Check if Number is Power of Two in DSA Java - Execution Trace

Concept Flow - Check if Number is Power of Two

Start with number n

↓

Check if n <= 0?

Yes→Return False

No↓

Compute n & (n-1)

↓

Is result 0?

Yes→Return True

No↓

Return False

Start with the number, check if positive, then use bitwise AND with n-1 to decide if it's a power of two.

Execution Sample

DSA Java

int n = 8;
boolean isPowerOfTwo = (n > 0) && ((n & (n - 1)) == 0);
System.out.println(isPowerOfTwo);

Check if 8 is a power of two using bitwise AND trick and print the result.

Execution Table

Step	Operation	Value of n	Calculation	Result	Output
1	Check if n > 0	8	8 > 0	true
2	Calculate n-1	8	8 - 1	7
3	Bitwise AND n & (n-1)	8 & 7	1000 & 0111 (binary)	0
4	Check if result == 0	0	0 == 0	true
5	Final result	8		true	true

💡 Bitwise AND result is 0 and n > 0, so 8 is a power of two.

Variable Tracker

Variable	Start	After Step 2	After Step 3	Final
n	8	8	8	8
n-1	N/A	7	7	7
n & (n-1)	N/A	N/A	0	0
isPowerOfTwo	N/A	N/A	N/A	true

Key Moments - 3 Insights

Why do we check if n is greater than 0 before the bitwise operation?

Because powers of two are positive numbers. If n is zero or negative, it cannot be a power of two. This is shown in Step 1 of the execution_table.

Why does n & (n-1) equal zero only for powers of two?

Because powers of two have only one '1' bit in binary. Subtracting 1 flips that bit and all lower bits, so ANDing them clears all bits resulting in zero. This is shown in Step 3 and Step 4.

What happens if n is not a power of two?

The bitwise AND n & (n-1) will not be zero, so the final check in Step 4 will be false, and the function returns false.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of n & (n-1) at Step 3?

A 8

B 7

C 0

D 1

💡 Hint

Check the 'Calculation' and 'Result' columns in Step 3 of the execution_table.

At which step does the code confirm that n is positive?

A Step 1

B Step 2

C Step 4

D Step 5

💡 Hint

Look at the 'Operation' column for the positivity check in the execution_table.

If n was 10 instead of 8, what would be the result of n & (n-1)?

A 0

B 8

C 2

D 10

💡 Hint

Try mentally calculating 10 & 9 in binary and compare with Step 3's calculation.

Concept Snapshot

Check if number n is power of two:
- Must be > 0
- Use bitwise AND: n & (n-1)
- If result is 0, n is power of two
- Efficient and fast bit trick
- Example: 8 & 7 = 0 means 8 is power of two

Full Transcript

To check if a number is a power of two, first ensure it is positive. Then perform a bitwise AND between the number and one less than the number. If the result is zero, the number is a power of two. For example, 8 in binary is 1000, and 7 is 0111. Their AND is 0000, so 8 is a power of two. This method is efficient and uses simple bit operations.

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 exactly one bit set to 1 in its binary form.

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, e.g., 4 is 100.

2. Step 2: Check other options

   Odd numbers and divisibility by 3 do not define powers of two. Having all bits set to 1 is not a power of two.

3. Final Answer:

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

4. Quick Check:

   Power of two = single 1 bit [OK]

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

Common Mistakes:

• Thinking all odd numbers are powers of two
• Confusing powers of two with numbers having all bits set
• Assuming divisibility by 3 means power of two

2. Which of the following Java code snippets 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;

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

Solution

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

   The expression n & (n - 1) clears the lowest set bit. For powers of two, this results in zero.

2. Step 2: Analyze other options

   Options A, C, and D use incorrect bitwise operations that do not correctly check power of two.

3. Final Answer:

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

4. Quick Check:

   Use n & (n-1) == 0 to check 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 in bitwise check
• Checking n & (n + 1) which is incorrect
• Not considering zero or negative numbers

3. What is the output of the following Java code?

int n = 16;
boolean result = (n & (n - 1)) == 0;
System.out.println(result);

medium

A. true

B. Compilation error

C. false

D. Runtime exception

Solution

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

   16 in binary is 10000, 15 is 01111. Bitwise AND is 00000 (0).

2. Step 2: Compare result to zero

   Since 16 & 15 == 0, the expression is true, so result 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, print true [OK]

Common Mistakes:

• Confusing bitwise AND with OR
• Expecting false for powers of two
• Ignoring parentheses in expression

4. The following Java method is intended to check if a number is a power of two. What is the error?

public static boolean isPowerOfTwo(int n) {
    return (n & n - 1) == 0;
}

medium

A. Method does not handle negative numbers.

B. Using bitwise OR instead of AND.

C. Incorrect return type.

D. Missing parentheses around (n - 1) causing wrong calculation.

Solution

1. Step 1: Identify issue with negative numbers

   For n = Integer.MIN_VALUE (-2147483648), which is 0x80000000 in two's complement, n-1 = 0x7FFFFFFF (Integer.MAX_VALUE), and 0x80000000 & 0x7FFFFFFF == 0, so it incorrectly returns true.

2. Step 2: Rule out other options

   Return type is correct, it uses AND (&), and operator precedence correctly parses as n & (n - 1) (no missing parentheses needed).

3. Final Answer:

   Method does not handle negative numbers. -> Option A

4. Quick Check:

   n > 0 && (n & (n-1)) == 0 [OK]

Hint: Include n > 0 to handle zero and negatives [OK]

Common Mistakes:

• Not checking n > 0 before bitwise operation
• Ignoring negative input handling
• Confusing bitwise AND with OR

5. You want to write a Java method that returns true if a given positive integer n is a power of two, and false otherwise. Which of the following implementations correctly handles all positive integers including 1, and excludes zero and negative numbers?

hard

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

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

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

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

Solution

1. Step 1: Check for positive number condition

   Zero and negative numbers are not powers of two, so n > 0 is required.

2. Step 2: Use bitwise check for power of two

   The expression (n & (n - 1)) == 0 correctly identifies powers of two including 1.

3. Step 3: Analyze other options

   return (n & (n - 1)) == 0; misses positive check, allowing zero. Options C and D use incorrect bitwise operations.

4. Final Answer:

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

5. Quick Check:

   Check positive and n & (n-1) == 0 [OK]

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

Common Mistakes:

• Not excluding zero and negative numbers
• Using incorrect bitwise operators
• Missing parentheses around (n - 1)