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:
Input: n = 3 Output: 5
Example 2:
Input: n = 1 Output: 1
Constraints:
1 <= n <= 19
Solution:
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class Solution {
public int numTrees(int n) {
this.n = n;
this.dp = new int[n + 1][n + 1];
return recurse(1, n);
}
/**
* Intuition: Since we just need to count the number of trees, we can use dp here.
* Number of trees for n i.e. f(n) = total trees with each of the numbers as root nodes from 1 to n.
* For a given node say 2 , total left sub tree * total right sub tree will result in total trees.
* Total = sum[1 to n](recurse(left ) * recurse(right))
**/
public int calculate(int n) {
int[] dp = new int[n + 1];
dp[0] = 1;
dp[1] = 1;
for (int i = 2; i <= n; i++) {
int res = 0;
for (int j = 1; j <= i; j++) {
res += (dp[j - 1] * dp[i - j]);
}
dp[i] = res;
}
return dp[n];
}
int n;
int[][] dp;
/**
* Recursive dynamic programming code. Total = sum[1 to n](recurse(left ) * recurse(right))
* Time Complexity: O(N^2)
* Space Complexity: O(N^2)
*/
public int recurse(int start, int end) {
if (start >= end) {
return 1;
}
System.out.println(start + ":" + end);
if (dp[start][end] != 0) {
return dp[start][end];
}
int res = 0;
for (int i = start; i <= end; i++) {
res += recurse(start, i - 1) * recurse(i + 1, end);
}
dp[start][end] = res;
return res;
}
}
