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.

Practice

1. What does the expression num & 1 check when determining if a number is even or odd?

easy

A. It checks if the least significant bit of num is 1, indicating odd number.

B. It checks if the number is greater than 1.

C. It shifts the bits of num to the right by 1.

D. It adds 1 to the number.

Solution

1. Step 1: Understand bitwise AND operation

   The expression num & 1 performs a bitwise AND between num and 1, which isolates the least significant bit (rightmost bit) of num.

2. Step 2: Interpret the least significant bit

   If the least significant bit is 1, the number is odd; if it is 0, the number is even.

3. Final Answer:

   It checks if the least significant bit of num is 1, indicating odd number. -> Option A

4. Quick Check:

   num & 1 = 1 means odd [OK]

Hint: Check last bit with & 1; 1 means odd, 0 means even [OK]

Common Mistakes:

• Thinking & 1 shifts bits instead of checking last bit
• Confusing bitwise AND with addition
• Assuming & 1 checks if number > 1

2. Which of the following Python code snippets correctly checks if a number n is even using bitwise operators?

easy

A. if n | 1 == 0:
print('Even')

B. if n & 0 == 1:
print('Even')

C. if n & 1 == 0:
print('Even')

D. if n ^ 1 == 0:
print('Even')

Solution

1. Step 1: Analyze each bitwise operation

   if n & 1 == 0: print('Even') uses n & 1 == 0, which checks if the least significant bit is 0, meaning even number. The other options use incorrect bitwise operations or comparisons.

2. Step 2: Confirm correct syntax and logic

   if n & 1 == 0: print('Even') correctly uses bitwise AND and compares to zero to detect even numbers.

3. Final Answer:

   if n & 1 == 0: print('Even') -> Option C

4. Quick Check:

   n & 1 == 0 means even [OK]

Hint: Use 'if n & 1 == 0' to check even numbers [OK]

Common Mistakes:

• Using bitwise OR or XOR instead of AND
• Comparing with wrong values like 1 instead of 0
• Misplacing parentheses causing syntax errors

3. What will be the output of the following code?

def check_even_odd(num):
    if num & 1:
        return 'Odd'
    else:
        return 'Even'

print(check_even_odd(10))
print(check_even_odd(7))

medium

A. Odd
Odd

B. Even
Odd

C. Odd
Even

D. Even
Even

Solution

1. Step 1: Evaluate check_even_odd(10)

   10 in binary ends with 0, so 10 & 1 = 0, condition is false, returns 'Even'.

2. Step 2: Evaluate check_even_odd(7)

   7 in binary ends with 1, so 7 & 1 = 1, condition is true, returns 'Odd'.

3. Final Answer:

   Even Odd -> Option B

4. Quick Check:

   10 even, 7 odd [OK]

Hint: num & 1 true means odd, false means even [OK]

Common Mistakes:

• Confusing return values for even and odd
• Misreading bitwise AND result
• Assuming & 1 returns boolean instead of int

4. Find the error in the following code that tries to print whether a number is even or odd using bits:

def is_even(num):
    if num & 1 = 0:
        return True
    else:
        return False

print(is_even(4))

medium

A. Using single '=' instead of '==' in the if condition

B. Using bitwise AND instead of bitwise OR

C. Missing colon after function definition

D. Incorrect indentation of return statements

Solution

1. Step 1: Identify syntax error in if condition

   The condition uses single '=' which is assignment, not comparison. It should be '==' to compare values.

2. Step 2: Confirm correct syntax for comparison

   Changing '=' to '==' fixes the syntax error and allows the function to work as intended.

3. Final Answer:

   Using single '=' instead of '==' in the if condition -> Option A

4. Quick Check:

   Use '==' for comparison, not '=' [OK]

Hint: Use '==' for comparison in if statements [OK]

Common Mistakes:

• Using '=' instead of '==' in conditions
• Confusing bitwise AND with OR
• Ignoring syntax errors causing runtime failure

5. You want to filter a list of integers and keep only the odd numbers using bitwise operations. Which of the following code snippets correctly achieves this?

hard

A. odds = [x for x in numbers if x ^ 1 == 1]

B. odds = [x for x in numbers if x & 1 == 0]

C. odds = [x for x in numbers if x | 1 == 1]

D. odds = [x for x in numbers if x & 1 == 1]

Solution

1. Step 1: Understand filtering condition for odd numbers

   Odd numbers have least significant bit 1, so x & 1 == 1 filters odd numbers.

2. Step 2: Check each option's logic

   odds = [x for x in numbers if x & 1 == 1] correctly filters odd numbers. odds = [x for x in numbers if x & 1 == 0] filters even numbers. Options C and D use incorrect bitwise operations for this purpose.

3. Final Answer:

   odds = [x for x in numbers if x & 1 == 1] -> Option D

4. Quick Check:

   Filter odds with x & 1 == 1 [OK]

Hint: Use list comprehension with 'x & 1 == 1' to get odd numbers [OK]

Common Mistakes:

• Using 'x & 1 == 0' which filters even numbers
• Using bitwise OR or XOR incorrectly
• Forgetting to compare with 1 explicitly