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

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

The 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 is given above.

C++

// Fibonacci Series using Recursion
#include <bits/stdc++.h>
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 <stdio.h>
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 */

Python3

# Fibonacci series using recursion
def fibonacci(n):
    if n <= 1:
        return n
    return fibonacci(n-1) + fibonacci(n-2)


if __name__ == "__main__":
    n = 9
    print(fibonacci(n))

 # This code is contributed by Manan Tyagi.

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

<?php
// Fibonacci Series
// using Recursion

// function returns
// the Fibonacci number
function fib($n)
{
    if ($n <= 1)
        return $n;
    return fib($n - 1) +
           fib($n - 2);
}

// Driver Code
$n = 9;
echo fib($n);

// This code is contributed by aj_36
?>

Javascript

<script>
//Fibonacci Series using Recursion
    let n = 9;

    // function returns the Fibonacci number
    function fib(n) {
    if (n <= 1)
        return n;
    return fib(n-1) + fib(n-2);
}

    //function call
    document.write(fib(n));
//This code is contributed by Surbhi Tyagi
</script>

Output

34

Time Complexity: Exponential, as every function calls two other functions.

If the original recursion tree were to be implemented then this would have been the tree but now for n times the recursion function is called

Original tree for recursion

                          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)

Optimized tree for recursion for code above

    fib(5) 

    fib(4)

    fib(3)

    fib(2)

    fib(1)

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 in method 1 by storing the Fibonacci numbers calculated so far. 

C++

// C++ program for Fibonacci Series
// using Dynamic Programming
#include<bits/stdc++.h>
using namespace std;

class GFG{

public:
int fib(int n)
{

    // Declare an array to store
    // Fibonacci numbers.
    // 1 extra to handle
    // case, n = 0
    int f[n + 2];
    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];
    }
};

// Driver code
int main ()
{
    GFG g;
    int n = 9;

    cout << g.fib(n);
    return 0;
}

// This code is contributed by SoumikMondal

C

//Fibonacci Series using Dynamic Programming
#include<stdio.h>

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] = 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];
}

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 */

Python3

# Fibonacci Series using Dynamic Programming
def fibonacci(n):

    # Taking 1st two fibonacci numbers as 0 and 1
    f = [0, 1]


    for i in range(2, n+1):
        f.append(f[i-1] + f[i-2])
    return f[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] = 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];
    }

    // Driver Code
    public static void Main ()
    {
        int n = 9;
        Console.WriteLine(fib(n));
    }
}

// This code is contributed by anuj_67.

PHP

<?php
//Fibonacci Series using Dynamic
// Programming

function fib( $n)
{

    /* Declare an array to store
    Fibonacci numbers. */

    // 1 extra to handle case,
    // n = 0
    $f = array();
    $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];
}

$n = 9;
echo fib($n);

// This code is contributed by
// anuj_67.
?>

Javascript

<script>

// Fibonacci Series using Dynamic Programming

    function  fib(n)
    {
        /* Declare an array to store Fibonacci numbers. */
        let f = new Array(n+2); // 1 extra to handle case, n = 0
        let 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];
    }
    let n=9;
    document.write(fib(n));

    // This code is contributed by avanitrachhadiya2155

</script>

Output

34

Time complexity: O(n) for given n

Auxiliary 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. 

C++

// Fibonacci Series using Space Optimized Method
#include<bits/stdc++.h>
using namespace std;

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;
}

// Driver code
int main()
{
    int n = 9;

    cout << fib(n);
    return 0;
}

// This code is contributed by Code_Mech

C

// Fibonacci Series using Space Optimized Method
#include<stdio.h>
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

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+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

<?php
// PHP program for Fibonacci Series
// using Space Optimized Method

function fib( $n)
{
    $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;
}

// Driver Code
$n = 9;
echo fib($n);

// This code is contributed by anuj_67.
?>

Javascript

<script>

// Javascript program for Fibonacci Series using Space Optimized Method

function fib(n)
{
    let 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;
}

// Driver code

    let n = 9;

    document.write(fib(n));

// This code is contributed by Mayank Tyagi

</script>

Output

34

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

Method 4: 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: 

