Check if Number is Power of Two in DSA Java

Mental Model

A number is a power of two if it has exactly one '1' bit in its binary form.

Analogy: Imagine a row of light switches where only one switch is turned on at a time; that means the number is a power of two.

Number: 8
Binary: 1 0 0 0
          ↑
Only one '1' bit means power of two.

Dry Run Walkthrough

Input: number = 8

Goal: Determine if 8 is a power of two

Step 1: Check if number is zero or negative

number = 8 (valid positive number)

Why: Only positive numbers can be powers of two

Step 2: Calculate number & (number - 1): 8 & 7

8 in binary: 1000
7 in binary:  0111
AND result:    0000

Why: If result is zero, number has only one '1' bit

Step 3: Since result is zero, conclude number is power of two

8 is power of two

Why: Only powers of two have this property

Result:

8 -> 1000
7 -> 0111
8 & 7 = 0 -> number is power of two

Annotated Code

DSA Java

public class PowerOfTwoChecker {
    public static boolean isPowerOfTwo(int number) {
        if (number <= 0) {
            return false; // non-positive numbers are not powers of two
        }
        return (number & (number - 1)) == 0; // true if only one '1' bit
    }

    public static void main(String[] args) {
        int number = 8;
        boolean result = isPowerOfTwo(number);
        System.out.println(number + " is power of two: " + result);
    }
}

if (number <= 0) {

Check for invalid input (non-positive numbers)

return (number & (number - 1)) == 0;

Check if number has exactly one '1' bit in binary

Output

8 is power of two: true

Complexity Analysis

Time: O(1) because bitwise operations take constant time

Space: O(1) because no extra space is used

vs Alternative: Compared to looping and dividing by 2 repeatedly (O(log n)), this bitwise method is faster and simpler

Edge Cases

number = 0

Returns false because zero is not a power of two

DSA Java

if (number <= 0) {

number = 1

Returns true because 1 is 2^0, a power of two

DSA Java

return (number & (number - 1)) == 0;

number = -4

Returns false because negative numbers are not powers of two

DSA Java

if (number <= 0) {

When to Use This Pattern

When you need to quickly check if a number is a power of two, use the bitwise AND trick number & (number - 1) == 0 because it efficiently tests the single '1' bit property.

Common Mistakes

Mistake: Checking if number % 2 == 0 repeatedly without stopping
Fix: Use the bitwise method instead of loops for constant time check

Mistake: Not handling zero or negative inputs
Fix: Add a check to return false if number <= 0

Summary

Checks if a number has exactly one '1' bit in binary to determine if it is a power of two.

Use when you need a fast and simple way to verify powers of two.

The key insight is that powers of two have only one bit set, so number & (number - 1) equals zero.

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)