README_EN.md

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

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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

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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);
}

C

int averageValue(int* nums, int numsSize) {
    int s = 0, n = 0;
    for (int i = 0; i < numsSize; ++i) {
        if (nums[i] % 6 == 0) {
            s += nums[i];
            ++n;
        }
    }
    return n == 0 ? 0 : s / n;
}

Rust

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
    }
}

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