Check if Number is Even or Odd Using Bits in DSA C

Mental Model

Every number in computers is stored in bits, and the last bit tells if a number is even or odd. If the last bit is 0, the number is even; if 1, it is odd.

Analogy: Think of a row of light switches representing bits. The last switch controls whether the number is even or odd. If the last switch is off (0), the number is even; if on (1), the number is odd.

Number bits:  ... b3 b2 b1 b0
                  ↑
               last bit

Dry Run Walkthrough

Input: number = 5

Goal: Determine if 5 is even or odd using bit operations

Step 1: Perform bitwise AND of number 5 with 1

5 in bits: 0101
1 in bits:  0001
AND result:  0001

Why: AND with 1 isolates the last bit to check if it is 0 or 1

Step 2: Check if AND result is 0 or 1

AND result = 1

Why: If result is 1, number is odd; if 0, number is even

Result:

5 is odd because last bit is 1

Annotated Code

DSA C

#include <stdio.h>

// Function to check even or odd using bitwise AND
const char* checkEvenOdd(int num) {
    if ((num & 1) == 0) {
        return "Even";
    } else {
        return "Odd";
    }
}

int main() {
    int number = 5;
    printf("%d is %s\n", number, checkEvenOdd(number));
    return 0;
}

if ((num & 1) == 0) {

Check last bit by ANDing with 1 to determine even or odd

return "Even";

Return Even if last bit is 0

return "Odd";

Return Odd if last bit is 1

Output

5 is Odd

Complexity Analysis

Time: O(1) because bitwise AND is a single operation regardless of number size

Space: O(1) because no extra memory is used besides input and output

vs Alternative: Compared to using modulo (%) operator which also runs in O(1), bitwise AND is often faster and more direct at hardware level

Edge Cases

number = 0

Returns Even because 0's last bit is 0

DSA C

if ((num & 1) == 0) {

number = -2

Returns Even because bitwise AND checks last bit regardless of sign

DSA C

if ((num & 1) == 0) {

number = 1

Returns Odd because last bit is 1

DSA C

if ((num & 1) == 0) {

When to Use This Pattern

When you need to quickly check if a number is even or odd without division, use bitwise AND with 1 because it directly inspects the last bit.

Common Mistakes

Mistake: Using modulo operator (%) instead of bitwise AND for checking even/odd
Fix: Use (num & 1) == 0 for faster and simpler check

Mistake: Checking if (num & 1) == 1 to decide even, which is reversed
Fix: Remember (num & 1) == 0 means even, else odd

Summary

Checks if a number is even or odd by looking at its last bit using bitwise AND.

Use when you want a fast, simple way to test evenness without division or modulo.

The last bit of a number tells if it is even (0) or odd (1).