Coding questions · ByteDance

Count Numbers with an Even Number of Digits

Count how many positive integers in an array have an even number of decimal digits. Respect ordinary representations without leading zeroes, count repeated values independently, and handle boundaries such as 10 and 10000 correctly.

Problem

Given an array of positive integers, count how many elements have an even number of decimal digits.

Function Contract

Implement count_even_digit_numbers(numbers) -> int.

Constraints

1 <= len(numbers) <= 1000.
1 <= numbers[i] <= 10000.
Use the ordinary base-10 representation without leading zeroes.

Examples

[12, 134, 111, 1111, 10] returns 3.
[7, 10000] returns 0.

Hint: Count decimal places

Converting to a decimal string is sufficient under these bounds; repeated division by ten is another valid approach.

Edge Cases

A one-digit number is not counted.
The upper-bound value 10000 has five digits.
Repeated values count separately.

Count Numbers with an Even Number of Decimal Digits

Hints

Count decimal places by converting to a decimal string or repeatedly dividing by ten.

Approach

For each positive number, repeatedly divide by ten and count how many divisions are needed to reach zero. This is exactly its number of decimal digits. Increment the answer when that count is even. Processing elements independently naturally preserves multiplicity.

Time complexity: O(n d), where d is at most 5 under the stated bounds.
Space complexity: O(1) auxiliary space.