Coding problem · Easy · Basics

Count the Even Numbers

Easy Algorithms & data structures

Return how many numbers in nums are even (divisible by 2). Negative numbers and zero count too.

Example 1

Input: nums = [1,2,3,4,6]
Output: 3

Example 2

Input: nums = [1,3,5]
Output: 0

Hints

Hint 1

A number is even when n % 2 === 0.

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Solution and explanation

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Time O(n) · Space O(1)

Loop and count elements where n % 2 == 0. In some languages -3 % 2 is -1, so compare with 0 rather than checking for 1.

Reference solution

function countEvens(nums) {
  return nums.filter((n) => n % 2 === 0).length;
}

function countEvens(nums: number[]): number {
  return nums.filter((n) => n % 2 === 0).length;
}

def count_evens(nums: list[int]) -> int:
    return sum(1 for n in nums if n % 2 == 0)

<?php

function countEvens(array $nums): int
{
    return count(array_filter($nums, fn ($n) => $n % 2 === 0));
}

class Solution {
    public int countEvens(int[] nums) {
        int count = 0;
        for (int n : nums) if (n % 2 == 0) count++;
        return count;
    }
}

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