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

    Fn = Fn-1 + Fn-2

with seed values

F0 = 0 and F1 = 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 ( Use recursion )
A simple method that is a direct recursive implementation mathematical recurrence relation given above.

# Function for nth Fibonacci number
 
def Fibonacci(n):
    if n<0:
        print("Incorrect input")
    # First Fibonacci number is 0
    elif n==1:
        return 0
    # Second Fibonacci number is 1
    elif n==2:
        return 1
    else:
        return Fibonacci(n-1)+Fibonacci(n-2)
 
# Driver Program
 
print(Fibonacci(9))

Output

34

Time Complexity: T(n) = T(n-1) + T(n-2) which is exponential.
We can observe that this implementation does a lot of repeated work (see the following recursion tree). So this is a bad implementation for nth Fibonacci number.

                         fib(5)   
                     /             \     
               fib(4)                fib(3)   
             /      \                /     \
         fib(3)      fib(2)         fib(2)    fib(1)
        /     \        /    \       /    \  
  fib(2)   fib(1)  fib(1) fib(0) fib(1) fib(0)
  /    \
fib(1) fib(0)

Extra Space: O(n) if we consider the function call stack size, otherwise O(1).

Method 2 ( Use Dynamic Programming )
We can avoid the repeated work done is the method 1 by storing the Fibonacci numbers calculated so far.

# Function for nth fibonacci number - Dynamic Programing
# Taking 1st two fibonacci nubers as 0 and 1
 
FibArray = [0,1]
 
def fibonacci(n):
    if n<0:
        print("Incorrect input")
    elif n<=len(FibArray):
        return FibArray[n-1]
    else:
        temp_fib = fibonacci(n-1)+fibonacci(n-2)
        FibArray.append(temp_fib)
        return temp_fib
 
# Driver Program
 
print(fibonacci(9))

Output:

34

Time Complexity: O(n)
Extra Space: O(n)

Method 3 ( Space Optimized Method 2 )
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.

# Function for nth fibonacci number - Space Optimisataion
# Taking 1st two fibonacci numbers as 0 and 1
 
def fibonacci(n):
    a = 0
    b = 1
    if n < 0:
        print("Incorrect input")
    elif n == 0:
        return a
    elif n == 1:
        return b
    else:
        for i in range(2,n):
            c = a + b
            a = b
            b = c
        return b
 
# Driver Program
 
print(fibonacci(9))

Output:

Time Complexity: O(n)
Extra Space: O(1)