Python Program for nth Catalan Number

Catalan numbers are a sequence of natural numbers that occurs in many interesting counting problems like following.

1) Count the number of expressions containing n pairs of parentheses which are correctly matched. For n = 3, possible expressions are ((())), ()(()), ()()(), (())(), (()()).

2) Count the number of possible Binary Search Trees with n keys (See this)
See this for more applications.



The first few Catalan numbers for n = 0, 1, 2, 3, … are 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, …

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Recursive Solution
Catalan numbers satisfy the following recursive formula.
C_0=1 \ and \ C_n_+_1=\sum_{i=0}^{n}C_iC_n_-_i \ for \ n\geq 0;
Following is the implementation of above recursive formula.

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# A recursive function to find nth catalan number
def catalan(n):
    # Base Case
    if n <= 1 :
        return 1 
  
    # Catalan(n) is the sum of catalan(i)*catalan(n-i-1)
    res = 0 
    for i in range(n):
        res += catalan(i) * catalan(n-i-1)
  
    return res
  
# Driver Program to test above function
for i in range(10):
    print catalan(i),
# This code is contributed by Nikhil Kumar Singh (nickzuck_007)

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Output:

1 1 2 5 14 42 132 429 1430 4862

Dynamic Programming Solution
We can observe that the above recursive implementation does a lot of repeated work (we can the same by drawing recursion tree). Since there are overlapping subproblems, we can use dynamic programming for this. Following is a Dynamic programming based implementation in C++.

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# A dynamic programming based function to find nth
# Catalan number
def catalan(n):
    if (n == 0 or n == 1):
        return 1
  
    # Table to store results of subproblems
    catalan = [0 for i in range(n + 1)]
  
    # Initialize first two values in table
    catalan[0] = 1
    catalan[1] = 1
  
    # Fill entries in catalan[] using recursive formula
    for i in range(2, n + 1):
        catalan[i] = 0
        for j in range(i):
            catalan[i] = catalan[i] + catalan[j] * catalan[i-j-1]
  
    # Return last entry
    return catalan[n]
  
# Driver code
for i in range (10):
    print (catalan(i))
# This code is contributed by Aditi Sharma

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Output:

1
1
2
5
14
42
132
429
1430
4862

Please refer complete article on Program for nth Catalan Number for more details!




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