The Even Checkpoint

An easy Python interview practice problem on DataDriven. Write the Python and run it against test cases, with instant feedback.

Domain
Python
Difficulty
easy
Seniority
L3
Asked in
i360

Problem

A hot parity check runs on every record in a tight loop, and profiling flagged the modulo it uses as the bottleneck. Given an integer n (which may be negative), return True if n is even and False otherwise, without using the modulo operator.

Is this number in the even club? Prove it the fast way.

Example 1

Input:

n = 4

Output:

true

Example 2

Input:

n = 7

Output:

false

Worked solution and explanation

What this problem really is

Under its 'check for even' costume this is really a question about where parity lives in a number: the ones-bit, and nothing else. Ban the modulo operator and the candidate has to reach for that bit directly. Anyone can write n & 1; what separates people is returning a real True/False instead of the raw 1 or 0, and knowing that in Python the low-bit test needs no special-casing for negatives. Return the bit itself and every downstream if is_even_bitwise(x): starts lying to you.

Break it down

Step 1: Isolate the low bit

n & 1 keeps only the ones-place bit and discards everything above it. That single bit carries the entire even/odd answer.

Step 2: Turn the bit into a real boolean

A raw bit is 0 or 1, an int, not a bool. Compare it and bind the result to a named variable: even = (n & 1) == 0 holds a genuine True/False, which is what callers branch on, and returning it keeps the intent readable.

The solution

Bitwise parity check
def is_even_bitwise(n: int) -> bool:
    even = (n & 1) == 0
    return even

Common follow-up questions