C++

#include<bits/stdc++.h>
using namespace std;

// 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[2][2], int M[2][2]);

// 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[2][2], int n);

int fib(int n)
{
    int F[2][2] = { { 1, 1 }, { 1, 0 } };

    if (n == 0)
        return 0;

    power(F, n - 1);

    return F[0][0];
}

void multiply(int F[2][2], int M[2][2])
{
    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;
}

void power(int F[2][2], int n)
{
    int i;
    int M[2][2] = { { 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 code
int main()
{
    int n = 9;

    cout << " " <<  fib(n);

    return 0;
}

// This code is contributed by shivanisinghss2110

C

#include <stdio.h>

/* 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[2][2], int M[2][2]);

/* 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[2][2], int n);

int fib(int n)
{
  int F[2][2] = {{1,1},{1,0}};
  if (n == 0)
      return 0;
  power(F, n-1);

  return F[0][0];
}

void multiply(int F[2][2], int M[2][2])
{
  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;
}

void power(int F[2][2], int n)
{
  int i;
  int M[2][2] = {{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 */

Python3

# 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#

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

<?php

function fib($n)
{
    $F = array(array(1, 1),
               array(1, 0));
    if ($n == 0)
        return 0;
    power($F, $n - 1);

    return $F[0][0];
}

function 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;
}

function power(&$F, $n)
{
    $M = array(array(1, 1),
               array(1, 0));

    // n - 1 times multiply the
    // matrix to {{1,0},{0,1}}
    for ($i = 2; $i <= $n; $i++)
        multiply($F, $M);
}

// Driver Code
$n = 9;
echo fib($n);

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>

// Note that this function is designed
// only for fib() and won't work as
// general power function


function fib( n)
{
     var F = [ [ 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[][]
  function 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;
}
// Helper function that calculates F[][]
// raise to the power n and puts the
// result in F[][]

function power( F, n)
{
    var i;
     var 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 code

    var   n = 9;

     document.write (" " +  fib(n));
 //This code is contributed by sweetyty
 </script>

Output

 34

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

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

C++

// Fibonacci Series using Optimized Method
#include <bits/stdc++.h>
using namespace std;

void multiply(int F[2][2], int M[2][2]);
void power(int F[2][2], int n);

// Function that returns nth Fibonacci number
int fib(int n)
{
    int F[2][2] = {{1, 1}, {1, 0}};
    if (n == 0)
        return 0;
    power(F, n - 1);

    return F[0][0];
}

// Optimized version of power() in method 4
void power(int F[2][2], int n)
{
    if(n == 0 || n == 1)
       return;
    int M[2][2] = {{1, 1}, {1, 0}};

    power(F, n / 2);
    multiply(F, F);

    if (n % 2 != 0)
        multiply(F, M);
}

void multiply(int F[2][2], int M[2][2])
{
    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;
}

// Driver code
int main()
{
    int n = 9;

    cout << fib(9);
    getchar();

    return 0;
}

// This code is contributed by Nidhi_biet

C

#include <stdio.h>

void multiply(int F[2][2], int M[2][2]);

void power(int F[2][2], int n);

/* function that returns nth Fibonacci number */
int fib(int n)
{
  int F[2][2] = {{1,1},{1,0}};
  if (n == 0)
    return 0;
  power(F, n-1);
  return F[0][0];
}

/* Optimized version of power() in method 4 */
void power(int F[2][2], int n)
{
  if( n == 0 || n == 1)
      return;
  int M[2][2] = {{1,1},{1,0}};

  power(F, n/2);
  multiply(F, F);

  if (n%2 != 0)
     multiply(F, M);
}

void multiply(int F[2][2], int M[2][2])
{
  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;
}

