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# Find Numbers with Even Number of Digits LeetCode Solution

## Problem – Find Numbers with Even Number of Digits LeetCode Solution

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`

### Find Numbers with Even Number of Digits LeetCode Solution in Java

``````class Solution {
public int findNumbers(int[] nums) {

int count=0;

for(int i =0 ; i< nums.length; i++){

if((nums[i]>9 && nums[i]<100) || (nums[i]>999 && nums[i]<10000) || nums[i]==100000){
count++;
}
}

return count;

}
}
``````

### Find Numbers with Even Number of Digits LeetCode Solution in C++

``````class Solution {
public:
int findNumbers(vector<int>& nums) {
int n,count=0;
for(int i=0;i<nums.size();i++)
{
n=nums[i];
if(( 10<=n && n<=99) || (1000<=n && n<=9999 ) || ( n==100000 ))
{
count++;
}
}
return count;
}
};
``````

### Find Numbers with Even Number of Digits LeetCode Solution in Python

``````class Solution:
def findNumbers(self, nums: List[int]) -> int:

counter = 0

for number in nums:

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

counter += 1

return counter
``````
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