Check if Number is Even or Not

Problem Statement:

Given a positive integer n, determine whether it is odd or even. Return true if the number is even and false if the number is odd.

• Example:

  Examples:

  Input: n = 15
  Output:false
  Explanation:The number is not divisible by 2, Odd number.

  Input: n = 44
  Output:true
  Explanation:The number is divisible by 2, Even number.

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✅ Solution: Bit Manipulation

class Solution {
  public:
    bool isEven(int n) {
        // If the least significant bit is 1, the number is odd
        // If it's 0, the number is even
        return !(n & 1);
    }
};

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📝 How It Works

• Every even number has 0 as its least significant bit (LSB).
• (n & 1) checks the LSB:
  • If n & 1 is 1, then the number is odd.
  • If n & 1 is 0, then the number is even.
• We use ! to negate the result, so:
  • !(n & 1) returns true for even numbers,
  • false for odd numbers.

Example:

n = 6  -> binary: 0110 → LSB = 0 → even → return true
n = 7  -> binary: 0111 → LSB = 1 → odd → return false

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🧩 Key Formula

!(n & 1)

This uses bitwise AND with 1 to check the parity.

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⏱️ Time & Space Complexity

Complexity	Value
Time	O(1)
Space	O(1)

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⚠️ Edge Cases

• Negative numbers are correctly handled by bitwise operations in C++.
• Zero is considered even (since LSB is 0).

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💡 Other Approaches

Approach	Example	Description
Modulo	n % 2 == 0	Simple and readable, but slightly slower due to division
Bitmasking	!(n & 1) ✅	Fast and efficient using LSB check

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🔁 Related Problems

• Check if number is odd or even
• Count number of even digits in a number
• Separate even and odd indexed characters
• Find the sum of even or odd indexed elements

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