0
$\begingroup$

I have been solving some algorithm questions recently and a pattern I have observed in some problems is as follows:

Given a string or a list, do an aggregation operation on each of its elements. Here in each of these elements we apply some recurrence to solve it.

An example of one such problem is below.


Problem: Given n integers return the total number of binary search trees that can be formed using the n integers

To solve this problem, I define a recurrence relation as follows:

f(n) = 1 // if n = 0
f(n) = ∑ f(i) * f(n-i-1) where 0 <= i <= n-1

This works and I get the correct answer however I want to modify the function a bit.

Instead of expressing the function in terms of f(n) I want to express it in terms of f(n, i) so I can remove the summation. However I am unable to do it correctly.


Code

My code to solve the problem by defining the recurrence in terms of f(n) is as follows: (I am aware it can be optimized by DP but that is not what I am trying to do here)

public int f(int n) {
    if(n == 0)
        return 1;

    int result = 0;
    for(int i = 0; i< n; i++)
        result += f(i) * f(n-i-1);
    return result;
}

I want to remove that for loop and instead express the function in terms of f(n,i) instead of f(n).


Question

  1. How to convert the recurrence shown above from f(n) to f(n,i) and remove the summation?
    • Here 'n' is the size of the list of element and 'i' is the ith element in the list that we choose to be the root of the tree.
$\endgroup$
0
$\begingroup$

You could do it by making the second parameter of your function function like the index of the summation, and recursively increment this parameter as long as $0 \leq i < n$.

$$f(n, i) = \begin{cases} 1 & \text{if } n =0\\ f(i, 0)\cdot f(n-i-1,0) + f(n, i+1) & \text{if } 0 \leq i < n\\ 0 & \text{otherwise}\\ \end{cases}$$

Then $f(n, 0)$ gives you the answer you want.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.