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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
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
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