# Program for Fibonacci numbers

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.` Given a number n, print n-th Fibonacci Number.
Examples:

```Input  : n = 2
Output : 1

Input  : n = 9
Output : 34
```

/div>

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

## C++

 `//Fibonacci Series using Recursion ` `#include ` `using` `namespace` `std; ` ` `  `int` `fib(``int` `n) ` `{ ` `    ``if` `(n <= 1) ` `        ``return` `n; ` `    ``return` `fib(n-1) + fib(n-2); ` `} ` ` `  `int` `main () ` `{ ` `    ``int` `n = 9; ` `    ``cout << fib(n); ` `    ``getchar``(); ` `    ``return` `0; ` `} ` ` `  `// This code is contributed ` `// by Akanksha Rai `

## C

 `//Fibonacci Series using Recursion ` `#include ` `int` `fib(``int` `n) ` `{ ` `   ``if` `(n <= 1) ` `      ``return` `n; ` `   ``return` `fib(n-1) + fib(n-2); ` `} ` ` `  `int` `main () ` `{ ` `  ``int` `n = 9; ` `  ``printf``(``"%d"``, fib(n)); ` `  ``getchar``(); ` `  ``return` `0; ` `} `

## Java

 `//Fibonacci Series using Recursion ` `class` `fibonacci ` `{ ` `    ``static` `int` `fib(``int` `n) ` `    ``{ ` `    ``if` `(n <= ``1``) ` `       ``return` `n; ` `    ``return` `fib(n-``1``) + fib(n-``2``); ` `    ``} ` `      `  `    ``public` `static` `void` `main (String args[]) ` `    ``{ ` `    ``int` `n = ``9``; ` `    ``System.out.println(fib(n)); ` `    ``} ` `} ` `/* This code is contributed by Rajat Mishra */`

## Python

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

## C#

 `// C# program for Fibonacci Series  ` `// using Recursion ` `using` `System;  ` ` `  `public` `class` `GFG  ` `{  ` `    ``public` `static` `int` `Fib(``int` `n)  ` `    ``{  ` `        ``if` `(n <= 1)  ` `        ``{  ` `            ``return` `n;  ` `        ``}  ` `        ``else` `        ``{  ` `            ``return` `Fib(n - 1) + Fib(n - 2);  ` `        ``}  ` `    ``}  ` `         `  `    ``// driver code ` `    ``public` `static` `void` `Main(``string``[] args)  ` `    ``{  ` `        ``int` `n = 9; ` `        ``Console.Write(Fib(n));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Sam007 `

## PHP

 ` `

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.

## C

 `//Fibonacci Series using Dynamic Programming ` `#include ` ` `  `int` `fib(``int` `n) ` `{ ` `  ``/* Declare an array to store Fibonacci numbers. */` `  ``int` `f[n+2];   ``// 1 extra to handle case, n = 0 ` `  ``int` `i; ` ` `  `  ``/* 0th and 1st number of the series are 0 and 1*/` `  ``f = 0; ` `  ``f = 1; ` ` `  `  ``for` `(i = 2; i <= n; i++) ` `  ``{ ` `      ``/* Add the previous 2 numbers in the series ` `         ``and store it */` `      ``f[i] = f[i-1] + f[i-2]; ` `  ``} ` ` `  `  ``return` `f[n]; ` `} ` ` `  `int` `main () ` `{ ` `  ``int` `n = 9; ` `  ``printf``(``"%d"``, fib(n)); ` `  ``getchar``(); ` `  ``return` `0; ` `} `

## Java

 `// Fibonacci Series using Dynamic Programming ` `class` `fibonacci ` `{ ` `   ``static` `int` `fib(``int` `n) ` `    ``{ ` `    ``/* Declare an array to store Fibonacci numbers. */` `    ``int` `f[] = ``new` `int``[n+``2``]; ``// 1 extra to handle case, n = 0 ` `    ``int` `i; ` `      `  `    ``/* 0th and 1st number of the series are 0 and 1*/` `    ``f[``0``] = ``0``; ` `    ``f[``1``] = ``1``; ` `     `  `    ``for` `(i = ``2``; i <= n; i++) ` `    ``{ ` `       ``/* Add the previous 2 numbers in the series ` `         ``and store it */` `        ``f[i] = f[i-``1``] + f[i-``2``]; ` `    ``} ` `      `  `    ``return` `f[n]; ` `    ``} ` `      `  `    ``public` `static` `void` `main (String args[]) ` `    ``{ ` `        ``int` `n = ``9``; ` `        ``System.out.println(fib(n)); ` `    ``} ` `} ` `/* This code is contributed by Rajat Mishra */`

