L1 · FoundationsNumbers & loops~3 min · 6 tests

Check If a Number Is a Power of Two

Test whether n is a power of two in O(1) with the bit trick n & (n - 1) == 0, and compare it with the O(log n) division loop. Python bit manipulation.

The problem

Return True if n is a power of two (1, 2, 4, 8, …). Zero and negative numbers are not.

Examples

Example 1
Input
```
is_power_of_two(16)
```
Expected output
```
True
```
Example 2
Input
```
is_power_of_two(18)
```
Expected output
```
False
```

+ 4 hidden tests on Submit.

Edge cases to ask about

Zero
Negative numbers
1 (2⁰)

Hints

0/3

How an interviewer scores this

0/9

1. Understand the problemRestate it, ask about edge cases and input size.
2. Write a brute-force solutionCorrect first, clever later.
3. Get the correct outputautoAll visible examples pass.
4. Analyse time complexityautoCommit to a Big-O in the Complexity Lab.
5. Analyse space complexityautoExtra memory, not the input.
6. Identify the bottleneckautoMeasure your code — where does it bend?
7. OptimiseUsually: a set or dict instead of a scan.
8. Explain why it worksSay it out loud to Solvi.
9. Test the edge casesautoSubmit passes every hidden test.

Python 3.13 · is_power_of_two

def is_power_of_two(n):
    """True if n is 1, 2, 4, 8, ..."""
    # Write your solution here
    pass

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Complexity Lab

What does this cost as n grows?

Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and read why.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

From brute force to optimal

The progression an interviewer wants to hear, one step at a time.

Approach	Time	Space	Idea
Divide by 2 while even	O(log n)	O(1)
bestn & (n − 1) == 0	O(1)	O(1)	A power of two has exactly one 1-bit; subtracting 1 flips it and everything below.

Walkthrough of the optimal approach (try it yourself first)

A power of two in binary is a single 1 followed by zeros (1000). Subtracting 1 turns it into all ones below that bit (0111), so n & (n - 1) is 0 only for powers of two. Guard n > 0 because 0 & -1 is also 0.

Complexity: O(1) time, O(1) space. A single bitwise AND and comparison.

Reveal the reference solution

def is_power_of_two(n):
    return n > 0 and n & (n - 1) == 0

The brute force, for comparison

def is_power_of_two(n):
    if n <= 0:
        return False
    while n % 2 == 0:
        n //= 2
    return n == 1

Follow-ups interviewers ask

Count the 1-bits in n.
Power of four?

Frequently asked interview questions

Core interview concepts, complexities, and follow-ups scored by hiring teams.

What is the time complexity of Check If a Number Is a Power of Two in Python?

The optimal solution runs in O(1) time and O(1) auxiliary space. A single bitwise AND and comparison.

What is the brute-force approach, and how do you optimise it?

Divide by 2 while even: O(log n) time, O(1) space. n & (n − 1) == 0: O(1) time, O(1) space. A power of two has exactly one 1-bit; subtracting 1 flips it and everything below.

What follow-up questions do interviewers ask about Check If a Number Is a Power of Two?

Count the 1-bits in n. Power of four?