Average Value of Even Numbers That Are Divisible by Three

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

Array Math Easy

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.

Input & Output

Example 1 — Mixed Numbers

$ Input: nums = [1,3,6,10,12,15]

› Output: 9

💡 Note: Numbers divisible by both 2 and 3: 6 (even, 6÷3=2) and 12 (even, 12÷3=4). Average = (6+12)/2 = 9

Example 2 — No Valid Numbers

$ Input: nums = [1,2,4,7,10]

› Output: 0

💡 Note: No numbers are both even AND divisible by 3. 2,4,10 are even but not divisible by 3. Return 0.

Example 3 — Single Valid Number

$ Input: nums = [4,4,9,6]

› Output: 6

💡 Note: Only 6 is both even and divisible by 3. Average = 6/1 = 6

Constraints

1 ≤ nums.length ≤ 1000
1 ≤ nums[i] ≤ 1000

Visualization

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Understanding the Visualization

Input Array

Array of positive integers

Filter

Find numbers that are both even AND divisible by 3

Average

Calculate sum/count, rounded down

Key Takeaway

🎯 Key Insight: Numbers divisible by both 2 and 3 are exactly those divisible by 6

Asked in

a Amazon 12 M Microsoft 8

The key insight is that a number divisible by both 2 and 3 must be divisible by 6. Best approach uses single modulo check: num % 6 == 0. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Single Pass Filter	O(n)	O(1)	Iterate through array, check each number for both conditions, calculate average
Optimized - Check Divisibility by 6	O(n)	O(1)	Use mathematical property: even AND divisible by 3 means divisible by 6

Brute Force - Single Pass Filter — Algorithm Steps

Iterate through each number in the array
Check if number is even AND divisible by 3
Add qualifying numbers to sum and increment count
Return sum divided by count (rounded down), or 0 if no valid numbers

Visualization

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Step-by-Step Walkthrough

Check Conditions

For each number, check if even AND divisible by 3

Track Valid Numbers

Keep sum and count of qualifying numbers

Calculate Average

Divide sum by count, rounded down

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int totalSum = 0;
    int count = 0;
    
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] % 2 == 0 && nums[i] % 3 == 0) {
            totalSum += nums[i];
            count++;
        }
    }
    
    return count > 0 ? totalSum / count : 0;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON array manually
    int nums[1000];
    int size = 0;
    char* ptr = line + 1; // Skip opening bracket
    
    while (*ptr != ']' && *ptr != '\0') {
        if (*ptr >= '0' && *ptr <= '9') {
            nums[size++] = strtol(ptr, &ptr, 10);
        } else {
            ptr++;
        }
    }
    
    int result = solution(nums, size);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array of n elements

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for sum and count variables

✓ Linear Space

12.0K Views

Medium Frequency

~8 min Avg. Time

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Single Pass Filter — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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