What is the time complexity of Average Value of Even Numbers That Are Divisible by Three?

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

What data structures are used to solve Average Value of Even Numbers That Are Divisible by Three?

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

What are the key insights for solving Average Value of Even Numbers That Are Divisible by Three?

Insight 1: The problem requires checking two conditions for each number: whether it's even (divisible by 2) and whether it's divisible by 3. This can be efficiently done using the modulo operator (%) in a single `if` condition. Insight 2: A simple iterative approach is sufficient. No complex data structures are needed.

Which companies ask LeetCode 2455. Average Value of Even Numbers That Are Divisible by Three in interviews?

This problem is frequently asked at Accenture, IBM.

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Average Value of Even Numbers That Are Divisible by Three - LeetCode 2455 Solution

Average Value of Even Numbers That Are Divisible by Three - Complete Solution Guide

Average Value of Even Numbers That Are Divisible by Three is LeetCode problem 2455, 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

Given an integer array nums of positive integers, return the average value of all even integers that are divisible by 3 . Note that the average of n elements is the sum of the n elements divided by n and rounded down to the nearest integer. Example 1: Input: nums = [1,3,6,10,12,15] Output: 9 Explanation: 6 and 12 are even numbers that are divisible by 3. (6 + 12) / 2 = 9. Example 2: Input: nums = [1,2,4,7,10] Output: 0 Explanation: There is no single number that satisfies the requirement, so ret

Detailed Explanation

The problem asks you to calculate the average of all even numbers in an input array that are also divisible by 3. The input is an array of positive integers (`nums`). The output is a single integer representing the average, rounded down to the nearest integer. If no such numbers exist, the output is 0. The constraints limit the array size and the range of numbers within the array.

Solution Approach

The provided solutions utilize a straightforward iterative approach. They iterate through the input array, checking each number for the specified conditions (even and divisible by 3). If a number meets both conditions, it's added to a running sum, and a counter is incremented. After iterating through the entire array, the average is calculated by dividing the sum by the count. A check for a zero count prevents division by zero errors.

Step-by-Step Algorithm

Step 1: Initialize two variables: `sum` (to accumulate the sum of eligible numbers) and `count` (to count the number of eligible numbers). Both are initialized to 0.
Step 2: Iterate through the input array `nums`.
Step 3: For each number `num`, check if it's even (`num % 2 == 0`) and divisible by 3 (`num % 3 == 0`).
Step 4: If both conditions are true, add `num` to `sum` and increment `count`.
Step 5: After iterating through all numbers, check if `count` is 0. If it is, return 0 (no eligible numbers).
Step 6: Otherwise, calculate the average by performing integer division (`sum / count`) and return the result.

Key Insights

Insight 1: The problem requires checking two conditions for each number: whether it's even (divisible by 2) and whether it's divisible by 3. This can be efficiently done using the modulo operator (%) in a single `if` condition.
Insight 2: A simple iterative approach is sufficient. No complex data structures are needed.
Insight 3: Handling the edge case where no numbers satisfy the conditions is crucial. The code should explicitly check if the count of eligible numbers is zero before calculating the average to prevent division by zero errors.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math.

Companies

Asked at: Accenture, IBM.

Average Value of Even Numbers That Are Divisible by Three | LeetCode 2455

Easy

Array Math

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

Solution Code

Available in:

class Solution {
    public int averageValue(int[] nums) {
        int sum = 0;
        int count = 0;
        for (int num : nums) {
            if (num % 2 == 0 && num % 3 == 0) {
                sum += num;
                count++;
            }
        }
        if (count == 0) {
            return 0;
        }
        return sum / count;
    }
}

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Problem Info

Description

Problem 2455: Average Value of Even Numbers That Are Divisible by Three - Detailed explanation and solution approaches.

Topics

Array Math

Companies

AccentureIBM

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:2455

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

6 steps

1
Step 1: Initialize two variables: `sum` (to accumulate the sum of eligible numbers) and `count` (to count the number of eligible numbers). Both are initialized to 0.
2
Step 2: Iterate through the input array `nums`.
3
Step 3: For each number `num`, check if it's even (`num % 2 == 0`) and divisible by 3 (`num % 3 == 0`).
4
Step 4: If both conditions are true, add `num` to `sum` and increment `count`.
5
Step 5: After iterating through all numbers, check if `count` is 0. If it is, return 0 (no eligible numbers).
6
Step 6: Otherwise, calculate the average by performing integer division (`sum / count`) and return the 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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because the algorithm iterates through the input array once. The space complexity is O(1) because only a constant amount of extra space is used to store the `sum` and `count` 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

Using streams (Java)

In Java, streams can be used to filter the array based on the conditions and then calculate the average in a more concise way.

Time: O(n)Space: O(1) or O(n) depending on the stream implementation

Trade-offs: Streams can improve code readability for those familiar with them but might introduce slightly higher overhead compared to the basic iterative approach, especially for smaller input sizes.

✨

Key Insights

1Insight 1: The problem requires checking two conditions for each number: whether it's even (divisible by 2) and whether it's divisible by 3. This can be efficiently done using the modulo operator (%) in a single `if` condition.
2Insight 2: A simple iterative approach is sufficient. No complex data structures are needed.
3Insight 3: Handling the edge case where no numbers satisfy the conditions is crucial. The code should explicitly check if the count of eligible numbers is zero before calculating the average to prevent division by zero errors.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: Empty input array. The solution handles this correctly by returning 0 because `count` will remain 0.
Edge case 2: Input array with no even numbers divisible by 3. The solution handles this correctly by returning 0 because `count` will remain 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check for the case where `count` is 0 before calculating the average, leading to a division by zero error.

✗Common mistake 2: Incorrectly implementing the divisibility check (using `==` instead of the modulo operator `%`).

✗Common mistake 3: Not using integer division (`//` in Python, `/` in Java, C++, C) resulting in a floating point average when an integer average is required.

✅

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: Data analysis tasks where you need to calculate statistics (like the average) on subsets of data meeting specific criteria. For example, finding the average price of even-numbered products that are on sale.

🏦Real-world scenario 2: Inventory management where you might need to calculate the average weight or cost of items with certain properties (e.g., even-numbered IDs and a specific weight range).

💡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.

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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: September 24, 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 7

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

@InterviewProJun 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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Average Value of Even Numbers That Are Divisible by Three"?

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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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