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Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.37

96\. Unique Binary Search Trees

Medium

Given an integer `n`, return _the number of structurally unique **BST'**s (binary search trees) which has exactly_ `n` _nodes of unique values from_ `1` _to_ `n`.

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/01/18/uniquebstn3.jpg)

**Input:** n = 3

**Output:** 5 

**Example 2:**

**Input:** n = 1

**Output:** 1 

**Constraints:**

*   `1 <= n <= 19`

To solve the "Unique Binary Search Trees" problem in Java with the Solution class, follow these steps:

1. Define a method `numTrees` in the `Solution` class that takes an integer `n` as input and returns the number of structurally unique BSTs (binary search trees) with exactly `n` nodes.
2. Implement a dynamic programming approach to solve the problem:
   - Create an array `dp` of size `n + 1` to store the number of unique BSTs for each number of nodes from 0 to `n`.
   - Initialize `dp[0] = 1` and `dp[1] = 1`, as there is only one unique BST for 0 and 1 node(s).
   - Use a nested loop to calculate `dp[i]` for each `i` from 2 to `n`.
   - For each `i`, calculate `dp[i]` by summing up the products of `dp[j]` and `dp[i - j - 1]` for all possible values of `j` from 0 to `i - 1`.
   - Return `dp[n]`, which represents the number of unique BSTs with `n` nodes.

Here's the implementation of the `numTrees` method in Java:

```java
class Solution {
    public int numTrees(int n) {
        int[] dp = new int[n + 1];
        dp[0] = 1;
        dp[1] = 1;
        for (int i = 2; i <= n; i++) {
            for (int j = 0; j < i; j++) {
                dp[i] += dp[j] * dp[i - j - 1];
            }
        }
        return dp[n];
    }
}
```

This implementation uses dynamic programming to compute the number of structurally unique BSTs with `n` nodes in O(n^2) time complexity.