Check if Number is Power of Two in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does it mean for a number to be a power of two?

A number is a power of two if it can be written as 2 raised to some integer exponent, like 1, 2, 4, 8, 16, etc.

intermediate

Which bitwise operation can quickly check if a number is a power of two?

Using (n & (n - 1)) == 0 checks if n is a power of two, because powers of two have only one bit set.

intermediate

Why must we check if the number is greater than zero before using (n & (n - 1))?

Because zero and negative numbers are not powers of two, and (n & (n - 1)) == 0 is true for zero, so we exclude zero by checking n > 0.

beginner

What is the output of the function if the input number is 16?

The function returns true because 16 is 2^4, a power of two.

beginner

Write the C code snippet to check if an integer n is a power of two.

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

Which of these numbers is NOT a power of two?

A 10

B 8

C 16

D 32

10 is not a power of two because it cannot be written as 2 raised to an integer.

What does the expression (n & (n - 1)) == 0 check for?

A If n is zero

B If n is a power of two

C If n is negative

D If n is odd

This expression is true only when n is a power of two (and n > 0).

Why do we check n > 0 before applying (n & (n - 1)) == 0?

A Because it improves performance

B Because negative numbers are powers of two

C Because zero also satisfies (n & (n - 1)) == 0 but is not a power of two

D Because it checks if n is odd

Zero satisfies the bitwise condition but is not a power of two, so we exclude it by checking n > 0.

What is the output of isPowerOfTwo(1)?

A True

B False

C Error

D Depends on compiler

1 is 2^0, so it is a power of two and the function returns true.

Which of these is a correct way to check if n is a power of two in C?

A return n % 2 == 0;

B return n & (n - 1);

C return n == 0 || n == 1;

D return n > 0 && (n & (n - 1)) == 0;

Option D correctly checks if n is positive and has only one bit set.

Explain how to check if a number is a power of two using bitwise operations.

Think about how powers of two look in binary.

Write a simple C function to check if a number is a power of two and explain why it works.

Use (n & (n - 1)) == 0 condition.

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