leetcode_gym/_problems/uniqueBST.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	Unique BST	2020-06-24	dp	https://leetcode.com/problems/unique-binary-search-trees/	Attempted

Problem

Given n, how many structurally unique BST's (binary search trees) that store values 1 ... n?

Example:

Input: 3
Output: 5
Explanation:
Given n = 3, there are a total of 5 unique BST's:

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

Solution

For n=1: you just have 1 pattern.
For n=2, you have 2 possible patterns
For n=3,
- if you put 1 at the root, you have 2 patterns for the remaining (calculated above)
- if you put 2 at the root, you just have 1 possible pattern
- if you put 3 at the root, you have 2 patterns avaiable for the remaining numbers

f(3) = g(2) + g(1) + g(2) = 2 + 1 + 2 = 5

For n=4:

1,2,3,4
- if 1 is at root, the remaining has g(3) combinations
- if 2 is at root, [1] will have 1 combination and [3,4] will have 2 combination, you need to multiply them
- if 3 is at root, [1,2] will have g(2) combo and [4] will have 1 combo, mulitply them
- if 4 is at root, [1,2,3] has 5 combo

f(4) = g(3) + (1*g(2)) + (g(2)* 1) + g(3)
     = 5 + (1*2) + (1*2) + 5
     = 14

{% highlight python %}

class Solution: def numTrees(self, n: int) -> int:

    @lru_cache(None)
    def helper(left, right):
        
        # base case, g(1)
        if left>=right:
            return 1
        
        count = 0
        for i in range(left, right+1):
            l_tree = helper(left, i-1)
            r_tree = helper(i+1, right)
            count += l_tree * r_tree
        
        return count

    return helper(1, n)

{% endhighlight %}

---

{% highlight python %}

def numTrees(self, n: int) -> int:

memo = {}

def helper(left, right):
    
    # base case, g(1)
    if left>=right:
        return 1
    
    count = 0
    if (left,right) in memo:
        return memo[(left,right)]
    else:
        for i in range(left, right+1):
            l_tree = helper(left, i-1)
            r_tree = helper(i+1, right)
            count += l_tree * r_tree
        
        memo[(left,right)] = count

        return count

return helper(1, n)

{% endhighlight %}