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Formatted question description: https://leetcode.ca/all/2455.html

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

  • Difficulty: Easy.
  • Related Topics: Array, Math.
  • Similar Questions: Binary Prefix Divisible By 5.

Problem

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

Solution (Java, C++, Python)

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

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

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


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

  • function averageValue(nums: number[]): number {
    let sum = 0;
    let n = 0;
    for (const num of nums) {
        if (num % 6 === 0) {
            sum += num;
            n++;
        }
    }

    if (n === 0) {
        return 0;
    }
    return Math.floor(sum / n);
}


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

        if n == 0 {
            return 0;
        }
        sum / n
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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