## Python

 `# Fibonacci Series using Dynamic Programming  ` `def` `fibonacci(n):  ` `     `  `    ``# Taking 1st two fibonacci nubers as 0 and 1  ` `    ``FibArray ``=` `[``0``, ``1``]  ` `     `  `    ``while` `len``(FibArray) < n ``+` `1``:  ` `        ``FibArray.append(``0``)  ` `     `  `    ``if` `n <``=` `1``:  ` `        ``return` `n  ` `    ``else``:  ` `        ``if` `FibArray[n ``-` `1``] ``=``=` `0``:  ` `            ``FibArray[n ``-` `1``] ``=` `fibonacci(n ``-` `1``)  ` ` `  `        ``if` `FibArray[n ``-` `2``] ``=``=` `0``:  ` `            ``FibArray[n ``-` `2``] ``=` `fibonacci(n ``-` `2``)  ` `             `  `    ``FibArray[n] ``=` `FibArray[n ``-` `2``] ``+` `FibArray[n ``-` `1``]  ` `    ``return` `FibArray[n]  ` `     `  `print``(fibonacci(``9``))  `

## C#

 `// C# program for Fibonacci Series  ` `// using Dynamic Programming ` `using` `System; ` `class` `fibonacci { ` `     `  `static` `int` `fib(``int` `n) ` `    ``{ ` `         `  `        ``// Declare an array to  ` `        ``// store Fibonacci numbers. ` `        ``// 1 extra to handle  ` `        ``// case, n = 0 ` `        ``int` `[]f = ``new` `int``[n + 2];  ` `        ``int` `i; ` `         `  `        ``/* 0th and 1st number of the  ` `           ``series are 0 and 1 */` `        ``f = 0; ` `        ``f = 1; ` `         `  `        ``for` `(i = 2; i <= n; i++) ` `        ``{ ` `            ``/* Add the previous 2 numbers ` `               ``in the series and store it */` `            ``f[i] = f[i - 1] + f[i - 2]; ` `        ``} ` `         `  `        ``return` `f[n]; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int` `n = 9; ` `        ``Console.WriteLine(fib(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67.  `

## PHP

 ` `

Output:

`34`

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.

## C/C++

 `// Fibonacci Series using Space Optimized Method ` `#include ` `int` `fib(``int` `n) ` `{ ` `  ``int` `a = 0, b = 1, c, i; ` `  ``if``( n == 0) ` `    ``return` `a; ` `  ``for` `(i = 2; i <= n; i++) ` `  ``{ ` `     ``c = a + b; ` `     ``a = b; ` `     ``b = c; ` `  ``} ` `  ``return` `b; ` `} ` ` `  `int` `main () ` `{ ` `  ``int` `n = 9; ` `  ``printf``(``"%d"``, fib(n)); ` `  ``getchar``(); ` `  ``return` `0; ` `} `

## Java

 `// Java program for Fibonacci Series using Space ` `// Optimized Method ` `class` `fibonacci ` `{ ` `    ``static` `int` `fib(``int` `n) ` `    ``{ ` `        ``int` `a = ``0``, b = ``1``, c; ` `        ``if` `(n == ``0``) ` `            ``return` `a; ` `        ``for` `(``int` `i = ``2``; i <= n; i++) ` `        ``{ ` `            ``c = a + b; ` `            ``a = b; ` `            ``b = c; ` `        ``} ` `        ``return` `b; ` `    ``} ` ` `  `    ``public` `static` `void` `main (String args[]) ` `    ``{ ` `        ``int` `n = ``9``; ` `        ``System.out.println(fib(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Mihir Joshi `

## Python

 `# 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``+``1``): ` `            ``c ``=` `a ``+` `b ` `            ``a ``=` `b ` `            ``b ``=` `c ` `        ``return` `b ` ` `  `# Driver Program ` ` `  `print``(fibonacci(``9``)) ` ` `  `#This code is contributed by Saket Modi `

