Merge pull request #669 from heman1/NewAlgo

added allPossibleBST program

Author: probot-auto-merge[bot]

Dynamic Programming/Count_all_possible_BST/c++/count_all_BSTs.cpp

@@ -0,0 +1,59 @@
+/* Given an integer N, find and return the count of unique Binary search trees (BSTs)
+   possible with nodes valued from 1 to N.
+   Output count can be very large, so return the count modulo 10^9+7.
+   Contraints : 1<= N <=1000
+   Time Complexity: O(n^2)
+*/
+
+
+/* Recurssive solution: 
+   BSTcount(for value n) = summationOf ( BSTcount( for node [k-1]) + BSTcount( for node [n-k]) );
+          (k)
+          / \
+        /    \
+      /       \
+(1 to k-1)  (k+1 to n)
+
+*/
+
+/* Input: 3
+   Output: 5
+   
+   Input: 8
+   Output: 1430
+*/			  	  				 	   			   
+   			  	  				 	   			   
+#include <bits/stdc++.h>
+#define mm 1000000007
+using namespace std;
+
+int countBST(int n, int *dp) {
+    //base condition
+    if(n==0 || n==1)
+        return 1;
+    if(n==2)
+        return 2;
+    //if dp[n] has some value, then answer is already calculated for that n.
+    if(dp[n]!=0) {
+        return dp[n];
+    }
+    //calculating sum for every value of n.
+    long long sum=0;
+    for(int k=1; k<=n; k++) {
+        sum = (sum%mm + (countBST(k-1, dp)%mm * countBST(n-k, dp))%mm)%mm;
+    }
+    //save the answer for n in the dp array.
+    dp[n] = sum;
+    return sum;
+}
+
+
+int main() {
+	int n ;
+	cin>>n;
+	//dp array to fill in the values which are calculated (memoization).
+	int *dp = new int[n+1] {0};
+	cout<<countBST(n, dp);
+	return 0;
+}
+