1295_find_nums_with_even_num_of_digits.cpp

#include <bits/stdc++.h>

using namespace std;

class Solution {
public:
    int findNumbers(vector<int>& nums) {
        int evenDigitNums = 0;
        for (int num : nums) {
            string str = to_string(num);
            evenDigitNums += !(str.length() % 2); 
        }
        return evenDigitNums;
    }
};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    vector<int> nums = {12, 345, 2, 6, 7896};
    Solution solve;
    int result = solve.findNumbers(nums);
    cout << result << '\n';
    return 0;
}

/*

    ------------------
    Problem Statement:
    ------------------
    
    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
*/