# Find Numbers with Even Number of Digits

This is a Leetcode Data Structure / Algorithm problem, to solve this problem you need to understand what an even number is.

Even Number:is a number that is divisible by two with no remainders left

**Problem**

Given an array `nums`

of integers, return how many of them contain an **even number** of digits.

**Example 1:**

**Input:** nums = [12,345,2,6,7896]

**Output:** 2

**Explanation: **

12 contains 2 digits (even number of digits).

345 contains 3 digits (odd number of digits).

2 contains 1 digit (odd number of digits).

6 contains 1 digit (odd number of digits).

7896 contains 4 digits (even number of digits).

Therefore only 12 and 7896 contain an even number of digits.

**First Step:** you add an Int variable that will act as a counter, which will hold the number of even digits.

`var count = 0`

**Second Step: **you would loop through the array to get the individual elements, then get the count of digits

This will get the count even with a negative number(val numlength = u.absoluteValue.toString().length)

`for (u in ty) {`

val numlength = u.toString().length

**Third Step: **check if the value is an even number, divide the length by 2. if the remainder is 0, it is even

`if (numlength % 2 == 0) count++`

**Fourth Step:** Return the count value

`return count`

Thanks for reading.