Power of Two, A Google Interview Question - Java 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 01:

Input: 4

Output: True (As 4 is 2^2)

Example 02:

Input: 12

Output: False

Problem statement

Write a program to check if a given number is a power of 2 or not.

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

Input = 64

Output: True

Explanation

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

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

Solution: 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

1. Read the input value.

2. 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:

Code

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));
  }
}

Complexity Analysis

Time complexity: O(logn)

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

Space complexity: O(1)

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

Coding Exercise

First, take a close look at the code snippets above and think of a solution. This problem is designed for your practice, so try to solve it on your own first. If you get stuck, you can always refer to the solution provided in the solution section. Good luck!

class Solution{
    public static boolean isPow2(int n){
        // Write - Your - Code- Here

        return false; // change this, return true/false based on inputs
    }
}

If you've got the answer, great! If not, it's normal, practice similar problems and you'll get a good hold of bit manipulation tricks.

I will explain the solution below.

Let's see how we make use of Brian Kernighan's algorithm to achieve this.

Solution review: Brian Kernighan’s algorithm

This is considered faster than the previous naive approach.

In this approach, we count the set bits. If a number is the power of 2, we know only one set bit is present in its binary representation.

In binary, we go from right to left with powers of 2.

For example:

2^0, 2^1, 2^2, 2^3, 2^4, and so on.

Algorithm

Before we talk about algorithmic steps, you should review the tabular form of steps that depicts the algorithm.

1. If (n & (n - 1) == 0), return True.

2. else, False.

Let’s visualize the values in the table below:

Let’s see a couple of examples:

        n   = 4    => 00000000 00000000 00000000 00000100
      n - 1 = 3    => 00000000 00000000 00000000 00000011
-----------------------------------------------------------
(n & (n - 1)) = 0   => 00000000 00000000 00000000 00000000   
-----------------------------------------------------------

(n&(n - 1)), here this becomes 0, which is true. Hence, the number 4 is a power of 2.

        n   = 6    => 00000000 00000000 00000000 00000110
      n - 1 = 5    => 00000000 00000000 00000000 00000101
-----------------------------------------------------------
(n & (n - 1)) = 4   => 00000000 00000000 00000000 00000100   
-----------------------------------------------------------

(n&(n - 1)) is 4, which is not equal to 0. Hence, the number 6 is not a power of 2.

Let us take a look at the optimized approach.

Code

Here is the reasoning behind this solution.

class IsPowerOf2 {
    public static void main(String[] args) {
        System.out.println(helper(6));
        System.out.println(helper(8));
    }

    private static boolean helper(int n) {
        if (n == 0) {
            return false;
        }
        return (n & (n - 1)) == 0;
    }
}

We can further simplify this code into a single line shown below.

class IsPowerOf2 {
    public static void main(String[] args) {
        System.out.println(helper(6));
        System.out.println(helper(8));
    }

    private static boolean helper(int n) {
        return n != 0 && (n & (n - 1)) == 0;
    }
}

Complexity analysis

Time complexity: O(1)

The run time depends on the number of 1-bits in n. In the worst case, all bits in n are 1-bits. With a 32-bit integer, the runtime is O(1).

Space complexity: O(1)

The space complexity is O(1). It allocated no additional space.

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Extras

If you are interested in mastering bit tricks, I've got a course that more than 100k+ programmers loved.

In this course, you will learn how to solve problems using bit manipulation, a powerful technique that can optimize your algorithmic and problem-solving skills. The course has a simple explanation with sketches, detailed step-by-step drawings, and various ways to solve it using bitwise operators.

These bit-tricks could help in competitive programming and coding interviews in running algorithms, mostly in O(1) time.

This is one of the most important/critical topics when someone prepares for coding interviews for FAANG(Facebook, Amazon, Apple, Netflix, and Google) companies.

To kick things off, you’ll start by learning about the number system and how it’s represented. Then you’ll move on to learn about the six different bitwise operators: AND, OR, NOT, XOR, and bit shifting. Throughout, you will get tons of hands-on experience working through practice problems to help sharpen your understanding.

By the time you’ve completed this course, you will solve problems faster with greater efficiency!! 🤩

Link to my course: Master Bit Manipulation for Coding Interviews.