Check if Number is Power of Two in DSA Javascript - Build from Scratch

Check if Number is Power of Two

📖 Scenario: Imagine you are working with a system that needs to quickly check if a number is a power of two. This is useful in many areas like computer memory allocation, graphics, and game development.

🎯 Goal: You will write a small program that checks if a given number is a power of two using a simple method.

📋 What You'll Learn

Create a variable called number with the value 16

Create a variable called result to store the check result

Use a condition to check if number is a power of two using bitwise AND operation

Print the result which should be true or false

💡 Why This Matters

🌍 Real World

Checking if a number is a power of two is important in computer memory management and optimizing algorithms.

💼 Career

This concept is useful for software developers working with low-level programming, game development, and performance optimization.

1

Create the number variable

Create a variable called number and set it to 16.

DSA Javascript

// Your code here

Hint

Use const number = 16; to create the variable.

2

Create the result variable

Create a variable called result and set it to false initially.

DSA Javascript

const number = 16;
// Your code here

Hint

Use let result = false; to create the variable.

3

Check if number is power of two

Use an if statement to check if number is greater than zero and if number & (number - 1) equals zero. If yes, set result to true.

DSA Javascript

const number = 16;
let result = false;
// Your code here

Hint

Use the bitwise AND operator & to check the power of two condition.

4

Print the result

Print the value of result using console.log(result).

DSA Javascript

const number = 16;
let result = false;
if (number > 0 && (number & (number - 1)) === 0) {
    result = true;
}
// Your code here

Hint

Use console.log(result); to print the result.

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