1295. Find Numbers with Even Number of Digits

Description

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] <= 10⁵

💡 Hint 1:
How to compute the number of digits of a number ?

💡 Hint 2:
Divide the number by 10 again and again to get the number of digits.

Submitted Code

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        result = 0

        for n in nums:
            digits = 0

            while n > 0:
                digits += 1
                n //= 10
            
            if digits % 2 == 0:
                result += 1

        return result

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        result = 0

        for n in nums:
            digits = len(str(n))
            if digits % 2 == 0:
                result += 1

        return result

Runtime: 0 ms | Beats 100.00%
Memory: 19.36 MB | Beats 25.90%

정수값을 문자열로 변경한 뒤 길이를 재는 방법이 가장 빨랐다. 다만 숫자가 0인 케이스가 있는 경우 에러가 발생한다.

Other Solutions

1st

class Solution(object):
    def findNumbers(self, nums):
        count = 0
        for i in nums:
            digits = int(math.log10(i)) + 1
            if digits % 2 == 0:
                count += 1
        return count

time complexity: 𝑂(𝑛)
space complexity: 𝑂(1)

밑이 10인 log를 통해 자릿수를 알 수 있다.