What is the best way to determine if a given number is a power of two?

Question

I need to return true if the n is a power of two and false otherwise. It should go something like:

function PowerOfTwo(n){
  //code here
}

This is the way I currently do it:

function isPowerOfTwo(n){
  var x = Math.pow(2, Math.round(Math.log(n) / Math.log(2)));
  return x;
}

Are there any more effective ways?

Answer 1

You can actually use ECMAScript5 Math.log :

function powerOfTwo(x) {
    return (Math.log(x)/Math.log(2)) % 1 === 0;
}

Remember, in math, to get a logarithm with an arbitrary base, you can just divide log ₁₀ of the operand ( x in this case) by log ₁₀ of the base. And then to see if the number is a regular integer (and not a floating point), just check if the remainder is 0 by using the modulus % operator.

In ECMAScript6 you can do something like this:

function powerOfTwo(x) {
    return Math.log2(x) % 1 === 0;
}

See the MDN docs for Math.log2 .

Answer 2

Source: Bit twiddling Hacks ,

function powerOf2(v) {
    return v && !(v & (v - 1));
}

You just bitwise AND the previous number with the current number. If the result is falsy, then it is a power of 2.

The explanation is in this answer .

Note:

• This will not be 100% true for programming, mathematical, [also read 'interviewing']. Some edge cases not handled by this are decimals (0.1, 0.2, 0.8…) or zero values (0, 0.0, …)

Answer 3

Using bitwise operators, this is by far the best way in terms of efficiency and cleanliness of your code:

function PowerofTwo(n){
    return ((x != 0) && !(x & (x - 1)));
}

what it does is checks the bits that make up the number, ie 8 looks like this:

1 0 0 0

x-1 or 7 in this case looks like this

0 1 1 1

when the bitwise operator & is used it invokes an && on each bit of the number (thus 1 & 1 = 1 , 1 & 0 = 0 , 0 & 1 = 0 , 0 & 0 = 1 ):

 1 0 0 0
-0 1 1 1
=========
 0 0 0 0

since the number turns into an exact 0 (or false when evaluted as a boolean) using the ! flag will return the correct answer

if you were to do this with a number like 7 it would look like this:

 0 1 1 1
-0 1 1 0
=========
 1 1 1 0

returning a number greater than zero causing the ! flag to take over and give the correct answer.

Answer 4

A number is a power of 2 if and only if log base 2 of that number is whole. The function below computes whether or not that is true:

function powerOfTwo(n){
    // Compute log base 2 of n using a quotient of natural logs
    var log_n = Math.log(n)/Math.log(2);
    // Round off any decimal component
    var log_n_floor = Math.floor(log_n);
    // The function returns true if and only if log_n is a whole number
    return log_n - log_n_floor == 0; 
}

Answer 5

Making use of ES6's Math.clz32(n) to count leading zeros of a 32-bit integer from 1 to 2³² - 1:

function isPowerOf2(n) {
  return Math.clz32(n) < Math.clz32(n - 1);
}

Answer 6

 /** * @param {number} n * @return {boolean} */ const isPowerOfTwo = function(n) { if(n == 0) return false; while(n % 2 == 0){ n = n/2 } return n === 1 };

Answer 7

function PowerOfTwo(n){
  // Exercise for reader: confirm that n is an integer
  return (n !== 0) && (n & (n - 1)) === 0;
}

Answer 8

without using libraries

so you need to create an inner loop using while

it should start with 1 keep multiplying it with 2 until i,j,k or whatever is greater than the current number(array) so it will have to do 2 4 6 8 10 12 14 16 18 until it is greater than the number then it will go to the outer loop then repeat again