## C#

 `// C# program for Fibonacci Series  ` `// using Space Optimized Method ` `using` `System; ` ` `  `namespace` `Fib  ` `{  ` `    ``public` `class` `GFG  ` `    ``{  ` `        ``static` `int` `Fib(``int` `n)  ` `        ``{  ` `            ``int` `a = 0, b = 1, c = 0;  ` `             `  `            ``// To return the first Fibonacci number  ` `            ``if` `(n == 0) ``return` `a;  ` `     `  `            ``for` `(``int` `i = 2; i <= n; i++)  ` `            ``{  ` `                ``c = a + b;  ` `                ``a = b;  ` `                ``b = c;  ` `            ``}  ` `     `  `            ``return` `b;  ` `        ``}  ` `         `  `    ``// Driver function ` `    ``public` `static` `void` `Main(``string``[] args)  ` `        ``{  ` `             `  `            ``int` `n = 9; ` `            ``Console.Write(``"{0} "``, Fib(n));  ` `        ``}  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Sam007. `

## PHP

 ` `

Output :

```34
```

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

Method 4 ( Using power of the matrix {{1,1},{1,0}} )
This another O(n) which 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: ## C

 `#include ` ` `  `/* Helper function that multiplies 2 matrices F and M of size 2*2, and ` `  ``puts the multiplication result back to F[][] */` `void` `multiply(``int` `F, ``int` `M); ` ` `  `/* 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 */` `void` `power(``int` `F, ``int` `n); ` ` `  `int` `fib(``int` `n) ` `{ ` `  ``int` `F = {{1,1},{1,0}}; ` `  ``if` `(n == 0) ` `      ``return` `0; ` `  ``power(F, n-1); ` ` `  `  ``return` `F; ` `} ` ` `  `void` `multiply(``int` `F, ``int` `M) ` `{ ` `  ``int` `x =  F*M + F*M; ` `  ``int` `y =  F*M + F*M; ` `  ``int` `z =  F*M + F*M; ` `  ``int` `w =  F*M + F*M; ` ` `  `  ``F = x; ` `  ``F = y; ` `  ``F = z; ` `  ``F = w; ` `} ` ` `  `void` `power(``int` `F, ``int` `n) ` `{ ` `  ``int` `i; ` `  ``int` `M = {{1,1},{1,0}}; ` ` `  `  ``// n - 1 times multiply the matrix to {{1,0},{0,1}} ` `  ``for` `(i = 2; i <= n; i++) ` `      ``multiply(F, M); ` `} ` ` `  `/* Driver program to test above function */` `int` `main() ` `{ ` `  ``int` `n = 9; ` `  ``printf``(``"%d"``, fib(n)); ` `  ``getchar``(); ` `  ``return` `0; ` `} `

## Java

 `class` `fibonacci ` `{ ` `     `  `    ``static` `int` `fib(``int` `n) ` `    ``{ ` `    ``int` `F[][] = ``new` `int``[][]{{``1``,``1``},{``1``,``0``}}; ` `    ``if` `(n == ``0``) ` `        ``return` `0``; ` `    ``power(F, n-``1``); ` `     `  `       ``return` `F[``0``][``0``]; ` `    ``} ` `      `  `     ``/* Helper function that multiplies 2 matrices F and M of size 2*2, and ` `     ``puts the multiplication result back to F[][] */` `    ``static` `void` `multiply(``int` `F[][], ``int` `M[][]) ` `    ``{ ` `    ``int` `x =  F[``0``][``0``]*M[``0``][``0``] + F[``0``][``1``]*M[``1``][``0``]; ` `    ``int` `y =  F[``0``][``0``]*M[``0``][``1``] + F[``0``][``1``]*M[``1``][``1``]; ` `    ``int` `z =  F[``1``][``0``]*M[``0``][``0``] + F[``1``][``1``]*M[``1``][``0``]; ` `    ``int` `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; ` `    ``} ` ` `  `    ``/* 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 */` `    ``static` `void` `power(``int` `F[][], ``int` `n) ` `    ``{ ` `    ``int` `i; ` `    ``int` `M[][] = ``new` `int``[][]{{``1``,``1``},{``1``,``0``}}; ` `     `  `    ``// n - 1 times multiply the matrix to {{1,0},{0,1}} ` `    ``for` `(i = ``2``; i <= n; i++) ` `        ``multiply(F, M); ` `    ``} ` `      `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main (String args[]) ` `    ``{ ` `    ``int` `n = ``9``; ` `    ``System.out.println(fib(n)); ` `    ``} ` `} ` `/* This code is contributed by Rajat Mishra */`

