Average Value of Even Numbers That Are Divisible by Three

Average Value of Even Numbers That Are Divisible by Three - Problem

Array Math Easy

You're given an array of positive integers and need to find the average value of numbers that satisfy two specific conditions:

The number must be even (divisible by 2)
The number must be divisible by 3

In other words, you're looking for numbers that are divisible by both 2 and 3 (which means divisible by 6!).

Return the integer average of all such numbers. If no numbers meet the criteria, return 0.

Note: The average should be rounded down to the nearest integer (floor division).

Example: In array [4, 6, 7, 9, 12], only 6 and 12 are both even and divisible by 3. Average = (6 + 12) / 2 = 9

Input & Output

example_1.py — Basic case

$ Input: [4, 6, 7, 9, 12, 18]

› Output: 12

💡 Note: Numbers divisible by both 2 and 3 (i.e., divisible by 6): 6, 12, 18. Average = (6 + 12 + 18) / 3 = 36 / 3 = 12

example_2.py — No valid numbers

$ Input: [1, 3, 5, 7, 9]

› Output: 0

💡 Note: None of these numbers are even, so none can be divisible by both 2 and 3. Return 0.

example_3.py — Single valid number

$ Input: [6]

› Output: 6

💡 Note: Only one number: 6. It's divisible by both 2 and 3. Average = 6 / 1 = 6

Constraints

1 ≤ nums.length ≤ 1000
1 ≤ nums[i] ≤ 1000
All integers are positive
Return 0 if no valid numbers exist

Visualization

Tap to expand

Understanding the Visualization

Inspect Each Candy

Check if candy meets both criteria: blue AND 3 ingredients

Keep Running Tally

Maintain count and sum of qualifying candies

Calculate Quality Score

Compute average quality (sum ÷ count) for premium shipment

Key Takeaway

🎯 Key Insight: Numbers divisible by both 2 and 3 are exactly those divisible by 6. This mathematical property allows us to solve the problem efficiently in a single pass with O(1) space.

Asked in

a Amazon 25 G Google 18 ⊞ Microsoft 12 Apple 8

The optimal solution uses the mathematical insight that a number divisible by both 2 and 3 must be divisible by 6. This allows us to solve the problem in O(n) time and O(1) space with a single pass through the array, maintaining a running sum and count of valid numbers.

Common Approaches

Approach	Time	Space	Notes
✓ One-Pass Optimal Solution	O(n)	O(1)	Calculate sum and count in a single pass through the array
Brute Force (Multiple Passes)	O(n)	O(k)	First collect all valid numbers, then calculate average

One-Pass Optimal Solution — Algorithm Steps

Initialize sum = 0 and count = 0
Iterate through the array once
For each number, check if it's divisible by 6 (equivalent to being even AND divisible by 3)
If valid, add to sum and increment count
After the loop, return sum // count (or 0 if count is 0)

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize Variables

Set sum = 0, count = 0

Process Each Element

Check if divisible by 6, update sum and count

Calculate Final Average

Return sum / count

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int averageValue(int* nums, int numsSize) {
    int totalSum = 0;
    int count = 0;
    
    // Single pass: check each number and update sum/count
    for (int i = 0; i < numsSize; i++) {
        // Key insight: divisible by both 2 and 3 means divisible by 6
        if (nums[i] % 6 == 0) {
            totalSum += nums[i];
            count++;
        }
    }
    
    // Return average or 0 if no valid numbers
    return count > 0 ? totalSum / count : 0;
}

int main() {
    int n;
    scanf("%d", &n);
    int* nums = (int*)malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        scanf("%d", &nums[i]);
    }
    printf("%d\n", averageValue(nums, n));
    free(nums);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array to check each element once

✓ Linear Growth

Space Complexity

O(1)

Only using a constant amount of extra variables (sum, count)

✓ Linear Space

28.4K Views

Medium Frequency

~8 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

One-Pass Optimal Solution — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler