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DSA Javascriptprogramming~5 mins

Check if Number is Power of Two in DSA Javascript - Time & Space Complexity

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Time Complexity: Check if Number is Power of Two
O(1)
Understanding Time Complexity

We want to understand how fast we can check if a number is a power of two.

How does the time taken change as the number gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


function isPowerOfTwo(n) {
  if (n <= 0) return false;
  return (n & (n - 1)) === 0;
}

This code checks if a number is a power of two using a bitwise trick.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: A single bitwise AND operation and subtraction.
  • How many times: Exactly once, no loops or recursion.
How Execution Grows With Input

The number of operations stays the same no matter how big the input number is.

Input Size (n)Approx. Operations
101
1001
10001

Pattern observation: The work does not increase with input size; it stays constant.

Final Time Complexity

Time Complexity: O(1)

This means the check takes the same amount of time regardless of the number size.

Common Mistake

[X] Wrong: "The time to check grows with the size of the number because we need to divide or loop through bits."

[OK] Correct: This method uses a single bitwise operation that works instantly, so no loops or divisions are needed.

Interview Connect

Knowing this quick check shows you understand efficient bitwise operations, a useful skill in many coding problems.

Self-Check

"What if we changed the method to repeatedly divide the number by 2 until it is 1? How would the time complexity change?"

Practice

(1/5)
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
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