Average Value of Even Numbers That Are Divisible by Three

Problem Statement

Problem Description

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.

Examples

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 return 0.

Constraints

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 1000

Solution of Given Problem

Intuition and Approach

The problem can be solved using a brute force approach or an optimized Technique.

• Brute Force
• Optimized approach

Approach 1:Brute Force (Naive)

Brute Force Approach: Iterate through the nums array. For each number, check if it is even and divisible by 3. If it meets both conditions, add it to a running sum and increment a count. If the count is greater than 0, calculate the average as the integer division of the sum by the count; otherwise, return 0.

Codes in Different Languages

C++

Written by @AmruthaPariprolu

#include <iostream>
#include <vector>

int averageValue(std::vector<int>& nums) {
    int sum = 0, count = 0;
    
    for (int num : nums) {
        if (num % 6 == 0) { // Even and divisible by 3 means divisible by 6
            sum += num;
            count++;
        }
    }
    
    return count > 0 ? sum / count : 0;
}

int main() {
    std::vector<int> nums = {1, 3, 6, 10, 12, 15};
    std::cout << averageValue(nums) << std::endl;  // Output: 9
    return 0;
}

Java

Written by @AmruthaPariprolu

public class AverageValue {
    public static int averageValue(int[] nums) {
        int sum = 0, count = 0;

        for (int num : nums) {
            if (num % 6 == 0) { // Even and divisible by 3 means divisible by 6
                sum += num;
                count++;
            }
        }
        
        return count > 0 ? sum / count : 0;
    }

    public static void main(String[] args) {
        int[] nums = {1, 3, 6, 10, 12, 15};
        System.out.println(averageValue(nums));  // Output: 9
    }
}

Python

Written by @AmruthaPariprolu

def average_value(nums):
    total_sum = 0
    count = 0

    for num in nums:
        if num % 6 == 0:  # Even and divisible by 3 means divisible by 6
            total_sum += num
            count += 1
    
    return total_sum // count if count > 0 else 0

nums = [1, 3, 6, 10, 12, 15]
print(average_value(nums))  # Output: 9

Complexity Analysis

• Time Complexity: $O(n)$
• where n is the number of elements in nums.
• Space Complexity: $O(1)$
• since we are only using a few extra variables.

Video Explanation of Given Problem

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