find-numbers-with-even-number-of-digits.py

# Given an array nums of integers, return how many of them contain an even number of digits.
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# Example 1:
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# 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.
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# Example 2:
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# Input: nums = [555,901,482,1771]
# Output: 1 
# Explanation: 
# Only 1771 contains an even number of digits.
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# Constraints:
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# 	1 <= nums.length <= 500
# 	1 <= nums[i] <= 105
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#


class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        ans = 0
        for num in nums:
            if len(str(num)) % 2 == 0:
                ans += 1
        return ans