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Sabarish  M S
Sabarish M S

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CODING JOURNEY DAY 1

🚀 Day 1 of My Coding Journey | GeeksforGeeks

Hello developers 👋
This is Sabarish M S from Government College of Engineering, Salem, Tamil Nadu, India 🇮🇳.
I’ve officially started my coding journey with GeeksforGeeks, and this is Day 1.

🔹 Problem Statement

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

Return true if the number is even

Return false if the number is odd

🔹 Examples

Input: n = 15
Output: false
Explanation: 15 is not divisible by 2, so it is an odd number.

Input: n = 44
Output: true
Explanation: 44 is divisible by 2, so it is an even number.

🔹 Constraints

1 ≤ n ≤ 10⁴

🔹 Solution Approach (Bitwise Logic)

Instead of using the modulus operator, I used a bitwise approach, which is both efficient and elegant.

💡 Key Insight:

In binary representation:

Even numbers always end with 0

Odd numbers always end with 1

🔧 How it works:

Perform a bitwise AND operation with 1

n & 1

If the result is 0 → the number is even

If the result is 1 → the number is odd

This works because the least significant bit (LSB) determines the parity of a number.

🔹 Why Bitwise?

✅ Faster than modulo
✅ Low-level understanding
✅ Frequently asked in interviews
✅ Clean and elegant logic

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