## Question

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:2Explanation: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:1Explanation:Only 1771 contains an even number of digits.

**Constraints:**

`1 <= nums.length <= 500`

`1 <= nums[i] <= 10`

^{5}

## Python Solution

class Solution: def findNumbers(self, nums: List[int]) -> int: s=0 for i in nums: if len(str(i))%2==0: s+=1 return s