Question

The following is a correct implementation of a non-recursive method to find the nth Fibonacci number in Java:

public static long fib(int n){

   long[ ] fibNums = new long[n + 1];

   fibNums[0] = 0;

   finNums[1] = 1;

   for(int i = 2; i <= n; i++){

      fibNums[i] = fibNums [i - 1] + finBums[i - 2];

}

return fibNums[n];

}

I am having trouble understanding the code. Why the creation of the long array make this code works and why you have to make it [n+1]? How is the for loop working here (I particularly have problem understanding the body and why returning fibNums[n] gives me the right answer)

Thank you.

Answer 1

Answer:

As we know that fibonacci series is as follows

0,1,1,2,3,5,8,13,21,34, 55, 89, 144…………

It is basically sum of previous number and current number

Here we first give 2 seed values F₀ = 0, F₁ = 1

F_n = F_n-1 + F_n-2

The Solution given here is a dynamic programming approach.

1. We have to create an array with n+1 size to handle 0th case.

I.e. to store the elements 0 to n

2. Here the loop is adding previous two values into the array that we have created.

for(int i=2;i<=n;i++){   

     fibNums[i] = fibNums [i-1] + fibNums[i-2];

}

iteration 1: i = 2

fibNums[2] = fibNums[2-1] + fibNums[2-2]

fibNums[2] = fibNums[1] + fibNums[0]

So now fibNums[2] = 1 + 0 = 1

Now i++ so i =3

iteration 2: i = 3

fibNums[3] = fibNums[3-1] + fibNums[3-2]

fibNums[3] = fibNums[2] + fibNums[1]

So now fibNums[3] = 1 + 1 = 2

Now i++ so i =4

iteration 3: i = 4

fibNums[4] = fibNums[4-1] + fibNums[4-2]

fibNums[4] = fibNums[3] + fibNums[2]

So now fibNums[4] = 2 + 1 = 3

Now i++ so i =5

iteration 4: i = 5

fibNums[5] = fibNums[5-1] + fibNums[5-2]

fibNums[5] = fibNums[4] + fibNums[3]

So now fibNums[5] = 3 + 2 = 5

Now i++ so i = 6

The iterations will continue till the i value become n value.

Suppose n value is 5, then return fibNums[n] values returns the required nth febinocci number.

I.e. fibNums[5] = 5.

The Full code and result screenshot has been attached below.