# [LeetCode]#1295. Find Numbers with Even Number of Digits

**Environment: Python 3.7**

**Key technique:** for loop

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] <= 10^5`

**Analysis:**

- Convert int to string.
- Use len() function to find is even number or not.

**Solution:**

`class Solution:`

def findNumbers(self, nums):

count=0

for i in nums:

if len(str(i))%2==0:

count+=1

return count

**Better Solution:**

`class Solution:`

def findNumbers(self, nums):

return sum([len(str(x))%2==0 for x in nums])

**Submitted result:**

**Reference:**