Why Check if Number is Power of Two in DSA Javascript?

The Big Idea

Discover a simple trick that turns a slow, boring check into a lightning-fast operation!

The Scenario

Imagine you have a big list of numbers and you want to find out which ones are powers of two, like 1, 2, 4, 8, 16, and so on. Doing this by hand means dividing each number by 2 again and again to see if you end up with 1.

The Problem

Doing this division repeatedly for many numbers is slow and tiring. It's easy to make mistakes, especially if the numbers are very large. Also, manually checking wastes a lot of time when you have thousands of numbers.

The Solution

Using a simple trick with bits, computers can quickly check if a number is a power of two without dividing. This method is fast and accurate, even for very large numbers, saving time and effort.

Before vs After

✗ Before

function isPowerOfTwo(number) {
  while (number > 1) {
    if (number % 2 !== 0) return false;
    number = number / 2;
  }
  return number === 1;
}

✓ After

function isPowerOfTwo(number) {
  return number > 0 && (number & (number - 1)) === 0;
}

What It Enables

This lets you instantly know if a number is a power of two, enabling faster programs and smarter decisions in tasks like memory allocation and game development.

Real Life Example

When a game engine needs to allocate memory for textures, it often requires sizes that are powers of two for better performance. Quickly checking this helps the game run smoothly.

Key Takeaways

Manual division to check powers of two is slow and error-prone.

Bitwise operations provide a fast and reliable solution.

This technique is useful in programming tasks involving memory and optimization.

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