# Find Numbers with Even Number of Digits using Java

## The challenge

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: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.

**Example 2:**

Input:nums = [555,901,482,1771]Output:1Explanation:Only 1771 contains an even number of digits.

**Constraints:**

`1 <= nums.length <= 500`

`1 <= nums[i] <= 10^5`

## The solution

```
class Solution {
public int findNumbers(int[] nums) {
// keep track of the amount to return
int evens = 0;
// loop through all the nums
for (int i=0; i<nums.length; i++) {
int digits = 0, num = nums[i];
// while we divide by 10
while (num!=0) {
num /= 10;
// increment our count
++digits;
}
// increment if even
if (digits%2==0) evens++;
}
return evens;
}
}
```