Check if Number is Power of Two in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of power of two check using bitwise AND

What is the output of the following JavaScript code snippet?

DSA Javascript

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

A true

B false

C undefined

D TypeError

💡 Hint

Remember that powers of two have only one bit set in binary.

The function returns true if n is greater than zero and n & (n-1) equals zero, which is true only for powers of two.

❓ Predict Output

intermediate

Output of power of two check with loop

What will be printed by this code?

DSA Javascript

function isPowerOfTwo(n) {
  if (n < 1) return false;
  while (n % 2 === 0) {
    n = n / 2;
  }
  return n === 1;
}
console.log(isPowerOfTwo(18));

A true

B false

C Infinity

D NaN

💡 Hint

Check if the number can be divided by 2 repeatedly until it becomes 1.

18 is not a power of two because dividing by 2 repeatedly does not end at 1.

🔧 Debug

advanced

Identify the error in power of two check

What error will this code produce when run?

DSA Javascript

function isPowerOfTwo(n) {
  return n && (n & (n - 1)) === 0;
}
console.log(isPowerOfTwo(8));

A true

B false

C TypeError

D SyntaxError

💡 Hint

Check operator precedence and how the expression is evaluated.

The expression is parsed as n && ((n & (n - 1)) === 0) because === has higher precedence than &&. For n=8, (8 & 7) = 0, 0 === 0 is true, then 8 && true is true. The code correctly returns and prints true.

❓ Predict Output

advanced

Output of recursive power of two check

What is the output of this recursive function call?

DSA Javascript

function isPowerOfTwo(n) {
  if (n === 1) return true;
  if (n < 1 || n % 2 !== 0) return false;
  return isPowerOfTwo(n / 2);
}
console.log(isPowerOfTwo(32));

A undefined

B false

C true

D RangeError

💡 Hint

Think about the base case and how the function reduces the problem size.

The function correctly checks if n is a power of two by recursively dividing by 2 until it reaches 1.

🧠 Conceptual

expert

Why does n & (n - 1) === 0 check power of two?

Which explanation best describes why the expression 'n & (n - 1) === 0' checks if n is a power of two?

A Because n-1 always equals zero for powers of two, so ANDing returns zero.

B Because powers of two are always even numbers, and ANDing with n-1 removes the last bit.

C Because the expression counts the number of set bits in n and returns zero if only one bit is set.

D Because powers of two have only one bit set, subtracting 1 flips all bits after that bit, so ANDing results in zero.

💡 Hint

Think about binary representation of powers of two and what subtracting 1 does.

Powers of two have exactly one bit set. Subtracting 1 flips that bit and all bits after it, so ANDing with n clears that bit and results in zero only for powers 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