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)

3) Count the number of full binary trees (A rooted binary tree is full if every vertex has either two children or no children) with n+1 leaves.

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, …



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.

C++

#include<iostream>
using namespace std;

// A recursive function to find nth catalan number
unsigned long int catalan(unsigned int n)
{
    // Base case
    if (n <= 1) return 1;

    // catalan(n) is sum of catalan(i)*catalan(n-i-1)
    unsigned long int res = 0;
    for (int i=0; i<n; i++)
        res += catalan(i)*catalan(n-i-1);

    return res;
}

// Driver program to test above function
int main()
{
    for (int i=0; i<10; i++)
        cout << catalan(i) << " ";
    return 0;
}

Java


class CatalnNumber {

    // A recursive function to find nth catalan number

    int catalan(int n) {
        int res = 0;
        
        // Base case
        if (n <= 1) {
            return 1;
        }
        for (int i = 0; i < n; i++) {
            res += catalan(i) * catalan(n - i - 1);
        }
        return res;
    }

    public static void main(String[] args) {
        CatalnNumber cn = new CatalnNumber();
        for (int i = 0; i < 10; i++) {
            System.out.print(cn.catalan(i) + " ");
        }
    }
}

Python

# 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)


Output : 
1 1 2 5 14 42 132 429 1430 4862

Time complexity of above implementation is equivalent to nth catalan number.
T(n)=\sum_{i=0}^{n-1}T(i)*T(n-i) \ for \ n\geq 0;

The value of nth catalan number is exponential that makes the time complexity exponential.



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++.

C++

#include<iostream>
using namespace std;

// A dynamic programming based function to find nth
// Catalan number
unsigned long int catalanDP(unsigned int n)
{
    // Table to store results of subproblems
    unsigned long int catalan[n+1];

    // Initialize first two values in table
    catalan[0] = catalan[1] = 1;

    // Fill entries in catalan[] using recursive formula
    for (int i=2; i<=n; i++)
    {
        catalan[i] = 0;
        for (int j=0; j<i; j++)
            catalan[i] += catalan[j] * catalan[i-j-1];
    }

    // Return last entry
    return catalan[n];
}

// Driver program to test above function
int main()
{
    for (int i = 0; i < 10; i++)
        cout << catalanDP(i) << " ";
    return 0;
}

Python

# 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 (11):
    print (catalanDP(i))
# This code is contributed by Aditi Sharma


Output: 
1 1 2 5 14 42 132 429 1430 4862

Time Complexity: Time complexity of above implementation is O(n2)



Using Binomial Coefficient
We can also use the below formula to find nth catalan number in O(n) time.
C_n=\frac{1}{n+1}\binom{2n}{n}

We have discussed a O(n) approach to find binomial coefficient nCr.

C++

#include<iostream>
using namespace std;

// Returns value of Binomial Coefficient C(n, k)
unsigned long int binomialCoeff(unsigned int n, unsigned int k)
{
    unsigned long int res = 1;

    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;

    // Calculate value of [n*(n-1)*---*(n-k+1)] / [k*(k-1)*---*1]
    for (int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }

    return res;
}

// A Binomial coefficient based function to find nth catalan
// number in O(n) time
unsigned long int catalan(unsigned int n)
{
    // Calculate value of 2nCn
    unsigned long int c = binomialCoeff(2*n, n);

    // return 2nCn/(n+1)
    return c/(n+1);
}

// Driver program to test above functions
int main()
{
    for (int i = 0; i < 10; i++)
        cout << catalan(i) << " ";
    return 0;
}

Python

# Returns value of Binomial Coefficient C(n, k)
def binomialCoefficient(n, k):

    # since C(n, k) = C(n, n - k)
    if (k > n - k):
        k = n - k

    # initialize result
    res = 1

    # Calculate value of [n * (n-1) *---* (n-k + 1)]
    # / [k * (k-1) *----* 1]
    for i in range(k):
        res = res * (n - i)
        res = res / (i + 1)
    return res

# A Binomial coefficient based function to
# find nth catalan number in O(n) time
def catalan(n):
    c = binomialCoefficient(2*n, n)
    return c/(n + 1)

for i in range (11):
    print (catalan(i))

# This code is contributed by Aditi Sharma


Output: 
1 1 2 5 14 42 132 429 1430 4862

Time Complexity: Time complexity of above implementation is O(n).

We can also use below formula to find nth catalan number in O(n) time.
C_n=\frac{(2n)!}{(n+1)!n!}=\prod_{k=2}^{n}\frac{n+k}{k} \ for \ n\geq 0

References:
http://en.wikipedia.org/wiki/Catalan_number

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