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.

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.

catalan

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;
}

 

Output :

1 1 2 5 14 42 132 429 1430 4862

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

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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++ Program 
#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;
}

Output:

1 1 2 5 14 42 132 429 1430 4862

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

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Using Binomial Coefficient
We can also use the below formula to find nth catalan number in O(n) time.

C++ Program 
#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;
}

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.

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