**Problem Link** :–

**Author** : Ranjan Kumar

**Editorialist** : Ranjan Kumar

**Difficulty** :–

Easy-Medium.

**Pre-Requisites**:

Maths, Bit Manipulation.

**Problem Statement** :–

The problem is to find all the even coefficient of a binomial expression (1+x)^N for a given value of N, the power raised to the expression (1+x).

**Quick Solution**:–

As we know that for the expression (1+x)^N has N+1 terms which are:—

C(N,0) + C(N,1)x + C(n,2)x^2 +………………+ C(N,N)x^N.

So in very short explanation I say that for finding the number even coefficient we can find the odd coefficient and subtract that from the N+1 so that we can get the number of even coefficients.

Now we come to a problem of finding the number of odd coefficient. For finding the number of odd coefficient we know that will count number of 1s in the Binary value of N and the final answer will be 2 to the power number of 1s.

**Detailed Approach**:–

We are provided with a value N that is the power of the term (1+x).

This will lead to the **N+1** terms which are

C(N,0) + C(N,1)x + C(n,2)x^2 +………………+ C(N,N)x^N.

In which the coefficients are C(N,0), C(N,1), C(n,2), ………………,C(N,N)

Suppose the number of odd coefficients is **o**.

Then number of even coefficients is **e=N+1-o**.

Now for finding the number of odd coefficients we find the binary form of N.

Let us take a counter c and initialize it by 0.

Now we perform subsequent division by 2 and increase the counter when the remainder is 1 and perform this operation until the value of **N** is not 0.

Then **o=power(2,c)**;

Now **e=N+1-o**.

Hence the required answer is **e.**

**Time Complexity** :–

**O(logN)**

**Solution** :–

Setter’s Solution can be found here.

**Feel free to post comments if anything is not clear to you.**