Find Numbers with Even Number of Digits using Java

The challenge

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

The solution

class Solution {
    public int findNumbers(int[] nums) {
        // keep track of the amount to return
        int evens = 0;
        
        // loop through all the nums
        for (int i=0; i<nums.length; i++) {
            
            int digits = 0, num = nums[i];
            // while we divide by 10
            while (num!=0) {
                num /= 10;
                // increment our count
                ++digits;
            }
            
            // increment if even
            if (digits%2==0) evens++;
            
        }
        
        return evens;
    }
}
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