Check if Number is Power of Two in DSA C++

Mental Model

A number is a power of two if it has exactly one '1' bit in its binary form.

Analogy: Imagine a row of light switches where only one switch is turned on at a time; that means the number is a power of two.

Number: 8
Binary: 1 0 0 0
          ↑
Only one '1' bit means power of two.

Dry Run Walkthrough

Input: number = 8

Goal: Determine if 8 is a power of two

Step 1: Check if number is zero or negative

number = 8

Why: Zero or negative numbers cannot be powers of two

Step 2: Calculate number & (number - 1): 8 & 7

8 (1000) & 7 (0111) = 0 (0000)

Why: If result is zero, number has only one '1' bit

Step 3: Since result is zero, conclude number is power of two

Result: true

Why: Only one bit set means power of two

Result:

8 is a power of two

Annotated Code

DSA C++

#include <iostream>
using namespace std;

bool isPowerOfTwo(int n) {
    if (n <= 0) return false; // zero or negative check
    return (n & (n - 1)) == 0; // check single '1' bit
}

int main() {
    int number = 8;
    if (isPowerOfTwo(number)) {
        cout << number << " is a power of two" << endl;
    } else {
        cout << number << " is NOT a power of two" << endl;
    }
    return 0;
}

if (n <= 0) return false; // zero or negative check

exclude zero and negative numbers as they can't be powers of two

return (n & (n - 1)) == 0; // check single '1' bit

bitwise trick to confirm only one bit is set

Output

8 is a power of two

Complexity Analysis

Time: O(1) because bitwise operations take constant time

Space: O(1) because no extra memory is used

vs Alternative: Compared to looping and dividing by 2 repeatedly (O(log n)), this bitwise method is faster and simpler

Edge Cases

number = 0

Returns false because zero is not a power of two

DSA C++

if (n <= 0) return false; // zero or negative check

number = 1

Returns true because 1 is 2^0, a power of two

DSA C++

return (n & (n - 1)) == 0; // check single '1' bit

number = negative (e.g., -4)

Returns false because negative numbers are not powers of two

DSA C++

if (n <= 0) return false; // zero or negative check

When to Use This Pattern

When you need to quickly check if a number is a power of two, use the bitwise AND trick n & (n-1) == 0 because it efficiently tests the single '1' bit condition.

Common Mistakes

Mistake: Checking if number % 2 == 0 repeatedly without stopping at 1
Fix: Use the bitwise method or ensure the loop stops when number becomes 1

Mistake: Not handling zero or negative inputs
Fix: Add a check to return false if number is less than or equal to zero

Summary

Checks if a number has exactly one '1' bit in binary to confirm power of two.

Use when you need a fast, simple way to verify powers of two.

The key insight is that powers of two have only one bit set, so n & (n-1) equals zero.

Practice

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

easy

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

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

C. It is divisible by 3.

D. It is always an odd number.

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

   Odd numbers can have many bits set; divisibility by 3 is unrelated; 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 B

4. Quick Check:

   Power of two = single 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. (n & (n + 1)) == 0

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

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

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

Solution

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

   The expression (n & (n - 1)) == 0 is true only if n has exactly one bit set, and n > 0 ensures positive numbers.

2. Step 2: Analyze other options

   (n | (n - 1)) == 0 uses OR which doesn't check power of two; (n & (n + 1)) == 0 uses n + 1 which is incorrect; (n ^ (n - 1)) == 0 uses XOR which is unrelated.

3. Final Answer:

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

4. Quick Check:

   Bitwise AND with n-1 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;
if ((n & (n - 1)) == 0 && n > 0) {
    std::cout << "Power of Two";
} else {
    std::cout << "Not Power of Two";
}

medium

A. Power of Two

B. Not Power of Two

C. Compilation Error

D. Runtime Error

Solution

1. Step 1: Evaluate the expression for n = 16

   16 in binary is 10000; 16 - 1 = 15 (01111). Bitwise AND: 10000 & 01111 = 0, condition true.

2. Step 2: Check if n > 0

   16 > 0 is true, so whole condition is true and "Power of Two" is printed.

3. Final Answer:

   Power of Two -> Option A

4. Quick Check:

   16 & 15 == 0 and 16 > 0 [OK]

Hint: Check bitwise AND with n-1 equals zero for power of two [OK]

Common Mistakes:

• Miscomputing bitwise AND
• Ignoring the n > 0 check
• Confusing output strings

4. The following code is intended to check if n is a power of two, but it has a bug. What is the bug?

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

medium

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

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

C. The function should return an integer, not bool.

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

Solution

1. Step 1: Analyze the expression without positivity check

   The expression returns true for n=0 because 0 & -1 = 0, which is incorrect since 0 is not a power of two.

2. Step 2: Identify missing condition

   Adding n > 0 fixes the bug by excluding zero and negative numbers.

3. Final Answer:

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

4. Quick Check:

   Check n > 0 to avoid zero/negative false positives [OK]

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

Common Mistakes:

• Ignoring zero and negative inputs
• Using OR instead of AND
• Wrong return type assumptions

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

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. Use a loop to shift 1 left until the value is >= n.

Solution

1. Step 1: Understand the problem

   We need the smallest power of two >= n, so we start from 1 and keep doubling (shifting left) until we reach or pass n.

2. Step 2: Analyze options

   Use a loop to shift 1 left until the value is >= n. uses a loop with left shift which correctly finds the smallest power of two >= n. Options B, C, D use bitwise operations that do not produce this result.

3. Final Answer:

   Use a loop to shift 1 left until the value is >= n. -> Option D

4. Quick Check:

   Shift 1 left until >= n to find smallest power of two [OK]

Hint: Shift 1 left repeatedly until >= n to find next power of two [OK]

Common Mistakes:

• Using bitwise AND/OR/XOR incorrectly for this problem
• Not handling cases where n is already a power of two
• Assuming bitwise ops alone give next power of two