Check if Number is Power of Two in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a number is a power of two using bitwise AND.

DSA Java

public boolean isPowerOfTwo(int n) {
    return n > 0 && (n & [1]) == 0;
}

Drag options to blanks, or click blank then click option'

An - 1

Bn + 1

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using n + 1 instead of n - 1

Not checking if n is greater than zero

Using bitwise OR instead of AND

The expression n & (n - 1) clears the lowest set bit. If the result is zero, then n is a power of two.

2fill in blank

medium

Complete the code to return false immediately if the number is not positive.

DSA Java

public boolean isPowerOfTwo(int n) {
    if ([1]) {
        return false;
    }
    return (n & (n - 1)) == 0;
}

Drag options to blanks, or click blank then click option'

An >= 0

Bn == 0

Cn <= 0

Dn < 0

Attempts:

3 left

💡 Hint

Common Mistakes

Only checking for zero but not negative numbers

Using n >= 0 which includes zero and positive numbers

Numbers less than or equal to zero cannot be powers of two, so we return false for those.

3fill in blank

hard

Fix the error in the code to correctly check if a number is a power of two.

DSA Java

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

Drag options to blanks, or click blank then click option'

A0

Bn + 1

Cn - 1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if the result equals 1 instead of 0

Using n + 1 instead of n - 1

The correct check is if (n & (n - 1)) == 0, not 1. This means only one bit is set.

4fill in blank

hard

Fill both blanks to complete the method that returns true if n is a power of two and false otherwise.

DSA Java

public boolean isPowerOfTwo(int n) {
    if ([1]) {
        return false;
    }
    return (n & [2]) == 0;
}

Drag options to blanks, or click blank then click option'

An <= 0

Bn == 0

Cn - 1

Dn + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n + 1 instead of n - 1

Not checking if n is positive

First, check if n is less than or equal to zero to return false. Then use n & (n - 1) to check if n is a power of two.

5fill in blank

hard

Fill all three blanks to complete the method that checks if a number is a power of two and prints the result.

DSA Java

public void checkPowerOfTwo(int n) {
    boolean result = [1];
    System.out.println("Is " + n + " a power of two? " + [2]);
    if ([3]) {
        System.out.println("Yes, it is.");
    } else {
        System.out.println("No, it is not.");
    }
}

Drag options to blanks, or click blank then click option'

AisPowerOfTwo(n)

Bresult

Dn > 0

Attempts:

3 left

💡 Hint

Common Mistakes

Printing the method call directly without storing

Using n > 0 in the if condition instead of the result

Call the method isPowerOfTwo(n) to get the result. Print the result variable. Use the result variable in the if condition.

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)