Description
https://leetcode.com/problems/find-numbers-with-even-number-of-digits/
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
Explanation
Simply count the numbers where length is even.
Python Solution
class Solution:
def findNumbers(self, nums: List[int]) -> int:
count = 0
for num in nums:
if len(str(num)) % 2 == 0:
count += 1
return count
- Time Complexity: O(N)
- Space Complexity: O(1)