# Nth Fibonacci Number

Last Updated : 03 Oct, 2023

Given a number n, print n-th Fibonacci Number

The Fibonacci numbers are the numbers in the following integer sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……..

Examples:

Input  : n = 1

Output : 1

Input  : n = 9

Output : 34

Input  : n = 10

Output : 55

Recommended Practice

## Iterative Approach to Find and Print Nth Fibonacci Numbers:

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation:  with seed values and  and .

## 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 codeint 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 Methodpublic 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 Methodusing 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.

## Javascript

 

## PHP

 

Output
34



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

## Recursion Approach to Find and Print Nth Fibonacci Numbers:

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation:  with seed values and  and .

The Nth Fibonacci Number can be found using the recurrence relation shown above:

• if n = 0, then return 0.
• If n = 1, then it should return 1.
• For n > 1, it should return Fn-1 + Fn-2

Below is the implementation of the above approach:

## 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 << n << "th Fibonacci Number: " << fib(n);    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("%dth Fibonacci Number: %d", n, fib(n));    return 0;}

## Java

 // Fibonacci Series using Recursionimport 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(            n + "th Fibonacci Number: " + fib(n));    }}/* This code is contributed by Rajat Mishra */

## Python3

 # Fibonacci series using recursiondef fibonacci(n):    if n <= 1:        return n    return fibonacci(n-1) + fibonacci(n-2)  if __name__ == "__main__":    n = 9    print(n, "th Fibonacci Number: ")    print(fibonacci(n))  # This code is contributed by Manan Tyagi.

## C#

 // C# program for Fibonacci Series// using Recursionusing 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(n + "th Fibonacci Number: " + Fib(n));    }} // This code is contributed by Sam007

## Javascript

 // Javascript program for Fibonacci Series// using Recursion function Fib(n) {  if (n <= 1) {    return n;  } else {    return Fib(n - 1) + Fib(n - 2);  }} // driver codelet n = 9;console.log(n + "th Fibonacci Number: " + Fib(n));

## PHP

 

Output
34



Time Complexity: Exponential, as every function calls two other functions.
Auxiliary space complexity: O(n), as the maximum depth of the recursion tree is n.

## Dynamic Programming Approach to Find and Print Nth Fibonacci Numbers:

Consider the Recursion Tree for the 5th Fibonacci Number from the above approach:

                          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)

If you see, the same method call is being done multiple times for the same value. This can be optimized with the help of Dynamic Programming. We can avoid the repeated work done in the Recursion approach by storing the Fibonacci numbers calculated so far.

Dynamic Programming Approach to Find and Print Nth Fibonacci Numbers:

Below is the implementation of the above approach:

## 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 codeint 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 Programmingpublic 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 Programmingdef 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 Programmingusing 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.

## Javascript

 

## PHP

 

Output
34



Time complexity: O(n) for given n
Auxiliary space: O(n)

## Nth Power of Matrix Approach to Find and Print Nth Fibonacci Numbers

This approach 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.

• If n is even then k = n/2:
• Therefore Nth Fibonacci Number = F(n) = [2*F(k-1) + F(k)]*F(k)
• If n is odd then k = (n + 1)/2:
• Therefore Nth Fibonacci Number = F(n) = F(k)*F(k) + F(k-1)*F(k-1)

How does this formula work?
The formula can be derived from the 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  ——–

( 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 approach

## 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 numberint 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 4void 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 codeint 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 Methodpublic 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 Codeif __name__ == "__main__":    n = 9    print(fib(n)) # This code is contributed# by ChitraNayal

## C#

 // Fibonacci Series using// Optimized Methodusing 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(Log n)
Auxiliary Space: O(Log n) if we consider the function call stack size, otherwise O(1).

Related Articles:
Large Fibonacci Numbers in Java

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