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# Program for Fibonacci numbers

• Difficulty Level : Medium
• Last Updated : 29 Jan, 2023

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
Recommended Practice

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

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  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 import java.io.*; 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

 

## Javascript

 

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 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   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 public 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.

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## Javascript

 

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 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 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 public 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

 

## Javascript

 

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 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 {{1,1},{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    /* 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

 public 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

 

## Javascript

 

Output

 34

Time Complexity: O(n)
Auxiliary 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  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    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 public 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

 

Output

34

Time Complexity: O(Logn)
Auxiliary 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  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.*;   public 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

 

## Javascript

 

Output

34

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

Method 7: (Another approach(Using Binet’s formula))
In this method, we directly implement the formula for the nth term in the Fibonacci series.
Fn = {[(âˆš5 + 1)/2] ^ n} / âˆš5

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 #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.*;   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(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

 

## Javascript

 

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  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.*;   public 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

 

Output

34

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

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.

### Method 9 : (Kartik’s K sequence) here K=3

1)  0,1,1,2,3,5,8,13,21,34,55,89,144,….. (Parallel 0 highlighted with Bold)

2)  0,1,1,2,3,5,8,13,21,34,55,89,144,….. (Parallel 1 highlighted with Bold)

3)  0,1,1,2,3,5,8,13,21,34,55,89,144,….. (Parallel 2 highlighted with Bold)

if you observed the bold Numbers

consider Parallel 0  bold Number form 1)

0,2,8,34,144,…

2 * 4  + 0 = 8  (7th)

8 * 4 + 2 = 34  (10th)

34 * 4 + 8 = 144  (13th)

#### N+1th * 4 + Nth = N+2th which can be applied to all three Parallel

and shifting rules

using for F1 and F2 it can be replicated to Lucas sequence as well

### in the below  image “Parallel 1” as r1 = 3 and r11=13

Shifting

AND HERE I USED  1,1,5,21,… Parallel 2

## Python3

 def nth_fibonnaci(n):     if n > 0:         n1, n2 = 1, 1         if n > 3:             for _ in range((n//3)):                 n1, n2 = n2, (n2 << 2)+n1  # << 2   is multiply by 4         if n % 3 == 0:             return n1         elif n % 3 == 1:             return (n2-n1) >> 1  # >> 1   is divide by 2  'F1'         elif n % 3 == 2:             return (n2+n1) >> 1  # >> 1   is divide by 2  'F2'     else:         return -1         for i in range(1, 9):     print(f"{nth_fibonnaci(i)} is {i}", end="th  | ") print("") for i in range(9, 30, 3):     print(f"{nth_fibonnaci(i)} is {i}", end="th  | ")

Output

0 is 1th  | 1 is 2th  | 1 is 3th  | 2 is 4th  | 3 is 5th  | 5 is 6th  | 8 is 7th  | 13 is 8th  |
21 is 9th  | 89 is 12th  | 377 is 15th  | 1597 is 18th  | 6765 is 21th  | 28657 is 24th  | 121393 is 27th  | 

Time Complexity: in between O(log n) and O(n) or  (n/3)

Auxiliary Space: O(1) (constant)

Related Articles: