# 1295. Find Numbers with Even Number of Digits

## Description

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] <= 105

## Solutions

• class Solution {
public int findNumbers(int[] nums) {
int ans = 0;
for (int v : nums) {
if (String.valueOf(v).length() % 2 == 0) {
++ans;
}
}
return ans;
}
}

• class Solution {
public:
int findNumbers(vector<int>& nums) {
int ans = 0;
for (int& v : nums) {
ans += to_string(v).size() % 2 == 0;
}
return ans;
}
};

• class Solution:
def findNumbers(self, nums: List[int]) -> int:
return sum(len(str(v)) % 2 == 0 for v in nums)


• func findNumbers(nums []int) (ans int) {
for _, v := range nums {
if len(strconv.Itoa(v))%2 == 0 {
ans++
}
}
return
}

• /**
* @param {number[]} nums
* @return {number}
*/
var findNumbers = function (nums) {
let ans = 0;
for (const v of nums) {
ans += String(v).length % 2 == 0;
}
return ans;
};