Program to count number of unique binary search tree can be formed with 0 to n values in Python

Suppose we have one number n, we have to find the number of unique BSTs we can generate with numbers from [0, n). If the answer is very large mod the result by 10^9+7

So, if the input is like n = 3, then the output will be 5

To solve this, we will follow these steps −

• numer := 1
• denom := n + 1
• for i in range 1 to n, do
  • numer := numer * n + i
  • numer := numer mod m
  • denom := denom * i
  • denom := denom mod m
• numer := numer * (denom^(m-2)) mod m
• return numer mod m

Let us see the following implementation to get better understanding −

Example

 Live Demo

class Solution:
   def solve(self, n):
      m = 10 ** 9 + 7
      numer = 1
      denom = n + 1
      for i in range(1, n + 1):
         numer *= n + i
         numer %= m
         denom *= i
         denom %= m
         numer *= pow(denom, m-2, m)
      return numer % m
ob = Solution()
print(ob.solve(4))

Input

4

Output

14