Problem

Given an integer N (which may be negative), return "Even" if it is divisible by 2 and "Odd" otherwise. Solve it without using the modulo or division operator as a follow-up.

Input / Output

• Input: integer N.
• Output: the string "Even" or "Odd".

Examples

N	Output
4	Even
7	Odd
0	Even
-3	Odd
-10	Even

Constraints

• -10^18 <= N <= 10^18

Approach

Modulo. N % 2 == 0 → even. Beware that in C, C++, Java and JavaScript the result of % for a negative operand can be -1, so test N % 2 == 0 rather than N % 2 == 1.

Bitwise (preferred). In two's-complement representation the least-significant bit is 1 for every odd number and 0 for every even number, positive or negative. So:

if (N & 1) == 0: Even else Odd

This is branch-free, works for negatives without special cases, and is what the follow-up is looking for.

Other tricks worth mentioning.

• (N >> 1) << 1 == N — shifting right then left clears the lowest bit; equality means it was already 0.
• N ^ 1 == N + 1 holds only for even N.

Complexity

• Time: O(1)
• Space: O(1)

Interview angle

The question is a warm-up used to move into bit manipulation: expect follow-ups such as checking whether a number is a power of two (N & (N-1) == 0), counting set bits, or finding the single odd-occurring element with XOR.

Reference solution (Python)

def parity(n: int) -> str:
    return "Even" if (n & 1) == 0 else "Odd"

Test cases

Input	Expected
4	Even
7	Odd
0	Even
-3	Odd