Find Numbers with Even Number of Digits | LeetCode # 1295

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

Constraints

• 1 <= nums.length <= 500
• 1 <= nums[i] <= 105

Example

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.

Solution

We can solve this coding problem by iterating through nums array and counting number of digits in each number. If number of digits in each number is even, increase ans by 1 else continue the loop.

C++ implementation of above solution is as follows.

class Solution {
public:
    int findNumbers(vector<int>& nums) {
        int n = nums.size();
        int ans = 0;

        for(int i=0; i<n; i++) {
            int x = nums[i];

            int digitCount = 0;

            while(x > 0) {
                x = x/10;
                digitCount += 1;
            }

            if(digitCount % 2 == 0) {
                ans += 1;
            }
        }

        return ans;
    }
};

Time Complexity

O(n) where n = number of elements in nums array.

Space Complexity

O(1) since we are not using any additional space.

Thanks for reading and have a nice day.