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

Time Complexity: Check if Number is Power of Two

O(1)

Understanding Time Complexity

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

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public bool IsPowerOfTwo(int 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

Since the code does only one check, the time does not grow with input size.

Input Size (n)	Approx. Operations
10	1
100	1
1000	1

Pattern observation: The number of operations stays the same no matter how big the number is.

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: "Checking if a number is a power of two requires looping or dividing repeatedly."

[OK] Correct: This bitwise method does the check in one step without loops, making it very fast.

Interview Connect

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

Self-Check

"What if we used a loop to divide the number by two repeatedly until it is 1? How would the time complexity change?"

Practice

1. Which of the following statements correctly describes a number that is a power of two?

easy

A. It is divisible by 3.

B. It is always an odd number.

C. It has all bits set to 1 in its binary form.

D. It has exactly one bit set to 1 in its binary form.

Solution

1. Step 1: Understand binary representation of powers of two

   A power of two in binary has exactly one '1' bit and the rest are '0's, e.g., 4 is 100.

2. Step 2: Check other options for correctness

   Odd numbers and divisibility by 3 do not define powers of two. Having all bits set to 1 means numbers like 7 (111), not powers of two.

3. Final Answer:

   It has exactly one bit set to 1 in its binary form. -> Option D

4. Quick Check:

   Power of two = one '1' bit [OK]

Hint: Power of two has only one '1' bit in binary [OK]

Common Mistakes:

• Thinking powers of two are always odd
• Confusing all bits set to 1 with power of two
• Assuming divisibility by 3 means power of two

2. Which of the following C# expressions correctly checks if an integer n is a power of two?

easy

A. return (n & (n + 1)) == 0;

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

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

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

Solution

1. Step 1: Understand bitwise check for power of two

   For a power of two, n & (n - 1) clears the only set bit, resulting in 0. Also, n must be positive.

2. Step 2: Analyze other expressions

   Other options use incorrect bitwise operations that do not correctly identify powers of two.

3. Final Answer:

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

4. Quick Check:

   Bitwise AND with n-1 equals zero means power of two [OK]

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

Common Mistakes:

• Using OR or XOR instead of AND
• Not checking if n is positive
• Using n + 1 instead of n - 1

3. What is the output of the following C# code?

int n = 16;
bool result = (n & (n - 1)) == 0 && n > 0;
Console.WriteLine(result);

medium

A. True

B. False

C. Compile-time error

D. Runtime exception

Solution

1. Step 1: Calculate bitwise AND of n and n-1

   For n=16 (binary 10000), n-1=15 (binary 01111). Bitwise AND is 0.

2. Step 2: Check if result is zero and n is positive

   Since (n & (n - 1)) == 0 and n > 0, the expression is true.

3. Final Answer:

   True -> Option A

4. Quick Check:

   16 is power of two, so output is True [OK]

Hint: If n & (n-1) == 0 and n>0, output is True [OK]

Common Mistakes:

• Miscomputing bitwise AND
• Ignoring n > 0 condition
• Confusing True and False outputs

4. The following code is intended to check if n is a power of two. What is the error?

bool IsPowerOfTwo(int n) {
    return (n & (n - 1)) == 0;
}

medium

A. The expression should be (n | (n - 1)) == 0 instead.

B. It does not check if n is positive, so zero or negative numbers return true incorrectly.

C. The function should return an integer, not a boolean.

D. The bitwise AND operator is used incorrectly; it should be OR.

Solution

1. Step 1: Analyze the bitwise condition

   The expression (n & (n - 1)) == 0 is true for powers of two but also for zero.

2. Step 2: Identify missing condition

   Zero and negative numbers are not powers of two, so the function must check n > 0.

3. Final Answer:

   It does not check if n is positive, so zero or negative numbers return true incorrectly. -> Option B

4. Quick Check:

   Check n > 0 to avoid false positives [OK]

Hint: Always check n > 0 when using bitwise power of two check [OK]

Common Mistakes:

• Omitting n > 0 check
• Using wrong bitwise operator
• Returning wrong data type

5. You want to write a function that returns the smallest power of two greater than or equal to a given positive integer n. Which approach correctly achieves this in C#?

hard

A. Use a loop doubling a variable starting at 1 until it is >= n.

B. Return n & (n - 1) to get the smallest power of two.

C. Return n | (n - 1) to get the smallest power of two.

D. Return n ^ (n - 1) to get the smallest power of two.

Solution

1. Step 1: Understand the problem requirement

   We need the smallest power of two >= n, so start from 1 and double until reaching or exceeding n.

2. Step 2: Analyze other options

   Bitwise operations like AND, OR, XOR with n-1 do not produce the smallest power of two >= n.

3. Final Answer:

   Use a loop doubling a variable starting at 1 until it is >= n. -> Option A

4. Quick Check:

   Doubling loop finds smallest power of two >= n [OK]

Hint: Double from 1 until >= n to find smallest power of two [OK]

Common Mistakes:

• Using bitwise AND/OR/XOR incorrectly for this task
• Not handling n less than 1
• Assuming bitwise ops give next power of two