[Solved] Identify the bug in the following code that attempts to check if a number is a power of two: def isPowerOfTwo(n: int) -> bool: while n % 2 == 0: n //= 2 return n == 1 What is the issue? | Bit Manipulation

Bit Manipulation - Power of Two

Identify the bug in the following code that attempts to check if a number is a power of two:

def isPowerOfTwo(n: int) -> bool:
    while n % 2 == 0:
        n //= 2
    return n == 1

What is the issue?

AThe return statement should check n == 0 instead of n == 1

BThe loop condition should be n > 1 instead of n % 2 == 0

CUsing integer division //= 2 causes infinite loops for odd numbers

DThe code does not handle zero or negative inputs, leading to incorrect results

Step-by-Step Solution

Solution:

Step 1: Analyze input handling
The code does not check if n is positive before the loop, so zero or negative inputs cause incorrect behavior.
Step 2: Consequence of missing check
For n=0 or negative, the loop condition n % 2 == 0 may cause infinite loop or wrong return value.
Final Answer:
Option D -> Option D
Quick Check:
Proper input validation is required to avoid incorrect results [OK]

Quick Trick: Always check n > 0 before division loops [OK]

Common Mistakes:

MISTAKES

Trap Explanation:

PITFALL

Interviewer Note:

CONTEXT

Master "Power of Two" in Bit Manipulation

3 interactive learning modes - each teaches the same concept differently

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