Sumant Kumar

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# The optimal approach to find the nth Fibonacci number using recursion in Java

If you are a programmer or computer science student, you must be aware of the Fibonacci sequence and how we calculate it using recursion. No?

Ok, then, let's see the typical recursive approach.

`````` private static int headRecursion(int n){
if (n==1)
return 0;
if (n==2)
return 1;
else{
}
}
``````

Well, the above approach is not an ideal approach, wondering why? Because it has exponential time and space complexities.

Time Complexity: O(2^N) (Exponential Time Complexity)
N -> index of nth fibonacci number

Since every value is made up of the previous 2 values, we need to find 2 values for finding any Fibonacci number. This gives rise to 2 calls every time in our recursion tree. The tree can have at most n levels. This gives us at most 2^n nodes in the tree.

Space Complexity: O(2^N) (Exponential Space Complexity)
N -> index of nth fibonacci number

All the function calls are pushed into the function call stack. At any point in time at max 2^n, function calls are there in the call stack. Therefore, space complexity will be O(2^n)

Now, let's see the optimized recursive approach:

``````private static int tailRecursion(int n,int a,int b){
if (n==1)
return a;
if (n==2)
return b;
else{
return tailRecursion(n-1,b,a+b);
}
}
``````

The initial values are a=0 and b=1. This is how we can optimally use recursion to find Fibonacci numbers!

In this case, there is just a single recursive function call which means that it is a faster approach - no need for two very similar recursive function calls (headRecursion(n-1) and headRecursion(n-2)). Java will not recalculate the same values several times.

Wondering what would be the time and space complexcities in this approach?

Time Complexity: O(N) (Linear Time Complexity)
N -> index of nth fibonacci number

For the nth number, there will be n function calls.

Space Complexity: O(1) (Constant Space Complexity), Isn't it amazing?
N -> index of nth fibonacci number

At any point in time, at max 1, function calls are there in the call stack. Therefore, space complexity will be O(1)

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`fib(n) = (Ψ ^n - (1 - Ψ)^n) / sqrt(5), where Ψ = 1.618`