Check if Number is Power of Two in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code snippet?

Given the function to check if a number is a power of two, what will be printed after calling print(is_power_of_two(16))?

DSA Python

def is_power_of_two(n):
    return n > 0 and (n & (n - 1)) == 0

print(is_power_of_two(16))

A True

B False

C 0

D TypeError

💡 Hint

Think about how binary AND works for powers of two.

A number that is a power of two has exactly one bit set in binary. The expression n & (n-1) clears the lowest set bit. If the result is zero, n was a power of two.

❓ Predict Output

intermediate

What is the output of this code snippet?

What will be printed after running this code?

DSA Python

def is_power_of_two(n):
    if n <= 0:
        return False
    while n % 2 == 0:
        n = n // 2
    return n == 1

print(is_power_of_two(18))

A True

B None

C False

D ZeroDivisionError

💡 Hint

Check if 18 can be divided by 2 repeatedly until 1.

18 is divisible by 2 once (18/2=9), but 9 is not divisible by 2, so the function returns False.

❓ Predict Output

advanced

What is the output of this code snippet?

What will be printed after running this code?

DSA Python

def is_power_of_two(n):
    return n > 0 and (n & (-n)) == n

print(is_power_of_two(32))

A True

B False

C TypeError

D 0

💡 Hint

Understand how n & (-n) isolates the lowest set bit.

For powers of two, n & (-n) equals n itself, so the function returns True.

🔧 Debug

advanced

What error does this code raise?

What error will occur when running this code?

DSA Python

def is_power_of_two(n):
    return n > 0 and (n & (n-1)) == 0

print(is_power_of_two(8))

A False

B SyntaxError

C TypeError

D True

💡 Hint

Check operator precedence and parentheses.

The expression (n & n-1) is parsed as (n & n) - 1 which is invalid syntax without parentheses.

🚀 Application

expert

How many numbers between 1 and 1000 (inclusive) are powers of two?

Count how many integers from 1 to 1000 are powers of two.

A 11

B 8

C 9

D 10

💡 Hint

List powers of two: 1, 2, 4, 8, ... up to 1000.

Powers of two up to 1000 are 1,2,4,8,16,32,64,128,256,512 (10 numbers).

Practice

1. Which of the following numbers is a power of two?

easy

A. 16

B. 18

C. 20

D. 22

Solution

1. Step 1: Understand powers of two

   Powers of two are numbers like 1, 2, 4, 8, 16, 32, etc., where each is 2 raised to an integer.

2. Step 2: Check each option

   16 = 2^4, so it is a power of two. Others like 18, 20, 22 are not powers of two.

3. Final Answer:

   16 -> Option A

4. Quick Check:

   16 = 2^4 [OK]

Hint: Only numbers with one '1' bit in binary are powers of two [OK]

Common Mistakes:

• Confusing even numbers with powers of two
• Thinking any multiple of 2 is a power of two

2. Which of the following Python expressions correctly checks if n is a power of two?

easy

A. n | (n - 1) == 0

B. n & (n - 1) == 0

C. n ^ (n - 1) == 0

D. n + (n - 1) == 0

Solution

1. Step 1: Understand bitwise AND check

   For powers of two, binary form has only one '1'. n & (n-1) clears the lowest set bit. If result is 0, n is power of two.

2. Step 2: Analyze options

   n & (n - 1) == 0 uses correct bitwise AND check. Others use OR, XOR, or addition which do not correctly check power of two.

3. Final Answer:

   n & (n - 1) == 0 -> Option B

4. Quick Check:

   Bitwise AND check is standard for power of two [OK]

Hint: Use n & (n-1) == 0 to check power of two [OK]

Common Mistakes:

• Using OR or XOR instead of AND
• Forgetting to check n > 0 before this test

3. What is the output of the following code?

def is_power_of_two(n):
    return n > 0 and (n & (n - 1)) == 0

print(is_power_of_two(32))
print(is_power_of_two(45))

medium

A. False\nFalse

B. False\nTrue

C. True\nFalse

D. True\nTrue

Solution

1. Step 1: Check 32

   32 is 2^5, so n > 0 and n & (n-1) == 0 is True.

2. Step 2: Check 45

   45 is not a power of two, so n & (n-1) != 0, returns False.

3. Final Answer:

   True\nFalse -> Option C

4. Quick Check:

   32 power of two = True, 45 not = False [OK]

Hint: Check output for known powers of two and non-powers [OK]

Common Mistakes:

• Ignoring n > 0 condition
• Assuming all even numbers are powers of two

4. Find the error in this code snippet that checks if a number is a power of two:

def check_power(n):
    if n & (n - 1) == 0:
        return True
    else:
        return False

print(check_power(0))

medium

A. Missing check for n > 0 before bitwise operation

B. No error, code works fine

C. Should use OR instead of AND

D. Return values should be strings 'True' or 'False'

Solution

1. Step 1: Analyze bitwise check

   n & (n-1) == 0 works only if n > 0. For n=0, it returns True incorrectly.

2. Step 2: Identify missing condition

   Code lacks n > 0 check, so 0 incorrectly returns True.

3. Final Answer:

   Missing check for n > 0 before bitwise operation -> Option A

4. Quick Check:

   Always check n > 0 before bitwise test [OK]

Hint: Always check n > 0 before bitwise power of two test [OK]

Common Mistakes:

• Not handling zero or negative inputs
• Using wrong bitwise operators

5. You want to filter a list to keep only numbers that are powers of two. Which code snippet correctly does this?

nums = [0, 1, 2, 3, 4, 5, 8, 16, 18]

A) result = [x for x in nums if x & (x - 1) == 0]
B) result = [x for x in nums if x > 0 and x & (x - 1) == 0]
C) result = [x for x in nums if x | (x - 1) == 0]
D) result = [x for x in nums if x > 0 or x & (x - 1) == 0]

hard

A. result = [x for x in nums if x & (x - 1) == 0]

B. result = [x for x in nums if x > 0 or x & (x - 1) == 0]

C. result = [x for x in nums if x | (x - 1) == 0]

D. result = [x for x in nums if x > 0 and x & (x - 1) == 0]

Solution

1. Step 1: Understand filtering condition

   We must check x > 0 to avoid zero or negative numbers passing the test.

2. Step 2: Analyze each option

   result = [x for x in nums if x > 0 and x & (x - 1) == 0] correctly uses x > 0 and bitwise check. result = [x for x in nums if x & (x - 1) == 0] misses x > 0, so 0 passes incorrectly. Options C and D use wrong operators or logic.

3. Final Answer:

   result = [x for x in nums if x > 0 and x & (x - 1) == 0] -> Option D

4. Quick Check:

   Filter with x > 0 and bitwise AND check [OK]

Hint: Filter with x > 0 and x & (x-1) == 0 for powers of two [OK]

Common Mistakes:

• Not filtering out zero
• Using OR instead of AND in condition
• Using bitwise OR instead of AND