## Python 3

 `# 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 ChitraNayal `

## C#

 `// C# program to find fibonacci number. ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``static` `int` `fib(``int` `n) ` `    ``{ ` `        ``int` `[,]F = ``new` `int``[,] {{1, 1}, ` `                               ``{1, 0} }; ` `        ``if` `(n == 0) ` `            ``return` `0; ` `        ``power(F, n-1); ` `         `  `        ``return` `F[0,0]; ` `    ``} ` `     `  `    ``/* Helper function that multiplies 2  ` `    ``matrices F and M of size 2*2, and puts ` `    ``the multiplication result back to F[][] */` `    ``static` `void` `multiply(``int` `[,]F, ``int` `[,]M) ` `    ``{ ` `        ``int` `x = F[0,0]*M[0,0] + F[0,1]*M[1,0]; ` `        ``int` `y = F[0,0]*M[0,1] + F[0,1]*M[1,1]; ` `        ``int` `z = F[1,0]*M[0,0] + F[1,1]*M[1,0]; ` `        ``int` `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; ` `    ``} ` ` `  `    ``/* 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 */` `    ``static` `void` `power(``int` `[,]F, ``int` `n) ` `    ``{ ` `        ``int` `i; ` `        ``int` `[,]M = ``new` `int``[,]{{1, 1}, ` `                              ``{1, 0} }; ` `         `  `        ``// n - 1 times multiply the matrix to ` `        ``// {{1,0},{0,1}} ` `        ``for` `(i = 2; i <= n; i++) ` `            ``multiply(F, M); ` `    ``} ` `     `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int` `n = 9; ` `        ``Console.WriteLine(fib(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67. `

## PHP

 ` `

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

Method 5 ( Optimized Method 4 )
The method 4 can be optimized to work in O(Logn) time complexity. We can do recursive multiplication to get power(M, n) in the prevous method (Similar to the optimization done in this post)

## C

 `#include ` ` `  `void` `multiply(``int` `F, ``int` `M); ` ` `  `void` `power(``int` `F, ``int` `n); ` ` `  `/* function that returns nth Fibonacci number */` `int` `fib(``int` `n) ` `{ ` `  ``int` `F = {{1,1},{1,0}}; ` `  ``if` `(n == 0) ` `    ``return` `0; ` `  ``power(F, n-1); ` `  ``return` `F; ` `} ` ` `  `/* Optimized version of power() in method 4 */` `void` `power(``int` `F, ``int` `n) ` `{ ` `  ``if``( n == 0 || n == 1) ` `      ``return``; ` `  ``int` `M = {{1,1},{1,0}}; ` ` `  `  ``power(F, n/2); ` `  ``multiply(F, F); ` ` `  `  ``if` `(n%2 != 0) ` `     ``multiply(F, M); ` `} ` ` `  `void` `multiply(``int` `F, ``int` `M) ` `{ ` `  ``int` `x =  F*M + F*M; ` `  ``int` `y =  F*M + F*M; ` `  ``int` `z =  F*M + F*M; ` `  ``int` `w =  F*M + F*M; ` ` `  `  ``F = x; ` `  ``F = y; ` `  ``F = z; ` `  ``F = w; ` `} ` ` `  `/* Driver program to test above function */` `int` `main() ` `{ ` `  ``int` `n = 9; ` `  ``printf``(``"%d"``, fib(9)); ` `  ``getchar``(); ` `  ``return` `0; ` `} `

