# Leetcode - Find Numbers with Even Number of Digits Solution

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.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`

## Solution in Python

```
class Solution:
def findNumbers(self, nums: List[int]) -> int:
return sum((len(str(num))+1)%2 for num in nums)
```