How to check if a number is a power of two for O(1)

Dart solution.

In JS looks something like this.

const isIntegerPowerOfTwo = number =>
    number > 0 && (number & (number - 1)) === 0;

Thanks to Luke Shiru for code.

Explanation

In binary representation, numbers consist of 0 and 1. And it just so happens that all powers of two are 10*.
For example, 2 -> 10, 8 -> 1000, and so on.
When we do a comparison of number & number - 1 == 0, we check 100 & 011, which will always yield 0.

3 → 11. 3 & 2 => 11 & 10 = 1.

Comments

[Andrei Iatsuk]

Sure, will update, thanks!

[Jon Randy 🎖️]

Surely any number is a power of 2?

$$2^{\log_2 x}=x$$

Your code is checking for integer powers of 2

[Andrei Iatsuk]

Hello, dont correctly understand about log. It can be simplified to x = x, and?

This method only about integer power numbers, check the input number type.

If you could found way to work with float , let us know)

[Jon Randy 🎖️]

The input number type does not tell you that this is checking for integer powers. To make the intent of your function clearer, it would be better named something like isIntegerPowOfTwo

[Andrei Iatsuk]

Thanks, let it be