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Akhil

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# Power of 2, solving a google interview question. Playing with bits.

Question : Given an integer, write a function to determine if it is a power of two. (Range: 1 - 2^31-1)

Eg: Input : 16, Output : true since 2^4 = 16.
Input : 18, Output : false.

### Brute Force

So the obvious brute force approach will be to just divide 2 and check if it reaches 1.

``````var powerOftwo = function(n){
if(n<=0) return false;
while(n%2 == 0) n = n/2;
return n == 1;
}
``````

### Percompute and Use Set

Since we're give that the range is in between 0 - 2^31-1, we could use this to precompute various possible output, save them into a set and cross check:

``````  let set = new Set();

for(let i=0;i<31;i++){
}

console.log(set);

var powerOfTwo = function(n){
if(set.has(n)) return true;
return false;
}
``````

Now that we know what we want to accomplish, let's think about what google might expect from us.

Tip : One common pattern I noticed is that whenever a company asks a question which looks easy on understand and easy to implement, that question has to be somehow related to bit manipulation.

So how do we relate power of 2 and bits

As we know, for a computer everything boils down to a combination of 0's and 1's including numbers. So let's see how the numbers which represent 2^n for n=0 to 31 are represented in bit form. So the common pattern here is that the numbers that are power of 2's have only 1 bit set to 1 and rest of it is 0.

So how do we use this to our advantage ? Let's look at what operations we have to work with.

``````   &: and operator          1&1 = 1, 0 & anything = 0
|: or  operator          0&0 = 0, 1 & anything = 1
^: xor operator          x^x = 0, x^n = 1
~: negate                ~1 = 0 | ~0 = 1
``````

Tip: if the question is regarding bit manipulation, first try solving with & and ^ operators

Let's try solving this with & operator. But what should we & it with ?

Let's generate a number, if a number is odd, +1/-1 will generate an even number and vice versa.

Now that we have "&" operator and 2 numbers let put them together.

``````   even number : 6  = 110
+1 : 7  = 111
-1 : 5  = 101

6&7 = 6   6&5 = 4

multiple of 2 : 16 = 10000
+1 : 17 = 10001
-1 : 15 = 01111

16&17 = 16  16&15 = 0
``````

Here comes our solution, 16&15 becomes 0 since as proved earlier that a number if it's power of 2, has only 1 bit set, so n-1 will set all previous bit's from current position to 1.

eg:

``````   8 : 1000       7: 0111      8&7 = 0
16: 10000     15: 01111    16&15 = 0
32: 100000    31: 011111   32&31 = 0
``````

But at the same time if we use an odd number

eg:

``````   17: 10001     16: 10000     17&16 = 16
``````

From this we can say that n&(n-1) = 0, if n is power of 2.

Due to this our code boiled down to :

``````   var powerOfTwo = function(n){
n = n&(n-1);
return n == 0;
}
``````

Amazing isn't it ? How few bit manipulation cause such drastic results aand lead us to a more concise readable code.

If you want to practice something similar, solve this :

Question : Count the number of ones in the binary representation of the given number.

Comment your solution for this question :)

I hope you liked my explanation comment down below if I messed up somewhere. Pankaj Tanwar

Super quick solution would be just count the no of 1 in binary form. If its 1. Return true.

In C++ just do __builtin_popcount(n) == 1 Akhil

Yep, it one of the solutions, you're in right direction now combine the solution of the original question with this.

hint : if n&n-1 = 0, it means that the rest of the bit is guaranteed set to 0, so no need to parse them.