Check if Number is Power of Two in DSA C# - Build from Scratch

Check if Number is Power of Two

📖 Scenario: You are working on a simple calculator app that needs to check if a given number is a power of two. This is useful for optimizing certain calculations and memory allocations.

🎯 Goal: Build a program that checks if a number is a power of two using a simple method and prints the result.

📋 What You'll Learn

Create an integer variable named number with the value 16

Create a boolean variable named isPowerOfTwo to store the check result

Use the bitwise operation (number & (number - 1)) == 0 to check if number is a power of two

Print "True" if number is a power of two, otherwise print "False"

💡 Why This Matters

🌍 Real World

Checking if a number is a power of two is useful in computer memory allocation, graphics programming, and optimizing algorithms.

💼 Career

Understanding bitwise operations and condition checks is important for software developers working on performance-critical applications.

1

Create the number variable

Create an integer variable called number and set it to 16.

DSA C#

# Your code here

Hint

Use int number = 16; to create the variable.

2

Create the boolean variable for the check

Create a boolean variable called isPowerOfTwo but do not assign it yet.

DSA C#

int number = 16;
// Your code here

Hint

Declare bool isPowerOfTwo; without assigning a value yet.

3

Check if number is power of two

Assign isPowerOfTwo the result of the expression (number & (number - 1)) == 0.

DSA C#

int number = 16;
bool isPowerOfTwo;
// Your code here

Hint

Use the bitwise AND operator & to check the power of two condition.

4

Print the result

Print the value of isPowerOfTwo using Console.WriteLine(isPowerOfTwo);.

DSA C#

int number = 16;
bool isPowerOfTwo;
isPowerOfTwo = (number & (number - 1)) == 0;
// Your code here

Hint

Use Console.WriteLine(isPowerOfTwo); to print the result.

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