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

Example 1:

`Input: nums = [12,345,2,6,7896]Output: 2Explanation: 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` 