LeetCode-in-Java
1295. Find Numbers with Even Number of Digits
Easy
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 <= 5001 <= nums[i] <= 105
Solution
public class Solution {
public int findNumbers(int[] nums) {
// initialising variable to hold number of digits and numbers having even number of digits
int digitCount = 0;
int evendigitCount = 0;
// traversing through the array
for (int i : nums) {
while (i != 0) {
// counting digits for each number
digitCount++;
i = i / 10;
}
// incrementing variable for numbers having even number of digits
if (digitCount % 2 == 0) {
evendigitCount++;
}
// reassigning the value to reset digits for next number in iteration
digitCount = 0;
}
return evendigitCount;
}
}