LeetCode in Kotlin
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
class Solution {
fun findNumbers(nums: IntArray): Int {
// initialising variable to hold number of digits and numbers having even number of digits
var digitCount = 0
var evendigitCount = 0
// traversing through the array
for (num in nums) {
var number = num
while (number != 0) {
// counting digits for each number
digitCount++
number /= 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
}
}