human
Solve in Python:
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
assistant
#Here is the Python code you requested.

class Solution:
  
    def findNumbers(self, nums: List[int]) -> int:
        def has_even_digits(number: int):
            if number < 10:
                return False
            elif number < 100:
                return True
            elif number < 1000:
                return False
            elif number < 10000:
                return True
            elif number < 100000:
                return False
            return True

        return sum([1 for num in nums if has_even_digits(num)])