Check if Number is Even or Odd Using Bits in DSA Python - Build from Scratch

Check if Number is Even or Odd Using Bits

📖 Scenario: Imagine you are working with a simple calculator program that needs to quickly decide if a number is even or odd. Instead of using slow division, you want to use a fast trick with bits.

🎯 Goal: You will build a small program that uses bit operations to check if a number is even or odd.

📋 What You'll Learn

Create a variable called number with the exact value 29.

Create a variable called mask with the exact value 1.

Use a bitwise AND operation between number and mask to find if the number is even or odd.

Print "Odd" if the number is odd, otherwise print "Even".

💡 Why This Matters

🌍 Real World

Bitwise operations are used in low-level programming and performance-critical applications like games and embedded systems.

💼 Career

Understanding bitwise operations helps in roles like systems programming, firmware development, and optimizing algorithms.

1

Create the number variable

Create a variable called number and set it to 29.

DSA Python

# Create a variable called number and set it to 29
# Your code here

Hint

Use number = 29 to create the variable.

2

Create the mask variable

Create a variable called mask and set it to 1.

DSA Python

number = 29
# Create a variable called mask and set it to 1
# Your code here

Hint

The mask helps us check the last bit of the number.

3

Check if number is even or odd using bitwise AND

Use a bitwise AND operation between number and mask and store the result in a variable called result.

DSA Python

number = 29
mask = 1
# Use bitwise AND between number and mask and store in result
# Your code here

Hint

Use & operator to do bitwise AND.

4

Print if number is Even or Odd

Write an if statement to print "Odd" if result is 1, otherwise print "Even".

DSA Python

number = 29
mask = 1
result = number & mask
# Print "Odd" if result is 1 else print "Even"
# Your code here

Hint

If the last bit is 1, the number is odd; otherwise, it is even.

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