Note the time complexity of str() convertion is O(len(S)) S: target string

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        """tc O(N*max(len(str(num))))  sc O(1)
        """
        res = 0
        for x in nums:
            res += 1 - len(str(x))%2 
        return res

Trick 2 : look at the value range

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        """tc O(N)  sc O(1)
        
        """
        cnt = 0
        for x in nums:
            if (x > 9 and x <100) or (x > 999 and x <10000) or x == 10**5:
                cnt += 1 
        return cnt