# 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: 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.
```

Example 2:

```Input: nums = [555,901,482,1771]
Output: 1
Explanation:
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;
}
}
``````