# 2455. Average Value of Even Numbers That Are Divisible by Three

## 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.

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

## Solutions

Solution 1: Simulation

We notice that an even number divisible by $3$ must be a multiple of $6$. Therefore, we only need to traverse the array, count the sum and the number of all multiples of $6$, and then calculate the average.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

• class Solution {
public int averageValue(int[] nums) {
int s = 0, n = 0;
for (int x : nums) {
if (x % 6 == 0) {
s += x;
++n;
}
}
return n == 0 ? 0 : s / n;
}
}

• class Solution {
public:
int averageValue(vector<int>& nums) {
int s = 0, n = 0;
for (int x : nums) {
if (x % 6 == 0) {
s += x;
++n;
}
}
return n == 0 ? 0 : s / n;
}
};

• class Solution:
def averageValue(self, nums: List[int]) -> int:
s = n = 0
for x in nums:
if x % 6 == 0:
s += x
n += 1
return 0 if n == 0 else s // n


• func averageValue(nums []int) int {
var s, n int
for _, x := range nums {
if x%6 == 0 {
s += x
n++
}
}
if n == 0 {
return 0
}
return s / n
}

• function averageValue(nums: number[]): number {
let s = 0;
let n = 0;
for (const x of nums) {
if (x % 6 === 0) {
s += x;
++n;
}
}
return n === 0 ? 0 : ~~(s / n);
}


• impl Solution {
pub fn average_value(nums: Vec<i32>) -> i32 {
let mut s = 0;
let mut n = 0;
for x in nums.iter() {
if x % 6 == 0 {
s += x;
n += 1;
}
}
if n == 0 {
return 0;
}
s / n
}
}