Find numbers with even number of digits

Given an array nums of integers, return how many of them contain an even number of digits. We will use a helper function to count the number of digits of each number. Then iterate over each number, count their digit, and check if it is even.

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        def numDigit(num):
            digit = 0
            while num:
                digit += 1
                num //= 10
            return digit

        res = 0
        for num in nums:
            if numDigit(num) % 2 == 0:
                res += 1
        return res

Time Complexity: O(n^2)
Space Complexity: O(1)