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

Check if Number is Power of Two

📖 Scenario: Imagine you are working with a system that needs to quickly check if certain numbers are powers of two. Powers of two are numbers like 1, 2, 4, 8, 16, and so on. This is useful in computer memory allocation and optimization.

🎯 Goal: You will build a simple TypeScript program that checks if a given number is a power of two using a clear step-by-step approach.

📋 What You'll Learn

Create a variable to store the number to check

Create a helper variable to hold the result

Write the core logic to check if the number is a power of two

Print the result as true or false

💡 Why This Matters

🌍 Real World

Checking if a number is a power of two is useful in computer memory management and optimization tasks.

💼 Career

Understanding bitwise operations and condition checks is important for software developers working with low-level programming and performance optimization.

1

Create the number variable

Create a variable called num and set it to 16.

DSA Typescript

# Your code here

Hint

Use const to create the variable num and assign the value 16.

2

Create the result variable

Create a variable called isPowerOfTwo and set it to false.

DSA Typescript

const num = 16;
# Your code here

Hint

Use let to create isPowerOfTwo and start with false.

3

Check if the number is a power of two

Write an if statement that sets isPowerOfTwo to true if num is greater than 0 and (num & (num - 1)) === 0.

DSA Typescript

const num = 16;
let isPowerOfTwo = false;
# Your code here

Hint

This bitwise trick checks if only one bit is set in num. Remember to check num & (num - 1) equals zero.

4

Print the result

Write console.log(isPowerOfTwo) to print the result.

DSA Typescript

const num = 16;
let isPowerOfTwo = false;
if (num > 0 && (num & (num - 1)) === 0) {
    isPowerOfTwo = true;
}
# Your code here

Hint

Use console.log to show the value of isPowerOfTwo.

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)