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

Mental Model

Every number in computers is stored in bits. The last bit tells if a number is even or odd: 0 means even, 1 means odd.

Analogy: Think of a parking lot where cars park in spots numbered 0,1,2,... The last spot number's last digit tells if the spot is even or odd. Checking the last digit is quick and easy.

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: Take the last bit of 5 using bitwise AND with 1

5 in bits: 0101
1 in bits: 0001
5 & 1 = 0001 (which is 1)

Why: The last bit shows if the number is odd (1) or even (0)

Step 2: Check if result is 0 or 1

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 Python

def is_even_or_odd(num: int) -> str:
    # Use bitwise AND to check last bit
    if (num & 1) == 0:
        return "Even"
    else:
        return "Odd"

# Driver code
number = 5
result = is_even_or_odd(number)
print(f"{number} is {result}")

if (num & 1) == 0:

Check last bit to decide 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 operations take constant time regardless of number size

Space: O(1) because no extra space is used

vs Alternative: Using modulo (%) operator also takes O(1) time but bitwise AND is faster and more direct

Edge Cases

number = 0

Returns Even because 0's last bit is 0

DSA Python

if (num & 1) == 0:

number = 1

Returns Odd because 1's last bit is 1

DSA Python

if (num & 1) == 0:

negative number, e.g., -3

Returns Odd because bitwise AND with 1 still checks last bit correctly

DSA Python

if (num & 1) == 0:

When to Use This Pattern

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

Common Mistakes

Mistake: Using modulo operator (%) instead of bitwise AND when bitwise is required for performance or learning
Fix: Replace num % 2 with num & 1 to check last bit directly

Mistake: Checking if num & 1 equals 1 explicitly instead of checking if equals 0 for even
Fix: Use if (num & 1) == 0 for even, else odd for clarity

Summary

It checks the last bit of a number to decide if it is even or odd.

Use it when you want a fast, simple way to check even or odd without division.

The last bit of a binary number is 0 for even and 1 for odd numbers.