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

Check if Number is Power of Two

📖 Scenario: Imagine you are working on a simple calculator program that needs to check if a given number is a power of two. This is useful in many computer applications like memory allocation and optimization.

🎯 Goal: You will write a program in C that checks if a number is a power of two using bitwise operations.

📋 What You'll Learn

Create an integer variable called number with the value 16

Create an integer variable called result to store the check result

Use a bitwise operation to check if number is a power of two and store the result in result

Print "Power of Two" if result is true, otherwise print "Not Power of Two"

💡 Why This Matters

🌍 Real World

Checking if a number is a power of two is important in computer memory management and optimization tasks.

💼 Career

Understanding bitwise operations and condition checks is useful for software developers working on low-level programming and performance optimization.

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 result variable

Create an integer variable called result to store the check result.

DSA C

int number = 16;
# Your code here

Hint

Declare int result; without initializing it yet.

3

Check if number is power of two

Use the expression number & (number - 1) to check if number is a power of two. Assign the result of the comparison (number != 0 && (number & (number - 1)) == 0) to the variable result.

DSA C

int number = 16;
int result;
# Your code here

Hint

Use bitwise AND & and check if number is not zero.

4

Print the result

Write a printf statement that prints "Power of Two" if result is true, otherwise prints "Not Power of Two".

DSA C

int number = 16;
int result;
result = (number != 0 && (number & (number - 1)) == 0);
# Your code here

Hint

Use if (result) to decide what to print.

Practice

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

easy

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

B. It is always an odd number.

C. It is divisible by 3.

D. It has all bits 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. For example, 4 is 100 in binary.

2. Step 2: Analyze the options

   It has exactly one bit set to 1 in its binary form. matches the binary property of powers of two. Options B, C, and D do not describe powers of two correctly.

3. Final Answer:

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

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
• Assuming powers of two have multiple '1' bits
• Confusing divisibility rules

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

easy

A. if (n & (n - 1)) { /* not power of two */ } else { /* power of two */ }

B. if (n | (n - 1)) { /* power of two */ } else { /* not power of two */ }

C. if (n & (n + 1)) { /* power of two */ } else { /* not power of two */ }

D. if (n ^ (n - 1)) { /* power of two */ } else { /* not power of two */ }

Solution

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

   The expression n & (n - 1) is zero only if n is a power of two.

2. Step 2: Analyze each option

   if (n & (n - 1)) { /* not power of two */ } else { /* power of two */ } uses the correct expression. Options A, C, and D use incorrect bitwise operations that do not correctly check for power of two.

3. Final Answer:

   if (n & (n - 1)) { /* not power of two */ } else { /* power of two */ } -> Option A

4. Quick Check:

   n & (n-1) == 0 means power of two [OK]

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

Common Mistakes:

• Using OR instead of AND in bitwise check
• Using n + 1 instead of n - 1
• Confusing XOR with AND

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

int n = 16;
if (n & (n - 1))
    printf("No\n");
else
    printf("Yes\n");

medium

A. Runtime error

B. No

C. Compilation error

D. Yes

Solution

1. Step 1: Calculate n & (n - 1) for n = 16

   16 in binary is 10000. 15 is 01111. Bitwise AND: 10000 & 01111 = 00000 (0).

2. Step 2: Check condition and output

   Since the result is 0, the else block executes, printing "Yes".

3. Final Answer:

   Yes -> Option D

4. Quick Check:

   16 & 15 = 0 means power of two [OK]

Hint: If n & (n-1) is zero, print Yes [OK]

Common Mistakes:

• Miscomputing bitwise AND
• Confusing output branches
• Assuming non-zero means power of two

4. Identify the error in the following C code that checks if n is a power of two:

int n = 10;
if (n & n - 1 == 0)
    printf("Power of two\n");
else
    printf("Not power of two\n");

medium

A. No error, code is correct

B. Missing parentheses around (n - 1) in bitwise operation

C. Incorrect variable initialization

D. Using bitwise OR instead of AND

Solution

1. Step 1: Analyze operator precedence in condition

   In C, == has higher precedence than &, so expression is evaluated as n & ((n - 1) == 0) which is incorrect.

2. Step 2: Correct the expression with parentheses

   Use (n & (n - 1)) == 0 to ensure correct evaluation.

3. Final Answer:

   Missing parentheses around (n - 1) in bitwise operation -> Option B

4. Quick Check:

   Use parentheses to fix precedence [OK]

Hint: Add parentheses: (n & (n - 1)) == 0 [OK]

Common Mistakes:

• Ignoring operator precedence
• Not using parentheses in bitwise expressions
• Assuming code compiles correctly without fix

5. You want to write a function in C that returns 1 if a number n is a power of two and 0 otherwise. Which of the following implementations correctly handles the case when n is zero or negative?

hard

A. int isPowerOfTwo(int n) { return n >= 0 && (n & (n + 1)) == 0; }

B. int isPowerOfTwo(int n) { return (n & (n - 1)) == 0; }

C. int isPowerOfTwo(int n) { return n > 0 && (n & (n - 1)) == 0; }

D. int isPowerOfTwo(int n) { return n != 0 || (n & (n - 1)) == 0; }

Solution

1. Step 1: Consider zero and negative inputs

   Zero and negative numbers are not powers of two, so the function must exclude them.

2. Step 2: Analyze each function

   int isPowerOfTwo(int n) { return n > 0 && (n & (n - 1)) == 0; } checks if n > 0 and then applies the bitwise check correctly. int isPowerOfTwo(int n) { return (n & (n - 1)) == 0; } misses the positive check. Options A, B, and D use incorrect logic or operators.

3. Final Answer:

   int isPowerOfTwo(int n) { return n > 0 && (n & (n - 1)) == 0; } -> Option C

4. Quick Check:

   Check n > 0 before bitwise test [OK]

Hint: Check n > 0 before bitwise test to avoid errors [OK]

Common Mistakes:

• Not handling zero or negative inputs
• Using wrong bitwise operators
• Using OR instead of AND in conditions