## Java

 `//Fibonacci Series using  Optimized Method ` `class` `fibonacci ` `{ ` `    ``/* function that returns nth Fibonacci number */` `    ``static` `int` `fib(``int` `n) ` `    ``{ ` `    ``int` `F[][] = ``new` `int``[][]{{``1``,``1``},{``1``,``0``}}; ` `    ``if` `(n == ``0``) ` `        ``return` `0``; ` `    ``power(F, n-``1``); ` `      `  `    ``return` `F[``0``][``0``]; ` `    ``} ` `      `  `    ``static` `void` `multiply(``int` `F[][], ``int` `M[][]) ` `    ``{ ` `    ``int` `x =  F[``0``][``0``]*M[``0``][``0``] + F[``0``][``1``]*M[``1``][``0``]; ` `    ``int` `y =  F[``0``][``0``]*M[``0``][``1``] + F[``0``][``1``]*M[``1``][``1``]; ` `    ``int` `z =  F[``1``][``0``]*M[``0``][``0``] + F[``1``][``1``]*M[``1``][``0``]; ` `    ``int` `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; ` `    ``} ` `      `  `    ``/* Optimized version of power() in method 4 */` `    ``static` `void` `power(``int` `F[][], ``int` `n) ` `    ``{ ` `    ``if``( n == ``0` `|| n == ``1``) ` `      ``return``; ` `    ``int` `M[][] = ``new` `int``[][]{{``1``,``1``},{``1``,``0``}}; ` `      `  `    ``power(F, n/``2``); ` `    ``multiply(F, F); ` `      `  `    ``if` `(n%``2` `!= ``0``) ` `       ``multiply(F, M); ` `    ``} ` `     `  `    ``/* Driver program to test above function */`  `    ``public` `static` `void` `main (String args[]) ` `    ``{ ` `         ``int` `n = ``9``; ` `     ``System.out.println(fib(n)); ` `    ``} ` `} ` `/* This code is contributed by Rajat Mishra */`

## Python 3

 `# Fibonacci Series using  ` `# Optimized Method ` ` `  `# function that returns nth  ` `# Fibonacci number  ` `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 ` `         `  `# Optimized version of ` `# power() in method 4  ` `def` `power(F, n): ` ` `  `    ``if``( n ``=``=` `0` `or` `n ``=``=` `1``): ` `        ``return``; ` `    ``M ``=` `[[``1``, ``1``], ` `         ``[``1``, ``0``]]; ` `         `  `    ``power(F, n ``/``/` `2``) ` `    ``multiply(F, F) ` `         `  `    ``if` `(n ``%` `2` `!``=` `0``): ` `        ``multiply(F, M) ` `     `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``n ``=` `9` `    ``print``(fib(n)) ` ` `  `# This code is contributed  ` `# by ChitraNayal `

## C#

 `// Fibonacci Series using  ` `// Optimized Method ` `using` `System; ` ` `  `class` `GFG ` `{ ` `/* function that returns  ` `nth Fibonacci number */` `static` `int` `fib(``int` `n) ` `{ ` `int``[,] F = ``new` `int``[,]{{1, 1},  ` `                      ``{1, 0}}; ` `if` `(n == 0) ` `    ``return` `0; ` `power(F, n - 1); ` ` `  `return` `F[0, 0]; ` `} ` ` `  `static` `void` `multiply(``int``[,] F,  ` `                     ``int``[,] M) ` `{ ` `int` `x = F[0, 0] * M[0, 0] +  ` `        ``F[0, 1] * M[1, 0]; ` `int` `y = F[0, 0] * M[0, 1] +  ` `        ``F[0, 1] * M[1, 1]; ` `int` `z = F[1, 0] * M[0, 0] +  ` `        ``F[1, 1] * M[1, 0]; ` `int` `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; ` `} ` ` `  `/* Optimized version of  ` `power() in method 4 */` `static` `void` `power(``int``[,] F, ``int` `n) ` `{ ` `if``( n == 0 || n == 1) ` `return``; ` `int``[,] M = ``new` `int``[,]{{1, 1},  ` `                      ``{1, 0}}; ` ` `  `power(F, n / 2); ` `multiply(F, F); ` ` `  `if` `(n % 2 != 0) ` `multiply(F, M); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main () ` `{ ` `    ``int` `n = 9; ` `    ``Console.Write(fib(n)); ` `} ` `} ` ` `  `// This code is contributed ` `// by ChitraNayal `

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

Method 6 (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 = Fn+1Fn-1 – Fn2
Moreover, since AnAm = An+m for any square matrix A, the following identities can be derived (they are obtained form two different coefficients of the matrix product)

FmFn + Fm-1Fn-1 = Fm+n-1

By putting n = n+1,

FmFn+1 + Fm-1Fn = Fm+n

Putting m = n

F2n-1 = Fn2 + Fn-12

F2n = (Fn-1 + Fn+1)Fn = (2Fn-1 + Fn)Fn (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 the implementation of above idea.

