low + (high - low) / 2;
            
            // Numbers that can go into arr1 only.
            // These are numbers not divisible by divisor1, but are divisible by divisor2.
            // Count = (divisible by divisor2) - (divisible by both divisor1 and divisor2)
            long arr1_only_slots = mid / divisor2 - mid / lcm;
            
            // Numbers that can go into arr2 only.
            // These are numbers not divisible by divisor2, but are divisible by divisor1.
            // Count = (divisible by divisor1) - (divisible by both divisor1 and divisor2)
            long arr2_only_slots = mid / divisor1 - mid / lcm;
            
            // Numbers that can go into either arr1 or arr2.
            // These are numbers not divisible by divisor1 and not divisible by divisor2.
            // Count = total - (divisible by divisor1 or divisor2)
            long common_slots = mid - (mid / divisor1 + mid / divisor2 - mid / lcm);
            
            long needed1 = 0;
            if (uniqueCnt1 > arr1_only_slots) {
                needed1 = uniqueCnt1 - arr1_only_slots;
            }
            
            long needed2 = 0;
            if (uniqueCnt2 > arr2_only_slots) {
                needed2 = uniqueCnt2 - arr2_only_slots;
            }
                
            if (common_slots >= needed1 + needed2) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
                
        return (int) ans;
    }

    private int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2513: Minimize the Maximum of Two Arrays - Detailed explanation and solution approaches.

Topics

Math Binary Search Number Theory

Companies

Goldman Sachs

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2513

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 `low` to 1 and `high` to a large value (e.g., 4 * 10^9) to define the search space for the minimum maximum integer.
2
Step 2: Calculate the Least Common Multiple (LCM) of `divisor1` and `divisor2` using the formula: `LCM(a, b) = (a * b) / GCD(a, b)`. The GCD (Greatest Common Divisor) can be found using the Euclidean algorithm.
3
Step 3: Perform binary search within the range `[low, high]`. In each iteration, calculate the `mid` value: `mid = low + (high - low) // 2`.
4
Step 4: Calculate the number of slots available for `arr1` only: `arr1_only_slots = mid // divisor2 - mid // lcm`. This represents numbers divisible by `divisor2` but not `divisor1`.
5
Step 5: Calculate the number of slots available for `arr2` only: `arr2_only_slots = mid // divisor1 - mid // lcm`. This represents numbers divisible by `divisor1` but not `divisor2`.
6
Step 6: Calculate the number of common slots (numbers that can be placed in either `arr1` or `arr2`): `common_slots = mid - (mid // divisor1 + mid // divisor2 - mid // lcm)`. This represents numbers not divisible by either `divisor1` or `divisor2`.
7
Step 7: Calculate the remaining counts `needed1` and `needed2`. `needed1` is the count left for arr1 after utilizing arr1_only_slots, and the same for `needed2` and arr2.
8
Step 8: Check if the current `mid` is a feasible solution. If `common_slots >= needed1 + needed2`, it means we can accommodate all the required numbers within the given `mid`. Update `ans = mid` and narrow the search space by setting `high = mid - 1`.
9
Step 9: If the current `mid` is not feasible, it means we need a larger `mid`. Update `low = mid + 1` to search in the upper half.
10
Step 10: Repeat steps 3-9 until `low > high`. The final `ans` value will be the minimum possible maximum integer.
11
Step 11: Cast ans to an integer and return it.

💡

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(log(n))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the binary search. The other operations within the loop (LCM calculation, integer divisions) take constant time. The space complexity is constant because we are only using a fixed number of variables regardless of the input size.

📖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

Iterative Approach

Instead of Binary Search, one could start from 1 and increment until we find a valid number that allows us to allocate the required uniqueCnt1 and uniqueCnt2 numbers without the divisibility and uniqueness constraints. This is not recommended due to extremely high time complexity

Time: O(n), where n could be as high as 4 * 10^9 in the worst-case scenario.Space: O(1)

Trade-offs: This approach has a significantly worse time complexity compared to the binary search approach, rendering it impractical for large inputs. Binary search is the optimal approach to reduce the search space efficiently.

✨

Key Insights

1Insight 1: Binary search is applicable because we are looking for the minimum possible maximum value from a range. We can check if a given maximum value is feasible or not.
2Insight 2: The core idea is to determine how many numbers are available for each array up to a given `mid` value (the potential maximum). This involves calculating the number of integers not divisible by `divisor1`, `divisor2`, or both.
3Insight 3: We need to handle the overlap (numbers divisible by both `divisor1` and `divisor2`) correctly using the Least Common Multiple (LCM) to avoid double-counting.

📚

Key Terms Explained

For Students

Binary Search

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

Edge Cases

Edge case 1: `divisor1` and `divisor2` are the same. The LCM will be equal to the divisor. Ensure that the algorithm correctly calculates the available slots.
Edge case 2: One of `uniqueCnt1` or `uniqueCnt2` is 0. This case is handled correctly since the needed variables can potentially be 0 and the binary search will still converge.
Edge case 3: `divisor1` and `divisor2` are relatively prime (their GCD is 1). The LCM will be their product. Again, the formula handles this correctly.
Edge case 4: Large values of `divisor1`, `divisor2`, `uniqueCnt1`, and `uniqueCnt2`. The `high` value is initialized to a large number to accommodate the constraint and prevent overflow during LCM calculation. The use of long long in C++, Java, and C avoids integer overflow during calculations.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow when calculating the LCM, especially when `divisor1` and `divisor2` are large. Using `long long` (or equivalent) is crucial.

✗Common mistake 2: Incorrectly calculating the number of available slots by not considering the overlap (numbers divisible by both divisors). The inclusion-exclusion principle with LCM is essential for accuracy.

✗Common mistake 3: Using `int` instead of `long` for `high` and `mid`. The maximum possible value can be close to 4 * 10^9, which exceeds the maximum value of a signed integer in many languages. Casting to int at the return is fine after finding the minimum result.

✗Common mistake 4: Failing to correctly implement binary search (e.g., infinite loops, incorrect update of `low` or `high`).

✅

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. Imagine allocating tasks to two different servers with different processing constraints. The problem can be generalized to minimizing the maximum amount of resources needed to fulfill the constraints of each server, where certain resources cannot be used by specific servers.

🏦Real-world scenario 2: Optimizing build processes. Consider optimizing the maximum time required to build two software components, where certain build tools can only be used for one component or the other, and there are dependencies and constraints based on divisible build tools.

💡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

Always test your solution with edge cases: empty input, single element, and boundary values.

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 21, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

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 "Minimize the Maximum of Two Arrays"?

For this medium-level Math 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?