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

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code snippet?

Given the following C# code, what will be printed to the console?

DSA C#

int n = 16;
bool isPowerOfTwo = (n > 0) && ((n & (n - 1)) == 0);
Console.WriteLine(isPowerOfTwo);

A False

B 1

C 0

D True

💡 Hint

Check how bitwise AND works between n and n-1 for powers of two.

The expression (n & (n - 1)) == 0 is true only if n is a power of two. Since 16 is 2^4, the output is True.

❓ Predict Output

intermediate

What is the output when n = 0?

Consider the same code but with n = 0. What will be printed?

DSA C#

int n = 0;
bool isPowerOfTwo = (n > 0) && ((n & (n - 1)) == 0);
Console.WriteLine(isPowerOfTwo);

A Exception

B False

C 0

D True

💡 Hint

Remember that zero is not a power of two.

Since n is zero, the first condition n > 0 is false, so the whole expression is false. It prints False.

🔧 Debug

advanced

Why does this code fail for negative numbers?

What is the output of this code when n = -8? Explain why.

DSA C#

int n = -8;
bool isPowerOfTwo = (n > 0) && ((n & (n - 1)) == 0);
Console.WriteLine(isPowerOfTwo);

A False

B True

C Exception

D Undefined behavior

💡 Hint

Check the first condition and how negative numbers behave with bitwise operations.

The first condition n > 0 is false for negative numbers, so the expression is false. Negative numbers are not powers of two.

🧠 Conceptual

advanced

Which property of powers of two does the bitwise check use?

The expression (n & (n - 1)) == 0 is used to check if n is a power of two. What property does this rely on?

A Powers of two have alternating bits set

B Powers of two are always even numbers

C Powers of two have only one bit set in binary representation

D Powers of two are prime numbers

💡 Hint

Think about how binary numbers look for powers of two.

Powers of two have exactly one bit set to 1 in their binary form. The bitwise check clears the lowest set bit, so if the result is zero, only one bit was set.

❓ Predict Output

expert

What is the output of this recursive function for n = 32?

Analyze the following C# code and determine what it prints when n = 32.

DSA C#

bool IsPowerOfTwo(int n) {
    if (n == 1) return true;
    if (n == 0 || n % 2 != 0) return false;
    return IsPowerOfTwo(n / 2);
}

int n = 32;
Console.WriteLine(IsPowerOfTwo(n));

A True

B False

C StackOverflowException

D Compile error

💡 Hint

Trace the recursive calls dividing n by 2 each time.

The function recursively divides n by 2. If n reaches 1, it returns true. Since 32 is a power of two, it returns True.

Practice

1. Which of the following statements correctly describes a number that is a power of two?

easy

A. It is divisible by 3.

B. It is always an odd number.

C. It has all bits set to 1 in its binary form.

D. It has exactly one bit set to 1 in its binary form.

Solution

1. Step 1: Understand binary representation of powers of two

   A power of two in binary has exactly one '1' bit and the rest are '0's, e.g., 4 is 100.

2. Step 2: Check other options for correctness

   Odd numbers and divisibility by 3 do not define powers of two. Having all bits set to 1 means numbers like 7 (111), not powers of two.

3. Final Answer:

   It has exactly one bit set to 1 in its binary form. -> Option D

4. Quick Check:

   Power of two = one '1' bit [OK]

Hint: Power of two has only one '1' bit in binary [OK]

Common Mistakes:

• Thinking powers of two are always odd
• Confusing all bits set to 1 with power of two
• Assuming divisibility by 3 means power of two

2. Which of the following C# expressions correctly checks if an integer n is a power of two?

easy

A. return (n & (n + 1)) == 0;

B. return (n | (n - 1)) == 0;

C. return (n & (n - 1)) == 0 && n > 0;

D. return (n ^ (n - 1)) == 0;

Solution

1. Step 1: Understand bitwise check for power of two

   For a power of two, n & (n - 1) clears the only set bit, resulting in 0. Also, n must be positive.

2. Step 2: Analyze other expressions

   Other options use incorrect bitwise operations that do not correctly identify powers of two.

3. Final Answer:

   return (n & (n - 1)) == 0 && n > 0; -> Option C

4. Quick Check:

   Bitwise AND with n-1 equals zero means power of two [OK]

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

Common Mistakes:

• Using OR or XOR instead of AND
• Not checking if n is positive
• Using n + 1 instead of n - 1

3. What is the output of the following C# code?

int n = 16;
bool result = (n & (n - 1)) == 0 && n > 0;
Console.WriteLine(result);

medium

A. True

B. False

C. Compile-time error

D. Runtime exception

Solution

1. Step 1: Calculate bitwise AND of n and n-1

   For n=16 (binary 10000), n-1=15 (binary 01111). Bitwise AND is 0.

2. Step 2: Check if result is zero and n is positive

   Since (n & (n - 1)) == 0 and n > 0, the expression is true.

3. Final Answer:

   True -> Option A

4. Quick Check:

   16 is power of two, so output is True [OK]

Hint: If n & (n-1) == 0 and n>0, output is True [OK]

Common Mistakes:

• Miscomputing bitwise AND
• Ignoring n > 0 condition
• Confusing True and False outputs

4. The following code is intended to check if n is a power of two. What is the error?

bool IsPowerOfTwo(int n) {
    return (n & (n - 1)) == 0;
}

medium

A. The expression should be (n | (n - 1)) == 0 instead.

B. It does not check if n is positive, so zero or negative numbers return true incorrectly.

C. The function should return an integer, not a boolean.

D. The bitwise AND operator is used incorrectly; it should be OR.

Solution

1. Step 1: Analyze the bitwise condition

   The expression (n & (n - 1)) == 0 is true for powers of two but also for zero.

2. Step 2: Identify missing condition

   Zero and negative numbers are not powers of two, so the function must check n > 0.

3. Final Answer:

   It does not check if n is positive, so zero or negative numbers return true incorrectly. -> Option B

4. Quick Check:

   Check n > 0 to avoid false positives [OK]

Hint: Always check n > 0 when using bitwise power of two check [OK]

Common Mistakes:

• Omitting n > 0 check
• Using wrong bitwise operator
• Returning wrong data type

5. You want to write a function that returns the smallest power of two greater than or equal to a given positive integer n. Which approach correctly achieves this in C#?

hard

A. Use a loop doubling a variable starting at 1 until it is >= n.

B. Return n & (n - 1) to get the smallest power of two.

C. Return n | (n - 1) to get the smallest power of two.

D. Return n ^ (n - 1) to get the smallest power of two.

Solution

1. Step 1: Understand the problem requirement

   We need the smallest power of two >= n, so start from 1 and double until reaching or exceeding n.

2. Step 2: Analyze other options

   Bitwise operations like AND, OR, XOR with n-1 do not produce the smallest power of two >= n.

3. Final Answer:

   Use a loop doubling a variable starting at 1 until it is >= n. -> Option A

4. Quick Check:

   Doubling loop finds smallest power of two >= n [OK]

Hint: Double from 1 until >= n to find smallest power of two [OK]

Common Mistakes:

• Using bitwise AND/OR/XOR incorrectly for this task
• Not handling n less than 1
• Assuming bitwise ops give next power of two