The Fibonacci numbers are the numbers in the following integer sequence.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……..

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation

F_{n}= F_{n-1}+ F_{n-2}

with seed values

F_{}= 0 and F_{1}= 1.

Write a function *int fib(int n)* that returns F_{n}. For example, if *n* = 0, then *fib()* should return 0. If n = 1, then it should return 1. For n > 1, it should return F_{n-1} + F_{n-2}

For n = 9 Output:34

Following are different methods to get the nth Fibonacci number.

**Method 1 ( Space Optimized )**

We can optimize the space used in method 2 by storing the previous two numbers only because that is all we need to get the next Fibonacci number in series.

*Time Complexity:* O(n)

*Extra Space: *O(1)

**Method 2 (O(Log n) Time)**

Below is one more interesting recurrence formula that can be used to find n’th Fibonacci Number in O(Log n) time.

If n is even then k = n/2: F(n) = [2*F(k-1) + F(k)]*F(k) If n is odd then k = (n + 1)/2 F(n) = F(k)*F(k) + F(k-1)*F(k-1)

**How does this formula work?**

The formula can be derived from above matrix equation.

Taking determinant on both sides, we get

(-1)^{n} = F_{n+1}F_{n-1} – F_{n}^{2}

Moreover, since A^{n}A^{m} = A^{n+m} for any square matrix A, the following identities can be derived (they are obtained form two different coefficients of the matrix product)

F_{m}F_{n} + F_{m-1}F_{n-1} = F_{m+n-1}

By putting n = n+1,

F_{m}F_{n+1} + F_{m-1}F_{n} = F_{m+n}

Putting m = n

F_{2n-1} = F_{n}^{2} + F_{n-1}^{2}

F_{2n} = (F_{n-1} + F_{n+1})F_{n} = (2F_{n-1} + F_{n})F_{n} (Source: Wiki)

To get the formula to be proved, we simply need to do following

If n is even, we can put k = n/2

If n is odd, we can put k = (n+1)/2

Below is C++ implementation of above idea.

Output :

34

Time complexity of this solution is O(Log n) as we divide the problem to half in every recursive call.