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AND Operator

Challenge 3: Power of 2

3 min read

We solve by making use of the & operator in computers. There are many ways to solve this, of which two approaches are simple, and one is a more complex but better solution.

In this lesson, we will try to check if the given number is a power of 2. We solve this by writing an efficient algorithm that takes an optimal amount of time.

Introduction

Let’s do another challenging question to check your understanding of Bitwise operators.

Example:

Input: 4
 
Output: True (As 4 is 2^2)
Input: 12
 
Output: False

Problem statement

Write a program to check whether a given number is a power of 2.

Let’s consider a number and find how the AND (&) operator does this.

Input = 64
 
Output: True

Explanation: We solve by making use of the & operator in computers. There are many ways to solve this, of which two approaches are simple, and one is a more complex but better solution.

Assume n is non-negative. Use the & operator to achieve this.

Brute-force/naive approach

✅
Hint: The exciting part of calculating the power of 2 is that they have a set-bit count equal to one.

Algorithm

  • Read the input value.
  • Repeatedly divide input with 2.
    • if n not equal to 1 and if it is odd, we will return false.
    • else true.

Here is what our algorithm will look like:

Solutions

Java

class IsPowerOf2 {
    private static boolean helper(int n) {
        if (n == 0) {
            return false;
        }

        while (n != 1) {
            if (n % 2 != 0) {
                return false;
            }
            n >>= 1;
        }
        return true;
    }

    public static void main(String[] args) {
        System.out.println(helper(6));
        System.out.println(helper(8));
    }
}

Python

def IsPowerOf2(n):
    if n == 0:
        return False
    while n != 1:
        if n %2 !=0:
            return False
        n >>= 1
    return True

print(IsPowerOf2(6))
print(IsPowerOf2(8))

JavaScript

const IsPowerOf2 = number => {

    function helper (n) {
        if(n === 0) {
            return false;
        }

        while (n !== 1) {
            if(n % 2 !== 0) {
                return false;
            }
            n >>= 1;
        }
        return true;
    }

    return helper (number);
}

console.log (IsPowerOf2 (6));
console.log (IsPowerOf2 (8));

C++

#include <iostream>

using namespace std;

bool helper(int n) {
    if (n == 0) {
        return false;
    }
    while (n != 1) {
        if (n % 2 != 0) {
            return false;
        }
        n >>= 1;
    }
    return true;
}

int main() {
    cout << helper(6) << endl;
    cout << helper(8) << endl;
    return 0;
}

TypeScript

export const IsPowerOf2 = (n: number): boolean => {
    function helper(value: number): boolean {
        if (value === 0) {
            return false;
        }

        while (value !== 1) {
            if (value % 2 !== 0) {
                return false;
            }
            value >>= 1;
        }
        return true;
    }

    return helper(n);
}

console.log(IsPowerOf2(6));
console.log(IsPowerOf2(8));

Complexity analysis

Time complexity

This takes log(n) complexity. We can do better in constant time using Brian Kernighan’s algorithm.

Space complexity

The space complexity is O(1). No additional space is allocated.

Coding exercise

Now, try to optimize the above solution/algorithm. There is a better way to solve this problem.

This problem is designed for your practice, so try to solve it yourself first. If you get stuck, you can always refer to the solution in the next lesson. Good luck!

Hint: Learn about "Brian Kernighan’s algorithm", because in next solution review, we are about to solve this with his approach.

First, look at these code snippets below and think of a solution.

// Java
// TODO: finish the challenge or check next lesson for solution
class Solution{
    public static boolean isPow2(int n){
        // Write - Your - Code- Here

        return false; // change this, return true/false based on inputs
    }
}
# Python
# TODO: finish the challenge or check next lesson for solution
def Solution(n):
	# Write - Your - Code- Here
    
	return False # change this, return True/False based on inputs

// JavaScript
// TODO: finish the challenge or check next lesson for solution
const isPow2 = n => {
    // Write - Your - Code- Here
    
    return false; // change this, return true/false based on inputs
}
// C++
// TODO: finish the challenge or check next lesson for solution
#include <iostream>
#include <string>
using namespace std;

bool isPowerOfTwo(int n) {
   // Write - Your - Code- Here

    return false; // change this, return true/false based on inputs
}
// TypeScript
// TODO: finish the challenge or check next lesson for solution
export const isPow2 = (n: number): boolean => {
    // Write - Your - Code- Here
    
    return false; // change this, return true/false based on inputs
}

The solution will be explained in the next lesson.

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