Find Numbers with Even Number of Digits

Find Numbers with Even Number of Digits

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

Solutions

🧠 Cpp

#include <limits>

class Solution
{
public:
    int findNumbers(vector<int>& nums)
    {
        int res = 0;
        for(int num : nums)
        for(int i = 1, is_even = 1;
            i < std::numeric_limits<int>::max();
            i*=10, is_even=!is_even)
        {
            if(num/i == 0)
            {
                if(is_even)
                    res++;
                break;
            }

        }

        return res;
    }
};