What is the time complexity of Find the Maximum Divisibility Score?

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

What data structures are used to solve Find the Maximum Divisibility Score?

This problem uses Array. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Find the Maximum Divisibility Score?

Iterate through each divisor and count how many numbers in `nums` are divisible by it. Maintain the maximum score encountered so far and the corresponding divisor.

Which companies ask LeetCode 2644. Find the Maximum Divisibility Score in interviews?

This problem is frequently asked at DE Shaw.

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

Find the Maximum Divisibility Score - LeetCode 2644 Solution

Find the Maximum Divisibility Score - Complete Solution Guide

Find the Maximum Divisibility Score is LeetCode problem 2644, a Easy 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 given two integer arrays nums and divisors . The divisibility score of divisors[i] is the number of indices j such that nums[j] is divisible by divisors[i] . Return the integer divisors[i] with the maximum divisibility score. If multiple integers have the maximum score, return the smallest one. Example 1: Input: nums = [2,9,15,50], divisors = [5,3,7,2] Output: 2 Explanation: The divisibility score of divisors[0] is 2 since nums[2] and nums[3] are divisible by 5. The divisibility score of

Detailed Explanation

The problem asks to find the divisor from the `divisors` array that has the maximum divisibility score. The divisibility score of a divisor is the number of elements in the `nums` array that are divisible by that divisor. If multiple divisors have the same maximum score, the smallest one should be returned.

Solution Approach

The solution uses a brute-force approach. It iterates through each divisor in the `divisors` array. For each divisor, it iterates through the `nums` array and counts the numbers divisible by the current divisor. This count becomes the divisibility score for that divisor. The algorithm keeps track of the maximum score seen so far and the corresponding divisor. If a divisor achieves a higher score, it updates the maximum score and the best divisor. If a divisor has the same maximum score, it selects the smaller divisor. Finally, it returns the divisor with the maximum divisibility score.

Step-by-Step Algorithm

Initialize `max_score` to -1 and `result` to positive infinity (or a very large number) to handle the initial comparison.
Iterate through each `divisor` in the `divisors` array.
For each `divisor`, iterate through each `num` in the `nums` array.
If `num` is divisible by `divisor` ( `num % divisor == 0`), increment the `score`.
After iterating through all `nums`, if `score` is greater than `max_score`, update `max_score` to `score` and `result` to `divisor`.
If `score` is equal to `max_score`, update `result` to the minimum of `result` and `divisor`.
After iterating through all divisors, return `result`.

Key Insights

Iterate through each divisor and count how many numbers in `nums` are divisible by it.
Maintain the maximum score encountered so far and the corresponding divisor.
If a divisor has the same maximum score as the current best, choose the smaller divisor.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: DE Shaw.

Find the Maximum Divisibility Score | LeetCode 2644

Easy

Array

Time: O(n*m)Space: O(1)

Solution Code

Available in:

class Solution {
    public int maxDivScore(int[] nums, int[] divisors) {
        int maxScore = -1;
        int result = -1;
        for (int divisor : divisors) {
            int score = 0;
            for (int num : nums) {
                if (num % divisor == 0) {
                    score++;
                }
            }
            if (score > maxScore) {
                maxScore = score;
                result = divisor;
            } else if (score == maxScore && score != -1) {
                result = Math.min(result, divisor);
            }
        }
        return result;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2644: Find the Maximum Divisibility Score - Detailed explanation and solution approaches.

Topics

Array

Companies

DE Shaw

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2644

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

7 steps

1
Initialize `max_score` to -1 and `result` to positive infinity (or a very large number) to handle the initial comparison.
2
Iterate through each `divisor` in the `divisors` array.
3
For each `divisor`, iterate through each `num` in the `nums` array.
4
If `num` is divisible by `divisor` ( `num % divisor == 0`), increment the `score`.
5
After iterating through all `nums`, if `score` is greater than `max_score`, update `max_score` to `score` and `result` to `divisor`.
6
If `score` is equal to `max_score`, update `result` to the minimum of `result` and `divisor`.
7
After iterating through all divisors, return `result`.

💡

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(n*m)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The code has nested loops. The outer loop iterates through `divisors` (m iterations), and the inner loop iterates through `nums` (n iterations for each divisor). Therefore, the time complexity is O(n*m). The code uses a constant amount of extra space to store variables like `max_score`, `result`, and `score`, regardless of the input size. Thus, the space complexity is O(1).

📖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

Using a HashMap

Instead of nested loops, a HashMap could store the divisors as keys and their scores as values. The iteration through `nums` would update the scores in the HashMap, eliminating the inner loop. After the HashMap is populated, find the maximum score and the corresponding smallest divisor.

Time: O(n + m)Space: O(m)

Trade-offs: This approach offers a better time complexity if m (the number of divisors) is significantly smaller than n (the number of elements in nums). However, it uses extra space for the HashMap.

✨

Key Insights

1Iterate through each divisor and count how many numbers in `nums` are divisible by it.
2Maintain the maximum score encountered so far and the corresponding divisor.
3If a divisor has the same maximum score as the current best, choose the smaller divisor.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Empty `nums` array: The code handles this correctly; the score will remain 0 for all divisors, and the smallest divisor will be returned.
Empty `divisors` array: This case is not explicitly handled in the provided code. It should throw an error or return an appropriate default value (e.g., -1).
All divisors have a score of 0: The code correctly returns the smallest divisor in this scenario.

⚠️

Common Mistakes to Avoid

✗Forgetting to handle the case where multiple divisors have the same maximum score: The solution must select the smallest among those with the maximum score.

✗Incorrect initialization of `max_score` or `result`: Improper initialization can lead to incorrect results, especially when all divisors have scores of 0.

✅

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?

💼Data analysis: Finding the most frequent factors in a dataset.

🏦Database optimization: Identifying the most efficient join keys based on divisibility patterns.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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: October 3, 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

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

@TechLeadJul 14

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

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Find the Maximum Divisibility Score"?

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

🤖 Need help with this problem?