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# Python Program for n-th Fibonacci number

• Difficulty Level : Easy
• Last Updated : 06 Mar, 2023

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.

Method 1 ( Use recursion ) :

## Python3

 # 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(10))   # This code is contributed by Saket Modi

Output

34

Time Complexity: O(2N)
Auxiliary Space: O(N)

Method 2 ( Use Dynamic Programming ) :

## Python3

 # Function for nth fibonacci number - Dynamic Programming # Taking 1st two fibonacci numbers 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))   # This code is contributed by Saket Modi

Output

21

Time Complexity: O(N)
Auxiliary Space: O(N)

Method 3 ( Use Dynamic Programming with Space Optimization) :

## Python3

 # Function for nth fibonacci number - Space Optimisation # 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))   # This code is contributed by Saket Modi

Output

21

Time Complexity: O(N)
Auxiliary Space: O(1)

Method 4 ( Using Arrays ) :

## Python3

 # creating an array in the function to find the #nth number in fibonacci series. [0, 1, 1, ...] def fibonacci(n):     if n <= 0:         return "Incorrect Output"     data = [0, 1]     if n > 2:         for i in range(2, n):             data.append(data[i-1] + data[i-2])     return data[n-1]   # Driver Program print(fibonacci(9))   # This Code is contributed by Prasun Parate (prasun_parate)

Output

21

Time Complexity: O(N)
Auxiliary Space: O(N)

Explanation:

[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144]
As we know that the Fibonacci series is the sum of the previous two terms, so if we enter 12 as the input in the program, so we should get 144 as the output. And that is what is the result.

Method 5 ( Using Direct Formula ) :

The formula for finding the n-th Fibonacci number is as follows:

## Python3

 # To find the n-th Fibonacci Number using formula from math import sqrt  # import square-root method from math library def nthFib(n):     res = (((1+sqrt(5))**n)-((1-sqrt(5)))**n)/(2**n*sqrt(5))     # compute the n-th fibonacci number     print(int(res),'is',str(n)+'th fibonacci number')     # format and print the number       # driver code nthFib(12)   # This code is contributed by Kush Mehta

Output

144 is 12th fibonacci number

Time Complexity: O(1)
Auxiliary Space: O(1)

### Method 6: Using power of the matrix {{1, 1}, {1, 0}}

This is another O(n) that relies on the fact that if we n times multiply the matrix M = {{1,1},{1,0}} to itself (in other words calculate power(M, n)), then we get the (n+1)th Fibonacci number as the element at row and column (0, 0) in the resultant matrix.
The matrix representation gives the following closed expression for the Fibonacci numbers:

## Python

 # Helper function that multiplies # 2 matrices F and M of size 2*2, # and puts the multiplication # result back to F[][]   # Helper function that calculates # F[][] raise to the power n and # puts the result in F[][] # Note that this function is # designed only for fib() and # won't work as general # power function def fib(n):     F = [[1, 1],          [1, 0]]     if (n == 0):         return 0     power(F, n - 1)     return F[0][0]     def multiply(F, M):   x = (F[0][0] * M[0][0] + F[0][1] * M[1][0])   y = (F[0][0] * M[0][1] + F[0][1] * M[1][1])   z = (F[1][0] * M[0][0] + F[1][1] * M[1][0])   w = (F[1][0] * M[0][1] + F[1][1] * M[1][1])   F[0][0] = x   F[0][1] = y   F[1][0] = z   F[1][1] = w     def power(F, n):     M = [[1, 1], [1, 0]]     # n - 1 times multiply the     # matrix to {{1,0},{0,1}}     for i in range(2, n + 1):         multiply(F, M)     # Driver Code if __name__ == "__main__":     n = 9     print(fib(n))   # This code is contributed by Yash Agarwal

Output

34


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

Please refer complete article on Program for Fibonacci numbers for more details!

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