Check if Number is Power of Two in DSA Javascript - 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 Javascript

function isPowerOfTwo(n) {
  return n > 0 && (n & (n [1] 1)) === 0;
}

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

A+

B|

C&

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition instead of subtraction.

Using bitwise OR instead of AND.

The expression n & (n - 1) checks if n is a power of two by clearing the lowest set bit. If the result is zero, n is a power of two.

2fill in blank

medium

Complete the code to return false for non-positive numbers.

DSA Javascript

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

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

A<=

B<

C==

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using equality check instead of less than or equal to.

Using greater than or equal to.

Numbers less than or equal to zero cannot be powers of two. Using n <= 0 excludes zero and negative numbers.

3fill in blank

hard

Fix the error in the bitwise check to correctly identify power of two.

DSA Javascript

function isPowerOfTwo(n) {
  return n > 0 && (n & (n [1] 1)) === 0;
}

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

A+

B-

C|

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition instead of subtraction.

Using bitwise OR or AND incorrectly.

The correct operation is n & (n - 1). Using subtraction clears the lowest set bit to check power of two.

4fill in blank

hard

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

DSA Javascript

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

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

A<=

B<

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition instead of subtraction.

Using greater than or equal to in the first check.

First, check if n is less than or equal to zero to exclude invalid inputs. Then use n & (n - 1) to check power of two.

5fill in blank

hard

Fill all three blanks to create a function that returns true if n is a power of two, false otherwise, and handles zero and negative inputs.

DSA Javascript

function isPowerOfTwo(n) {
  if (n [1] 0) return false;
  return (n [2] (n [3] 1)) === 0;
}

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

A<=

B>

C-

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of <= in the first check.

Using + instead of - in the subtraction.

Using | instead of & for bitwise operation.

Check if n is less than or equal to zero to exclude invalid inputs. Then use bitwise AND between n and (n - 1) to check power of two.

Practice

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

easy

A. It is any even number greater than zero.

B. It is any number divisible by 3.

C. It can be written as 2 raised to an integer exponent (e.g., 2^3 = 8).

D. It is a number that ends with digit 2.

Solution

1. Step 1: Understand the definition of power of two

   A power of two is a number that can be expressed as 2 raised to some integer exponent, like 2^0=1, 2^1=2, 2^2=4, etc.

2. Step 2: Compare options with the definition

   Only It can be written as 2 raised to an integer exponent (e.g., 2^3 = 8). correctly describes this. Options A, C, and D do not match the definition.

3. Final Answer:

   It can be written as 2 raised to an integer exponent (e.g., 2^3 = 8). -> Option C

4. Quick Check:

   Power of two = 2^n [OK]

Hint: Power of two means 2 multiplied by itself n times [OK]

Common Mistakes:

• Thinking all even numbers are powers of two
• Confusing divisibility by 3 with powers of two
• Assuming number ending with 2 is power of two

2. Which JavaScript expression correctly checks if a number n is a power of two using bitwise operations?

easy

A. n & (n - 1) === 0

B. n | (n - 1) === 0

C. n ^ (n - 1) === 0

D. n & (n + 1) === 0

Solution

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

   A number that is a power of two has only one bit set. The expression n & (n - 1) clears the lowest set bit. If the result is zero, it means only one bit was set.

2. Step 2: Analyze each option

   n & (n - 1) === 0 uses the correct expression. Options B, C, and D use incorrect bitwise operators or operands.

3. Final Answer:

   n & (n - 1) === 0 -> Option A

4. Quick Check:

   Bitwise AND with n-1 equals zero means 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
• Using n + 1 instead of n - 1
• Not comparing result to zero

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

function isPowerOfTwo(n) {
  return n > 0 && (n & (n - 1)) === 0;
}
console.log(isPowerOfTwo(16));
console.log(isPowerOfTwo(18));

medium

A. false\nfalse

B. false\ntrue

C. true\ntrue

D. true\nfalse

Solution

1. Step 1: Evaluate isPowerOfTwo(16)

   16 in binary is 10000. 16 & 15 (01111) = 0, so returns true.

2. Step 2: Evaluate isPowerOfTwo(18)

   18 in binary is 10010. 18 & 17 (10001) = 16 (not zero), so returns false.

3. Final Answer:

   true\nfalse -> Option D

4. Quick Check:

   16 is power of two: true, 18 is not: false [OK]

Hint: Check bitwise AND with n-1 for zero to confirm power of two [OK]

Common Mistakes:

• Assuming all even numbers return true
• Forgetting to check n > 0
• Misunderstanding bitwise AND result

4. Find the error in this JavaScript function that checks if a number is a power of two:

function checkPower(n) {
  return (n & n - 1) === 0;
}

medium

A. Using bitwise OR instead of AND

B. Not checking if n is positive

C. Missing parentheses around (n & (n - 1))

D. Using subtraction instead of addition

Solution

1. Step 1: Test with edge case n=0

   0 & (0 - 1) = 0 & (-1) = 0 === 0 returns true, but 0 is not a power of two.

2. Step 2: Analyze options

   The function lacks n > 0 check. Not checking if n is positive identifies this. No OR used (B wrong), parentheses not required due to operator precedence (C wrong), subtraction is correct (D wrong).

3. Final Answer:

   Not checking if n is positive -> Option B

4. Quick Check:

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

Hint: Include n > 0 check for power of two validation [OK]

Common Mistakes:

• Ignoring operator precedence
• Not checking if number is positive
• Using wrong bitwise operators

5. You want to filter an array nums to keep only numbers that are powers of two. Which code snippet correctly does this in JavaScript?

hard

A. nums.filter(n => n > 0 && (n & (n - 1)) === 0)

B. nums.filter(n => n % 2 === 0)

C. nums.filter(n => n & (n + 1) === 0)

D. nums.filter(n => n > 0 && (n | (n - 1)) === 0)

Solution

1. Step 1: Understand filtering condition for power of two

   We need to keep numbers greater than zero where n & (n - 1) === 0 is true.

2. Step 2: Evaluate each filter condition

   nums.filter(n => n > 0 && (n & (n - 1)) === 0) correctly uses this condition. nums.filter(n => n % 2 === 0) filters even numbers, which is incorrect. Options C and D use wrong bitwise operations.

3. Final Answer:

   nums.filter(n => n > 0 && (n & (n - 1)) === 0) -> Option A

4. Quick Check:

   Filter with bitwise AND and n > 0 [OK]

Hint: Filter with n > 0 and n & (n-1) === 0 to find powers of two [OK]

Common Mistakes:

• Filtering only even numbers
• Using wrong bitwise operators
• Not checking for positive numbers