/* 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 */

Python3

# 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

Javascript

<script>

// Fibonacci Series using Optimized Method

// Function that returns nth Fibonacci number
function fib(n)
{
    var F = [ [ 1, 1 ], [ 1, 0 ] ];
    if (n == 0)
        return 0;

    power(F, n - 1);

    return F[0][0];
}

function multiply(F, M)
{
    var x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
    var y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
    var z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
    var 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 */
function power(F, n)
{
    if (n == 0 || n == 1)
        return;

    var M = [ [ 1, 1 ], [ 1, 0 ] ];

    power(F, n / 2);
    multiply(F, F);

    if (n % 2 != 0)
        multiply(F, M);
}

// Driver code
var n = 9;

document.write(fib(n));

// This code is contributed by gauravrajput1

</script>

Output

34

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 the 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 
from two different coefficients of the matrix product)

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

By putting n = n+1 in equation(1),
FmFn+1 + Fm-1Fn = Fm+n             --------------------------(2)

Putting m = n in equation(1).
F2n-1 = Fn2 + Fn-12
Putting m = n in equation(2)

F2n = (Fn-1 + Fn+1)Fn = (2Fn-1 + Fn)Fn (Source: Wiki)   --------
( By putting Fn+1 = Fn + Fn-1 )
To get the formula to be proved, we simply need to do the 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 the above idea.  

C++

// C++ Program to find n'th fibonacci Number in
// with O(Log n) arithmetic operations
#include <bits/stdc++.h>
using namespace std;

const int MAX = 1000;

// Create an array for memoization
int f[MAX] = {0};

// Returns n'th fibonacci 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;

    // Applying 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;

        // Applying 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.

Python3

# Python3 Program to find n'th fibonacci Number in
# with O(Log n) arithmetic operations
MAX = 1000

# Create an array for memoization
f = [0] * MAX

# Returns n'th fibonacci 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

    # Applying 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;

    // Applying 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

<?php
// PHP Program to find n'th
// fibonacci Number in with
// O(Log n) arithmetic operations

$MAX = 1000;

// Returns n'th fibonacci
// number using table f[]
function fib($n)
{
    global $MAX;

    // Create an array for memoization
    $f = array_fill(0, $MAX, NULL);

    // 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];

    $k = ($n & 1) ? ($n + 1) / 2 : $n / 2;

    // Applying 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 Code
$n = 9;
echo fib($n);

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>
      // JavaScript Program to find n'th fibonacci Number in
      // with O(Log n) arithmetic operations

      const MAX = 1000;

      // Create an array for memoization
      var f = [...Array(MAX)];
      f.fill(0);

      // Returns n'th fibonacci number using table f[]
      function fib(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];

        var k = n & 1 ? (n + 1) / 2 : n / 2;

        // Applying 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 */

      var n = 9;
      document.write(fib(n));

      // This code is contributed by rdtank.
    </script>

Output

34 

Time complexity: O(Log n), as we divide the problem in half in every recursive call.

Method 7: (Another approach(Using Binet’s formula))
In this method, we directly implement the formula for the nth term in the Fibonacci series. 
F_n = {[(√5 + 1)/2] ^ n} / √5 

Note: Above Formula gives correct result only upto for n<71. Because as we move forward from n>=71 , rounding error becomes significantly large . Although , using floor function instead of round function will give correct result for n=71 . But after from n=72 , it also fails.

Example: For N=72 , Correct result is 498454011879264 but above formula gives 498454011879265.

Reference: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html 

C++

// C++ Program to find n'th fibonacci Number
#include<iostream>
#include<cmath>

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<stdio.h>
#include<math.h>
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

Python3

# Python3 program to find n'th
# fibonacci Number
import math

def fibo(n):
    phi = (1 + math.sqrt(5)) / 2

    return round(pow(phi, n) / math.sqrt(5))

# Driver code
if __name__ == '__main__':

    n = 9

    print(fibo(n))

# This code is contributed by prasun_parate

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

<?php
// PHP Program to find n'th
// fibonacci Number

function fib($n)
{
    $phi = (1 + sqrt(5)) / 2;
    return round(pow($phi, $n) / sqrt(5));
}

// Driver Code
$n = 9;
echo fib($n) ;

// This code is contributed by Ryuga
?>

Javascript

<script>
    // Javascript Program to find n'th fibonacci Number
    function fib(n) {
      let phi = (1 + Math.sqrt(5)) / 2;
      return Math.round(Math.pow(phi, n) / Math.sqrt(5));
    }

    let n = 9;
      document.write(fib(n));

// This code is contributed by mukesh07.
</script>

Output

34

Time Complexity: O(logn), this is because calculating phi^n takes logn time
Auxiliary Space: O(1)

Method 8: DP using memoization(Top down approach)

We can avoid the repeated work done in method 1 by storing the Fibonacci numbers calculated so far. We just need to store all the values in an array.

