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Find Numbers With Even Number Of Digits in C++

 Find Numbers With Even Number Of Digits

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

Example1:
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.

Example2:
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

C++ Code:

    int findNumbers(vector<int>& nums) {
        int n=0,count=0;
        for(int x:nums){
            n=0;
            while(x){
                x/=10;
                n++;
            }
            if(n%2==0) count++;
        }
        return count;
    }

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