The Bitwise Judge

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

Given an integer n (possibly negative), return True if n is even, False if odd. Solve using bitwise operations only - no %, no /, no //.

No division, no modulo - just a single bit tells you everything.

Example 1

Input:

n = 4

Output:

true

Example 2

Input:

n = 7

Output:

false

Worked solution and explanation

Why this problem exists in real interviews

Checking even/odd with bitwise AND tests the most basic bit manipulation concept: the least significant bit determines parity.

Break down the requirements

Step 1: Check the least significant bit

If n & 1 is 0, the number is even. If 1, it is odd.

The solution

def is_even(n):
    result = (n & 1) == 0
    return result

Common follow-up questions

• How would you check if a number is a power of 2? (Tests `n > 0 and (n & (n - 1)) == 0`, which checks that only one bit is set.)
• What if n is negative? (Tests that bitwise AND on negative numbers in Python works with two's complement, and the LSB check still determines parity.)
• Can you check divisibility by 4 using bitwise operations? (Tests `(n & 3) == 0`, checking the two least significant bits.)