Find Numbers with Even Number of Digits

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] <= 105

SOLUTION:

import math

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        ctr = 0
        for num in nums:
            if int(math.log10(num)) & 1:
                ctr += 1
        return ctr