Coding Dynamic Programming JAVA

Java Programming – Program for Fibonacci numbers

Java Programming - Program for Fibonacci numbers - Dynamic Programming The Fibonacci numbers are the numbers in the following integer sequence.

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

F = 0 and F1 = 1.

Write a function int fib(int n) that returns Fn. For example, if n = 0, then fib() should return 0. If n = 1, then it should return 1. For n > 1, it should return Fn-1 + Fn-2

For n = 9
Output:34

Following are different methods to get the nth Fibonacci number.

Method 1 ( Use recursion )
A simple method that is a direct recursive implementation mathematical recurrence relation given above.

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));
    }
}

Output

34

Time Complexity: T(n) = T(n-1) + T(n-2) which is exponential.
We can observe that this implementation does a lot of repeated work (see the following recursion tree). So this is a bad implementation for nth Fibonacci number.

                         fib(5)   
                     /             \     
               fib(4)                fib(3)   
             /      \                /     \
         fib(3)      fib(2)         fib(2)    fib(1)
        /     \        /    \       /    \  
  fib(2)   fib(1)  fib(1) fib(0) fib(1) fib(0)
  /    \
fib(1) fib(0)

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

Method 2 ( Use Dynamic Programming )
We can avoid the repeated work done is the method 1 by storing the Fibonacci numbers calculated so far.

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+1];
    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));
    }
}

Output:

34

Time Complexity: O(n)
Extra 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.

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));
    }
}

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

READ  C Programming - Longest Common Subsequence

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:

Problems For Fibonacci Series

 

 

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));
    }
}

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

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

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));
    }
}

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

About the author

Venkatesan Prabu

Venkatesan Prabu

Wikitechy Founder, Author, International Speaker, and Job Consultant. My role as the CEO of Wikitechy, I help businesses build their next generation digital platforms and help with their product innovation and growth strategy. I'm a frequent speaker at tech conferences and events.

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