## Algorithm/Insights

Recursive Solution:

1. Boundary Condition: If n = 0 or n = 1, return n

2. Return F(n-1) + F(n-2)

F(n-1) = F(n-2) + F(n-3)

We see that F(n-2) will be computed again for F(n-1). So, a lot of redundant calculations.

Time Complexity: Exponential

Efficient Solution:

The problem can be solved by Dynamic Programming because it has both the properties:

1. Overlapping Subproblems -

As F(n) = F(n-1) + F(n-2), computation of nth fibonacci number requires (n-1) and (n-2) fibonacci numbers.

Further F(n-1) = F(n-2) + F(n-3), and so on

So, we see that there are overlapping sub problems.

2. Optimal Substructure -

If subproblems F(n-1) and F(n-2) are computed optimally, then the solution to F(n) is also optimal. Hence it satisfies optimal substructure property.

So, if we store the solutions to subproblems here, we can find next fibonacci number efficiently without computing the values again.

Also, as F(n) = F(n-1) + F(n-2), we need only the last 2 values of the series to find the next value. Hence, we do not need to store all subproblem solutions but only the previous 2 i.e. F(i-1) and F(i-2). Once we have F(i), for calculating F(i+1) we just need F(i) and F(i-1), and therefore we can discard F(i-2) as it is not needed further.

So starting with F(0) = 0, F(1) = 1, we can use bottom up approach to calculate F(n)

1. Boundary Condition: If n = 0 or n = 1, return n

2. Take a = 0, b = 1

3. Loop n-2 times, and calculate

c = a+b

a = b

b = c

4. Return c