0315EasyArrayMath

Even-Length Part Codes

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Trains the technique from

LeetCode 1295Find Numbers with Even Number of Digits

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A hardware shop stocks parts identified by positive whole-number codes, listed in codes. The stock system prints a code onto a two-column label, which only lines up when the code has an even number of digits.

Return how many of the codes in codes have an even number of digits. Codes never carry leading zeros, and the same code may appear more than once in the list, in which case each appearance counts separately.

Examples

Example 1

Input: codes = [7, 42, 903, 1250, 10000]
Output: 2

The codes have 1, 2, 3, 4 and 5 digits. Only 42 and 1250 have an even digit count.

Example 2

Input: codes = [9, 99, 999, 9999]
Output: 2

The digit counts are 1, 2, 3 and 4, so 99 and 9999 qualify.

Example 3

Input: codes = [100000]
Output: 1

The largest code the constraints allow has 6 digits, which is even.

Example 4

Input: codes = [42, 42, 903]
Output: 2

Both copies of 42 are counted, and 903 has 3 digits.

Example 5

Input: codes = [1]
Output: 0

A single-digit code has an odd digit count.

Constraints

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

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def count_even_length(codes: list[int]) -> int:

Java

public int countEvenLength(int[] codes)

The intended approach

Time O(n * d), where d is the number of digits of the largest codeSpace O(1)

Count the digits of each code arithmetically: start a counter at 1 and divide the code by 10 until it drops below 10, adding one each time. Add one to the running total whenever that counter is even.

Pythonentry point: count_even_length

def count_even_length(codes):
    # O(n * d) time, at most 6 divisions per code. O(1) space, two integers.
    total = 0
    for code in codes:
        digits = 1
        rest = code
        while rest >= 10:
            rest //= 10
            digits += 1
        if digits % 2 == 0:
            total += 1
    return total

The obvious version

Render each code as text and count the ones whose text length is even.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_count_even_length(codes):
    hits = 0
    for code in codes:
        if len(str(code)) % 2 == 0:
            hits += 1
    return hits

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
5 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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