From 0968d73de88e667cb779c3f6b435c1096449cdf9 Mon Sep 17 00:00:00 2001
From: SjxSubham <sjxsubham@gmail.com>
Date: Mon, 20 Oct 2025 23:22:55 +0530
Subject: [PATCH] Add countBSTs method to calculate unique BSTs

Implement a method to count unique BSTs for given elements.
---
 October_2025/potd_20_10_2025.java | 44 +++++++++++++++++++++++++++++++
 1 file changed, 44 insertions(+)
 create mode 100644 October_2025/potd_20_10_2025.java

diff --git a/October_2025/potd_20_10_2025.java b/October_2025/potd_20_10_2025.java
new file mode 100644
index 0000000..a8822a6
--- /dev/null
+++ b/October_2025/potd_20_10_2025.java
@@ -0,0 +1,44 @@
+class Solution {
+    public ArrayList<Integer> countBSTs(int[] arr) {
+        // Code here
+        
+        int[] cloned = arr.clone();
+        ArrayList<Integer> ans = new ArrayList<>();
+        HashMap<Integer,Integer> h = new HashMap<>();
+        
+        
+        Arrays.sort(arr);
+        int n = arr.length;
+        
+        // Step 1: find the catalan numbers
+        long[] catalan = new long[n+1];
+        catalan[0] = catalan[1] = 1;
+        
+        for(int i = 2;i<=n;i++){
+            catalan[i] = 0;
+            for(int j = 0; j<i;j++){
+                catalan[i] += (catalan[j] * catalan[i-j-1]);
+            }
+        }
+        
+        // Step 2:  Find unique bsts for each distinct element
+        
+        for(int i = 0;i < n; i++){
+            int L = i; // number elements in left subtree (elements are <= arr[i])
+            int R = n-i-1; // number of elements in right subtree (elements are > arr[i])
+            
+            long bstsPossible = catalan[L] * catalan[R];
+            // bstsPossible = number of valid left subtrees * number of valid right subtrees
+            
+            h.put(arr[i],(int) bstsPossible);
+        }
+        
+        // Step 3: Form the answer list
+        
+        for(int num: cloned){
+            ans.add(h.get(num));
+        }
+        
+        return ans;
+    }
+}