C++

#include <bits/stdc++.h>
using namespace std;
int dp[10];
int fib(int n)
{
    if (n <= 1)
        return n;

    // temporary variables to store
    //  values of fib(n-1) & fib(n-2)
    int first, second;

    if (dp[n - 1] != -1)
        first = dp[n - 1];
    else
        first = fib(n - 1);

    if (dp[n - 2] != -1)
        second = dp[n - 2];
    else
        second = fib(n - 2);

    // memoization
    return dp[n] = first + second;
}

// Driver Code
int main()
{
    int n = 9;

    memset(dp, -1, sizeof(dp));

    cout << fib(n);
    getchar();
    return 0;

    // This code is contributed by Bhavneet Singh
}

Java

import java.util.*;

class GFG{

// Initialize array of dp
static int[] dp = new int[10];

static int fib(int n)
{
    if (n <= 1)
        return n;

    // Temporary variables to store
    // values of fib(n-1) & fib(n-2)
    int first, second;

    if (dp[n - 1] != -1)
        first = dp[n - 1];
    else
        first = fib(n - 1);

    if (dp[n - 2] != -1)
        second = dp[n - 2];
    else
        second = fib(n - 2);

    // Memoization
    return dp[n] = first + second;
}

// Driver Code
public static void main(String[] args)
{
    int n = 9;

    Arrays.fill(dp, -1);

    System.out.print(fib(n));
}
}

// This code is contributed by sujitmeshram

Python3

# Initialize array of dp
dp = [-1 for i in range(10)]

def fib(n):
    if (n <= 1):
        return n;
    global dp;

    # Temporary variables to store
    # values of fib(n-1) & fib(n-2)
    first = 0;
    second = 0;

    if (dp[n - 1] != -1):
        first = dp[n - 1];
    else:
        first = fib(n - 1);
    if (dp[n - 2] != -1):
        second = dp[n - 2];
    else:
        second = fib(n - 2);
    dp[n] = first + second;

    # Memoization
    return dp[n] ;

# Driver Code
if __name__ == '__main__':
    n = 9;
    print(fib(n));

# This code contributed by Rajput-Ji

C#

using System;
class GFG {

    // Initialize array of dp
    static int[] dp = new int[10];
    static int fib(int n)
    {
        if (n <= 1)
            return n;

        // Temporary variables to store
        // values of fib(n-1) & fib(n-2)
        int first, second;

        if (dp[n - 1] != -1)
            first = dp[n - 1];
        else
            first = fib(n - 1);

        if (dp[n - 2] != -1)
            second = dp[n - 2];
        else
            second = fib(n - 2);

        // Memoization
        return dp[n] = first + second;
    }

  // Driver code
  static void Main()
  {
    int n = 9;
    Array.Fill(dp, -1);
    Console.Write(fib(n));
  }
}

// This code is contributed by divyeshrabadiya07.

Javascript

<script>
// Initialize array of dp
dp = Array.from({length: 10}, (_, i) => -1);

function fib(n)
{
    if (n <= 1)
        return n;

    // Temporary variables to store
    // values of fib(n-1) & fib(n-2)
    var first, second;

    if (dp[n - 1] != -1)
        first = dp[n - 1];
    else
        first = fib(n - 1);

    if (dp[n - 2] != -1)
        second = dp[n - 2];
    else
        second = fib(n - 2);

    // Memoization
    return dp[n] = first + second;
}

// Driver Code
var n = 9;
document.write(fib(n));

// This code is contributed by Amit Katiyar
</script>

Output

34

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

References:
http://en.wikipedia.org/wiki/Fibonacci_number
http://www.ics.uci.edu/~eppstein/161/960109.html