Check If an Integer is Even or Odd

Problem Statement

You are given an integer, and need to find whether the integer is even or odd.

Examples

Example 1:

Input: n = 4
Output: Even

Explanation: 4 is divisible by 2 -> Even

Example 2:

Input: n = 7
Output: Odd

Explanation: 7 is not divisible by 2 -> Odd

Example 3:

Input: n = 0
Output: Even

Explanation: 0 is considered an even number

Different Approaches

1️⃣ Traditional Approach

The traditional solution for checking if an integer is even or odd involves using the modulus operator (%). We divide the number with 2, if remainder output is 0, the number is even, otherwise it is odd.

Approach:

• If n % 2 == 0, it’s even
• Else, it’s odd

Code:

#include <iostream>

bool isEven(int num) {
    return num % 2 == 0;
}

int main() {
    int num = 77;

    if (isEven(num)) {
        std::cout << num << " is even." << std::endl;
    } else {
        std::cout << num << " is odd." << std::endl;
    }

    return 0;
}

// Output
77 is odd.

2️⃣ Using Bitwise AND (&1) LSB Check

This solution checks if the least significant bit (LSB) of the binary representation of the integer. If the LSB is 0, the number is even; otherwise, it is odd.

Intuition:

• Even numbers have 0 as the least significant bit.
• Odd numbers have 1 as the least significant bit.
• n & 1 checks whether the last bit is 1 (odd) or 0 (even)

Approach:

• n & 1 == 0 → even
• n & 1 == 1 → odd

#include <iostream>

bool isEvenLSB(int num) {
    // Using bitwise AND to check the LSB
    return (num & 1) == 0;
}

int main() {
    int num = 44;

    if (isEvenLSB(num)) {
        std::cout << num << " is even." << std::endl;
    } else {
        std::cout << num << " is odd." << std::endl;
    }

    return 0;
}

// Output
44 is even.