Find Numbers with Even Number of Digits in Python

Given a list of numbers, we need to count how many numbers have an even number of digits. For example, if the array is [12, 345, 2, 6, 7896], the output will be 2, since 12 (2 digits) and 7896 (4 digits) have even digit counts.

Approach

To solve this problem, we will follow these steps ?

• Convert each integer to a string to easily count digits
• Check if the length of the string is even using modulo operation
• Increment counter for numbers with even digit count
• Return the final count

Using Class-based Solution

Here's the implementation using a class method ?

class Solution:
    def findNumbers(self, nums):
        count = 0
        for num in nums:
            # Convert to string and check if length is even
            if len(str(num)) % 2 == 0:
                count += 1
        return count

# Test the solution
ob1 = Solution()
result = ob1.findNumbers([12, 345, 2, 6, 7896])
print(f"Numbers with even digits: {result}")

Numbers with even digits: 2

Using Simple Function

A more straightforward approach without using a class ?

def count_even_digits(numbers):
    count = 0
    for num in numbers:
        if len(str(num)) % 2 == 0:
            count += 1
    return count

# Test with different examples
test_cases = [
    [12, 345, 2, 6, 7896],
    [555, 901, 482, 1771],
    [1, 22, 333, 4444, 55555]
]

for i, nums in enumerate(test_cases, 1):
    result = count_even_digits(nums)
    print(f"Test case {i}: {nums}")
    print(f"Count of numbers with even digits: {result}")
    print()

Test case 1: [12, 345, 2, 6, 7896]
Count of numbers with even digits: 2

Test case 2: [555, 901, 482, 1771]
Count of numbers with even digits: 3

Test case 3: [1, 22, 333, 4444, 55555]
Count of numbers with even digits: 2

Using List Comprehension

A more concise solution using list comprehension ?

def count_even_digits_compact(numbers):
    return sum(1 for num in numbers if len(str(num)) % 2 == 0)

# Test the compact version
numbers = [12, 345, 2, 6, 7896]
result = count_even_digits_compact(numbers)
print(f"Input: {numbers}")
print(f"Numbers with even digits: {result}")

# Show which numbers have even digits
even_digit_nums = [num for num in numbers if len(str(num)) % 2 == 0]
print(f"The numbers with even digits are: {even_digit_nums}")

Input: [12, 345, 2, 6, 7896]
Numbers with even digits: 2
The numbers with even digits are: [12, 7896]

Comparison

Conclusion

To count numbers with even digits, convert each number to a string and check if its length is even. The list comprehension approach provides the most concise solution, while the simple function offers the best readability.