# Number of Unique BST with a given key | Dynamic Programming

Given N, Find the total number of unique BSTs that can be made using values from 1 to N.

Examples:

```Input: n = 3
Output: 5
For n = 3, preorder traversal of Unique BSTs are:
1. 1 2 3
2. 1 3 2
3. 2 1 3
4. 3 1 2
5. 3 2 1

Input: 4
Output: 14
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

In the previous post a O(n) solution has been discussed. In this post we will discuss a solution based on Dynamic Programming. For all possible values of i, consider i as root, then [1….i-1] numbers will fall in the left subtree and [i+1….n] numbers will fall in the right subtree. So, add (i-1)*(n-i) to the answer. The summation of the products will be the answer to the number of unique BST.

Below is the implementation for above approach:

## C++

 `// C++ code to find number of unique BSTs ` `// Dynamic Programming solution ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find number of unique BST ` `int` `numberOfBST(``int` `n) ` `{ ` ` `  `    ``// DP to store the number of unique BST with key i ` `    ``int` `dp[n + 1]; ` `    ``fill_n(dp, n + 1, 0); ` ` `  `    ``// Base case ` `    ``dp = 1; ` `    ``dp = 1; ` ` `  `    ``// fill the dp table in top-down approach. ` `    ``for` `(``int` `i = 2; i <= n; i++) { ` `        ``for` `(``int` `j = 1; j <= i; j++) { ` ` `  `            ``// n-i in right * i-1 in left ` `            ``dp[i] = dp[i] + (dp[i - j] * dp[j - 1]); ` `        ``} ` `    ``} ` ` `  `    ``return` `dp[n]; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` `    ``cout << ``"Number of structurally Unique BST with "` `<<  ` `    ``n << ``" keys are : "` `<< numberOfBST(n) << ``"\n"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java code to find number ` `// of unique BSTs Dynamic  ` `// Programming solution ` `import` `java.io.*; ` `import` `java.util.Arrays; ` ` `  `class` `GFG ` `{ ` `    ``static` `int` `numberOfBST(``int` `n) ` `    ``{ ` ` `  `    ``// DP to store the number  ` `    ``// of unique BST with key i ` `    ``int` `dp[] = ``new` `int``[n + ``1``]; ` `    ``Arrays.fill(dp, ``0``); ` ` `  `    ``// Base case ` `    ``dp[``0``] = ``1``; ` `    ``dp[``1``] = ``1``; ` ` `  `    ``// fill the dp table in ` `    ``// top-down approach. ` `    ``for` `(``int` `i = ``2``; i <= n; i++)  ` `    ``{ ` `        ``for` `(``int` `j = ``1``; j <= i; j++)  ` `        ``{ ` ` `  `            ``// n-i in right * i-1 in left ` `            ``dp[i] = dp[i] + (dp[i - j] *  ` `                             ``dp[j - ``1``]); ` `        ``} ` `    ``} ` ` `  `    ``return` `dp[n]; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `n = ``3``; ` `    ``System.out.println(``"Number of structurally "` `+  ` `                           ``"Unique BST with "``+ n + ` `                                  ``" keys are : "` `+  ` `                                  ``numberOfBST(n)); ` `} ` `} ` ` `  `// This code is contributed ` `// by shiv_bhakt. `

## Python3

 `# Python3 code to find number of unique  ` `# BSTs Dynamic Programming solution  ` ` `  `# Function to find number of unique BST  ` `def` `numberOfBST(n):  ` ` `  `    ``# DP to store the number of unique ` `    ``# BST with key i  ` `    ``dp ``=` `[``0``] ``*` `(n ``+` `1``)  ` ` `  `    ``# Base case  ` `    ``dp[``0``], dp[``1``] ``=` `1``, ``1` ` `  `    ``# fill the dp table in top-down  ` `    ``# approach.  ` `    ``for` `i ``in` `range``(``2``, n ``+` `1``):  ` `        ``for` `j ``in` `range``(``1``, i ``+` `1``):  ` ` `  `            ``# n-i in right * i-1 in left  ` `            ``dp[i] ``=` `dp[i] ``+` `(dp[i ``-` `j] ``*` `                             ``dp[j ``-` `1``])  ` ` `  `    ``return` `dp[n]  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"``:  ` ` `  `    ``n ``=` `3` `    ``print``(``"Number of structurally Unique BST with"``,  ` `                   ``n, ``"keys are :"``, numberOfBST(n))  ` ` `  `# This code is contributed  ` `# by Rituraj Jain `

## C#

 `// C# code to find number ` `// of unique BSTs Dynamic  ` `// Programming solution ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``static` `int` `numberOfBST(``int` `n) ` `    ``{ ` ` `  `    ``// DP to store the number  ` `    ``// of unique BST with key i ` `    ``int` `[]dp = ``new` `int``[n + 1]; ` `     `  `    ``// Base case ` `    ``dp = 1; ` `    ``dp = 1; ` ` `  `    ``// fill the dp table in ` `    ``// top-down approach. ` `    ``for` `(``int` `i = 2; i <= n; i++)  ` `    ``{ ` `        ``for` `(``int` `j = 1; j <= i; j++)  ` `        ``{ ` ` `  `            ``// n-i in right * i-1 in left ` `            ``dp[i] = dp[i] + (dp[i - j] *  ` `                             ``dp[j - 1]); ` `        ``} ` `    ``} ` `    ``return` `dp[n]; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main ()  ` `{ ` `    ``int` `n = 3; ` `    ``Console.Write(``"Number of structurally "` `+  ` `                      ``"Unique BST with "``+ n + ` `                             ``" keys are : "` `+  ` `                             ``numberOfBST(n)); ` `} ` `} ` ` `  `// This code is contributed ` `// by shiv_bhakt. `

## PHP

 ` `

Output:

```Number of structurally Unique BST with 3 keys are : 5
```

Time Complexity: O(n2)

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