DSA interview question

Check Whether a Number Is a Power of Two

Easy · Algorithms · Topics: bit manipulation

The problem

A memory allocator only accepts block sizes that are powers of two (1, 2, 4, 8, …), because it splits and merges blocks by halving and doubling. Every allocation request must be validated instantly. A power of two has exactly one set bit, and subtracting 1 flips that bit off and turns every lower bit on, so n & (n − 1) is zero exactly for powers of two. Zero and negative numbers must be rejected separately.

Example: 16 → true; 3 → false; 0 → false.

Function signature

bool isPowerOfTwo(int n);

Test cases

Input	Output	Explanation
`1`	`true`	2⁰
`16`	`true`	10000
`3`	`false`	11 has two set bits
`0`	`false`	No set bits
`-16`	`false`	Negative numbers are never powers of two
`1073741824`	`true`	2³⁰, the largest in a 32-bit int

(Confirm: n is a 32-bit signed integer; only positive powers of two, 2⁰ to 2³⁰, return true.)

How interviewers phrase it

"Given an integer n, return true if it is a power of two" (LeetCode 231), "without loops or recursion" as a follow-up.

Clarifying questions to ask

Is 1 (2⁰) a power of two? (Yes.)
Can n be zero or negative? (Yes, and they return false.)
Are loops allowed?

Follow-ups to expect

"Why does n & (n − 1) work?"
"Why check n > 0?"
"Another bit trick?"
"How would you test for a power of four?"
"Complexity?"

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