Returning array values that are not odd in JavaScript

Problem

We are required to write a JavaScript function that takes in an array of numbers.

Our function should construct and return a new array that contains all the numbers of the input array that are not odd (i.e., even numbers).

Example

Following is the code ?

const arr = [5, 32, 67, 23, 55, 44, 23, 12];
const findNonOdd = (arr = []) => {
    const res = [];
    for(let i = 0; i < arr.length; i++){
        const el = arr[i];
        if(el % 2 !== 1){
            res.push(el);
        }
    }
    return res;
};
console.log(findNonOdd(arr));

Output

[ 32, 44, 12 ]

Alternative Methods

Using filter() Method

A more concise approach using the built-in filter() method:

const arr = [5, 32, 67, 23, 55, 44, 23, 12];
const findNonOdd = (arr = []) => arr.filter(num => num % 2 === 0);

console.log(findNonOdd(arr));

[ 32, 44, 12 ]

Using forEach() Method

Another approach using forEach() for iteration:

const arr = [5, 32, 67, 23, 55, 44, 23, 12];
const findNonOdd = (arr = []) => {
    const res = [];
    arr.forEach(num => {
        if(num % 2 === 0) {
            res.push(num);
        }
    });
    return res;
};

console.log(findNonOdd(arr));

[ 32, 44, 12 ]

How It Works

The key concept is using the modulo operator (%) to check if a number is even:

• num % 2 === 0 - checks if number is even
• num % 2 !== 1 - alternative way to check for even numbers
• Even numbers have no remainder when divided by 2

Comparison

Conclusion

The filter() method provides the most readable and concise solution for finding non-odd (even) numbers. For performance-critical applications, the traditional for loop remains the fastest approach.