findEvenDigitNumbers.py

 # 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
   # Hide Hint #1  
# How to compute the number of digits of a number ?
   # Hide Hint #2  
# Divide the number by 10 again and again to get the number of digits.

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        counter = 0
        for el in nums:
            mods=[el/10,el/100,el/1000,el/10000,el/100000,el/1000000]
            for mod in mods:
                if mod < 1:
                    break
            if mods.index(mod) % 2 != 0:
                counter += 1
        return counter