Bit wise odd or even

Problem Link

Odd or Even – GFG

Problem Statement

Given an integer n, determine whether it is even or odd.

Return:

• true if the number is even

• false if the number is odd

Examples

Example 1:

Input: n = 1
Output: false
Explanation: 1 is odd.

Example 2:

Input: n = 2
Output: true
Explanation: 2 is even.

Constraints

• 0 <= n <= 10⁶

Intuition

In binary representation:

• Even numbers always end with a 0 as the least significant bit (LSB)

• Odd numbers always end with a 1 as the LSB

So, we can check the parity of a number by checking if the LSB is set using a bitwise AND with 1.

Approach: Bitwise AND with 1

To check whether a number is even:

• If (n & 1) == 0 → Even

• Else → Odd

This is because:

• & 1 extracts the least significant bit

• That bit determines whether the number is odd or even

Java Code

class Solution {
    static boolean isEven(int n) {
       return (n & 1) == 0; 
    }
}

Time and Space Complexity

Dry Run

Input: n = 2

• Binary of 2: 10

• (2 & 1) = 10 & 01 = 00 → 0 → Even → Returns true

Input: n = 3

• Binary of 3: 11

• (3 & 1) = 11 & 01 = 01 → 1 → Odd → Returns false

Conclusion

This problem is a basic yet powerful example of how bitwise operations can optimize simple checks like odd/even determination.

Using (n & 1) is faster than using % 2 and is widely used in low-level or performance-critical code.