## C++

 `// C++ Program to find n'th fibonacci Number in ` `// with O(Log n) arithmatic operations ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `MAX = 1000; ` ` `  `// Create an array for memoization ` `int` `f[MAX] = {0}; ` ` `  `// Returns n'th fuibonacci number using table f[] ` `int` `fib(``int` `n) ` `{ ` `    ``// Base cases ` `    ``if` `(n == 0) ` `        ``return` `0; ` `    ``if` `(n == 1 || n == 2) ` `        ``return` `(f[n] = 1); ` ` `  `    ``// If fib(n) is already computed ` `    ``if` `(f[n]) ` `        ``return` `f[n]; ` ` `  `    ``int` `k = (n & 1)? (n+1)/2 : n/2; ` ` `  `    ``// Applyting above formula [Note value n&1 is 1 ` `    ``// if n is odd, else 0. ` `    ``f[n] = (n & 1)? (fib(k)*fib(k) + fib(k-1)*fib(k-1)) ` `           ``: (2*fib(k-1) + fib(k))*fib(k); ` ` `  `    ``return` `f[n]; ` `} ` ` `  `/* Driver program to test above function */` `int` `main() ` `{ ` `    ``int` `n = 9; ` `    ``printf``(``"%d "``, fib(n)); ` `    ``return` `0; ` `} `

## Java

 `// Java Program to find n'th fibonacci  ` `// Number with O(Log n) arithmetic operations ` `import` `java.util.*; ` ` `  `class` `GFG { ` `     `  `    ``static` `int` `MAX = ``1000``; ` `    ``static` `int` `f[]; ` `     `  `    ``// Returns n'th fibonacci number using  ` `    ``// table f[] ` `    ``public` `static` `int` `fib(``int` `n) ` `    ``{ ` `        ``// Base cases ` `        ``if` `(n == ``0``) ` `            ``return` `0``; ` `             `  `        ``if` `(n == ``1` `|| n == ``2``) ` `            ``return` `(f[n] = ``1``); ` `      `  `        ``// If fib(n) is already computed ` `        ``if` `(f[n] != ``0``) ` `            ``return` `f[n]; ` `      `  `        ``int` `k = (n & ``1``) == ``1``? (n + ``1``) / ``2`  `                            ``: n / ``2``; ` `      `  `        ``// Applyting above formula [Note value ` `        ``// n&1 is 1 if n is odd, else 0. ` `        ``f[n] = (n & ``1``) == ``1``? (fib(k) * fib(k) +  ` `                        ``fib(k - ``1``) * fib(k - ``1``)) ` `                       ``: (``2` `* fib(k - ``1``) + fib(k))  ` `                       ``* fib(k); ` `      `  `        ``return` `f[n]; ` `    ``} ` `     `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `n = ``9``; ` `        ``f= ``new` `int``[MAX]; ` `        ``System.out.println(fib(n)); ` `    ``} ` `} ` `     `  `// This code is contributed by Arnav Kr. Mandal. `

## Python

 `# Python 3 Program to find n'th fibonacci Number in ` `# with O(Log n) arithmatic operations ` `MAX` `=` `1000` ` `  `# Create an array for memoization ` `f ``=` `[``0``] ``*` `MAX` ` `  `# Returns n'th fuibonacci number using table f[] ` `def` `fib(n) : ` `    ``# Base cases ` `    ``if` `(n ``=``=` `0``) : ` `        ``return` `0` `    ``if` `(n ``=``=` `1` `or` `n ``=``=` `2``) : ` `        ``f[n] ``=` `1` `        ``return` `(f[n]) ` ` `  `    ``# If fib(n) is already computed ` `    ``if` `(f[n]) : ` `        ``return` `f[n] ` ` `  `    ``if``( n & ``1``) : ` `        ``k ``=` `(n ``+` `1``) ``/``/` `2` `    ``else` `:  ` `        ``k ``=` `n ``/``/` `2` ` `  `    ``# Applyting above formula [Note value n&1 is 1 ` `    ``# if n is odd, else 0. ` `    ``if``((n & ``1``) ) : ` `        ``f[n] ``=` `(fib(k) ``*` `fib(k) ``+` `fib(k``-``1``) ``*` `fib(k``-``1``)) ` `    ``else` `: ` `        ``f[n] ``=` `(``2``*``fib(k``-``1``) ``+` `fib(k))``*``fib(k) ` ` `  `    ``return` `f[n] ` ` `  ` `  `# Driver code ` `n ``=` `9` `print``(fib(n)) ` ` `  ` `  `# This code is contributed by Nikita Tiwari. `

