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

The Big Idea

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

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 checking each number by dividing it repeatedly by 2 until you reach 1 or find a remainder.

The Problem

This manual method is slow and boring, especially if the list is long. It's easy to make mistakes when dividing many times, and it wastes a lot of time that could be used for other tasks.

The Solution

Using a simple check with bitwise operations, we can quickly tell if a number is a power of two without dividing. This method is fast, clean, and works perfectly even for very large numbers.

Before vs After

✗ Before

function isPowerOfTwo(num: number): boolean {
  if (num < 1) return false;
  while (num % 2 === 0) {
    num = num / 2;
  }
  return num === 1;
}

✓ After

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

What It Enables

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

Real Life Example

In computer games, checking if a number is a power of two helps in optimizing textures and memory, making games run smoother and look better.

Key Takeaways

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

Bitwise check is a fast and reliable solution.

This technique is useful in many programming and real-world scenarios.

Practice

1. What does it mean if a number is a power of two?

easy

A. It is divisible by 3

B. It is always an odd number

C. It can be written as 2 raised to some integer exponent

D. It is a prime number

Solution

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

   A number is a power of two if it equals 2 raised to an integer exponent, like 2^0=1, 2^1=2, 2^2=4, etc.

2. Step 2: Check other options

   Odd numbers, divisibility by 3, or primality do not define powers of two.

3. Final Answer:

   It can be written as 2 raised to some integer exponent -> 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 powers of two are odd
• Confusing powers of two with prime numbers
• Assuming divisibility by 3 means power of two

2. Which of the following TypeScript expressions correctly checks if a number n is a power of two using bitwise operations?

easy

A. return n > 0 && (n & (n - 1)) === 0;

B. return n > 0 && (n | (n - 1)) === 0;

C. return n > 0 && (n & (n + 1)) === 0;

D. return n > 0 && (n ^ (n - 1)) === 0;

Solution

1. Step 1: Understand the 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 each option

   return n > 0 && (n & (n - 1)) === 0; uses the correct bitwise AND with (n-1). Options A, B, and C use incorrect bitwise operators or operands.

3. Final Answer:

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

4. Quick Check:

   Bitwise AND with (n-1) = 0 means power of two [OK]

Hint: Use n & (n-1) === 0 to check power of two [OK]

Common Mistakes:

• Using bitwise OR instead of AND
• Using (n + 1) instead of (n - 1)
• Forgetting to check n > 0

3. What will be the output of this TypeScript code?

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

medium

A. true true

B. false true

C. false false

D. true false

Solution

1. Step 1: Check isPowerOfTwo(16)

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

2. Step 2: Check isPowerOfTwo(18)

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

3. Final Answer:

   true false -> Option D

4. Quick Check:

   16 is power of two, 18 is not [OK]

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

Common Mistakes:

• Assuming 18 is power of two because it's even
• Ignoring the n > 0 condition
• Misreading bitwise AND results

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

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

medium

A. Missing check for n > 0 to avoid false positives

B. Incorrect bitwise operator used

C. Function should return number, not boolean

D. Parentheses are misplaced

Solution

1. Step 1: Analyze the function logic

   The function uses bitwise AND correctly but does not check if n is positive.

2. Step 2: Understand why n > 0 is needed

   Without n > 0, input 0 returns true incorrectly because 0 & -1 = 0.

3. Final Answer:

   Missing check for n > 0 to avoid false positives -> Option A

4. Quick Check:

   Check n > 0 before bitwise test [OK]

Hint: Always check n > 0 before bitwise power of two test [OK]

Common Mistakes:

• Ignoring zero or negative inputs
• Assuming bitwise alone is enough
• Misunderstanding bitwise AND behavior

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

const numbers = [0, 1, 2, 3, 4, 5, 8, 16, 18];
const powersOfTwo = numbers.filter(n => ???);
console.log(powersOfTwo);

hard

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

B. n > 0 && (n & (n - 1)) === 0

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

D. n !== 0 && (n ^ (n - 1)) === 0

Solution

1. Step 1: Understand the filter condition

   We want to keep numbers that are powers of two, so the condition must check n > 0 and bitwise AND with (n-1) equals zero.

2. Step 2: Evaluate each option

   n > 0 && (n & (n - 1)) === 0 correctly uses the standard power of two check. Options B, C, and D use incorrect bitwise operations or conditions.

3. Final Answer:

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

4. Quick Check:

   Filter with bitwise AND and n > 0 [OK]

Hint: Filter with n > 0 and n & (n-1) === 0 [OK]

Common Mistakes:

• Using bitwise OR or XOR instead of AND
• Not checking n > 0
• Using (n + 1) instead of (n - 1)