Program for Fibonacci numbers

Difficulty Level : Medium
Last Updated : 25 Nov, 2020

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

F_n = F_n-1 + F_n-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

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<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 */

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

<?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 
?> 

Output

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 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 */

Python

# Fibonacci Series using Dynamic Programming  
def fibonacci(n):  
      
    # Taking 1st two fibonacci nubers 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.  
?> 

Output

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 

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

<?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. 
?> 

Output

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:

fibonaccimatrix

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 
?> 

Output

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 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 

Output

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.

fibonaccimatrix

Taking determinant on both sides, we get
(-1)ⁿ = F_n+1F_n-1 – F_n²
Moreover, since AⁿA^m = A^n+m for any square matrix A, the following identities can be derived (they are obtained form two different coefficients of the matrix product)
F_mF_n + F_m-1F_n-1 = F_m+n-1
By putting n = n+1,
F_mF_n+1 + F_m-1F_n = F_m+n
Putting m = n
F_2n-1 = F_n² + F_n-1²
F_2n = (F_n-1 + F_n+1)F_n = (2F_n-1 + F_n)F_n (Source: Wiki)
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 above idea.

C++

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

Python

# Python3 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
  
    # 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) arithmatic operations 
  
$MAX = 1000; 
  
// Returns n'th fuibonacci  
// 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 
?> 

Output

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.
F_n = {[(√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<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 
?> 

Output

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

Method 8:

DP using memoization(Top down approach)

We can avoid the repeated work done is 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 
}

Output

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

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