What is the brute force approach and complexity for Power of Two?

Divide by 2 while even; the answer is yes only if you end at 1 with time complexity O(log N) and space complexity O(1).

What is the optimal approach and complexity for Power of Two?

A power of two has one 1 bit, so n & (n − 1) is 0 exactly for powers of two (n > 0) with optimal time complexity O(1) and space complexity O(1).

How do you identify the pattern for Power of Two?

Look for key signals in the prompt: "is a power of two" indicates Exactly one 1 bit in binary, because 2^x has a single bit set.; "return true or false" indicates A yes/no answer, so one bit test is enough and no loop is needed.; "range starts below zero" indicates Zero and negative numbers must be rejected before the bit trick..

Power of Two

Easy·LeetCode #231 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

Given an integer n, decide whether it is a power of two. That means n = 2^x for some whole number x ≥ 0. Return true if it is, and false if it is not.

Follow-up: can you solve it without loops or recursion?

Input

An integer n. It can be zero or negative.

Output

true if n is a power of two, otherwise false.

Example 1

Input: n = 1

Output: true

1 = 2^0.

Example 2

Input: n = 16

Output: true

16 = 2^4.

Example 3

Input: n = 3

Output: false

No whole power of 2 is 3.

Constraints

• −2^31 ≤ n ≤ 2^31 − 1

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“is a power of two”

→ Exactly one 1 bit in binary, because 2^x has a single bit set.

“return true or false”

→ A yes/no answer, so one bit test is enough and no loop is needed.

“range starts below zero”

→ Zero and negative numbers must be rejected before the bit trick.

Approach Complexity Ladder

Brute Force — halve until you reach 1O(log N)O(1)

Divide by 2 while even; the answer is yes only if you end at 1

Optimal — a single 1 bit testO(1)O(1)

A power of two has one 1 bit, so n & (n − 1) is 0 exactly for powers of two (n > 0)

Brute Force Intuition

A power of two can be halved again and again, and it always ends at 1. Any other number gets stuck on an odd number that is not 1. So halve while the number is even, then check whether you reached 1.

Recognition rule for this sub-pattern:

n & (n−1) removes the lowest set bit; shifts read bits one at a time.

Remember this · Bit Manipulation

Trigger

"Every element appears twice except…", power of two, counting set bits, subsets as bitmasks, O(1) space tricks.

Think

What happens to this problem if I look at each bit separately?

Key question

Which bit operation (XOR, AND, OR, shift) cancels or isolates what I need?

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