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; } }