Given an array nums of integers, return how many of them contain an even number of digits.

&nbsp;
Example 1:

Input: nums = [12,345,2,6,7896]
Output: 2
Explanation: 
12 contains 2 digits (even number of digits).&nbsp;
345 contains 3 digits (odd number of digits).&nbsp;
2 contains 1 digit (odd number of digits).&nbsp;
6 contains 1 digit (odd number of digits).&nbsp;
7896 contains 4 digits (even number of digits).&nbsp;
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.


&nbsp;
Constraints:


	1 &lt;= nums.length &lt;= 500
	1 &lt;= nums[i] &lt;= 105


class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        even_count = 0
        for elem in nums:
            if(len(str(elem))%2 == 0):
                even_count += 1
        return even_count
        
class Solution 
{
    public int findNumbers(int[] nums) 
    {
        int count = 0;
        for(int val : nums)
        {
            if((val>9 && val<100)  ||  (val>999 && val<10000)  || val==100000  )
                count++;
        }
        return count;
    }
}
class Solution {
public:
    int findNumbers(vector<int>& nums) {
        int count = 0;

        for(auto it:nums)
        {
            int amount = 0;
            while(it>0)
            {
                amount++;
                it /= 10;
            }
            if (amount % 2 == 0)
            {
                count++;
            }
        }
        return count;
    }
};
var findNumbers = function(nums) {
	let count = 0;
	for(let num of nums){
		if(String(num).length % 2 === 0) count++
	}
	return count;
};
Find Numbers with Even Number of Digits