Find Numbers with Even Number of Digits in Python

Problem: 

Given an array nums of integers, return how many of them contain an even number of digits.

Example:

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.

Constraints:

1 <= nums.length <= 500
1 <= nums[i] <= 10^5

Solution:

class Solution:
  def findNumbers(nums):
    return len([num for num in nums if len(str(num))%2 == 0])

num_list = [12,345,2,6,7896]
print(Solution.findNumbers(num_list))

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