## C#

 `// C# Program to find n'th  ` `// fibonacci Number with  ` `// O(Log n) arithmetic operations ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `static` `int` `MAX = 1000; ` `static` `int``[] f; ` ` `  `// Returns n'th fibonacci  ` `// number using table f[] ` `public` `static` `int` `fib(``int` `n) ` `{ ` `    ``// Base cases ` `    ``if` `(n == 0) ` `        ``return` `0; ` `         `  `    ``if` `(n == 1 || n == 2) ` `        ``return` `(f[n] = 1); ` ` `  `    ``// If fib(n) is already  ` `    ``// computed ` `    ``if` `(f[n] != 0) ` `        ``return` `f[n]; ` ` `  `    ``int` `k = (n & 1) == 1 ? (n + 1) / 2 ` `                         ``: n / 2; ` ` `  `    ``// Applyting above formula  ` `    ``// [Note value n&1 is 1 if  ` `    ``// n is odd, else 0. ` `    ``f[n] = (n & 1) == 1 ? (fib(k) * fib(k) +  ` `                           ``fib(k - 1) * fib(k - 1)) ` `                        ``: (2 * fib(k - 1) + fib(k)) *  ` `                                            ``fib(k); ` ` `  `    ``return` `f[n]; ` `} ` ` `  `// Driver Code ` `static` `void` `Main()  ` `{ ` `    ``int` `n = 9; ` `    ``f = ``new` `int``[MAX]; ` `    ``Console.WriteLine(fib(n)); ` `} ` `} ` ` `  `// This code is contributed by mits `

## PHP

 ` `

Output :

`34`

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

Method 7
Another approach:(Using formula)
In this method we directly implement the formula for nth term in the fibonacci series.
Fn = {[(5 + 1)/2] ^ n} / 5
Reference: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html

## C++

 `// C++ Program to find n'th fibonacci Number ` `#include ` `#include ` ` `  `int` `fib(``int` `n) { ` `  ``double` `phi = (1 + ``sqrt``(5)) / 2; ` `  ``return` `round(``pow``(phi, n) / ``sqrt``(5)); ` `} ` ` `  `// Driver Code ` `int` `main () ` `{ ` `  ``int` `n = 9; ` `  ``std::cout << fib(n) << std::endl; ` `  ``return` `0; ` `} ` `//This code is contributed by Lokesh Mohanty. `

## C

 `// C Program to find n'th fibonacci Number ` `#include ` `#include ` `int` `fib(``int` `n) { ` `  ``double` `phi = (1 + ``sqrt``(5)) / 2; ` `  ``return` `round(``pow``(phi, n) / ``sqrt``(5)); ` `} ` `int` `main () ` `{ ` `  ``int` `n = 9; ` `  ``printf``(``"%d"``, fib(n)); ` `  ``return` `0; ` `} `

## Java

 `// Java Program to find n'th fibonacci Number ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `static` `int` `fib(``int` `n) { ` `double` `phi = (``1` `+ Math.sqrt(``5``)) / ``2``; ` `return` `(``int``) Math.round(Math.pow(phi, n)  ` `                        ``/ Math.sqrt(``5``)); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) { ` `        ``int` `n = ``9``; ` `    ``System.out.println(fib(n)); ` `    ``} ` `} ` `// This code is contributed by PrinciRaj1992 `

## C#

 `// C# Program to find n'th fibonacci Number ` `using` `System; ` ` `  `public` `class` `GFG  ` `{ ` `    ``static` `int` `fib(``int` `n)  ` `    ``{ ` `    ``double` `phi = (1 + Math.Sqrt(5)) / 2; ` `    ``return` `(``int``) Math.Round(Math.Pow(phi, n)  ` `                            ``/ Math.Sqrt(5));  ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `n = 9; ` `        ``Console.WriteLine(fib(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## PHP

 ` `

Output:

`34`

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

This method is contributed by Chirag Agarwal.

Related Articles:
Large Fibonacci Numbers in Java

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.