Data Modelling & AI Data Structure & Algorithm

Program for nth Catalan Number

31 July 2024

1

Catalan numbers are defined as a mathematical sequence that consists of positive integers, which can be used to find the number of possibilities of various combinations.

The nth term in the sequence denoted C_n, is found in the following formula: $\frac{(2n)!}{(n + 1)! n!)}$

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

Catalan numbers occur in many interesting counting problems like the following.

Count the number of expressions containing n pairs of parentheses that are correctly matched. For n = 3, possible expressions are ((())), ()(()), ()()(), (())(), (()()).
Count the number of possible Binary Search Trees with n keys (See this)
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.
Given a number n, return the number of ways you can draw n chords in a circle with 2 x n points such that no 2 chords intersect.

See this for more applications.

Examples:

Input: n = 6
Output: 132

Input: n = 8
Output: 1430

Recursive Solution for Catalan number:

Catalan numbers satisfy the following recursive formula: $C_0=1 \ and \ C_{n+1}=\sum_{i=0}^{n}C_iC_{n-i} \ for \ n\geq 0$

Follow the steps below to implement the above recursive formula

Base condition for the recursive approach, when n <= 1, return 1
Iterate from i = 0 to i < n
- Make a recursive call catalan(i) and catalan(n – i – 1) and keep adding the product of both into res.
Return the res.

Following is the implementation of the 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 code
int main()
{
    for (int i = 0; i < 10; i++)
        cout << catalan(i) << " ";
    return 0;
}

Java

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

Python3

# 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 Code
for i in range(10):
    print(catalan(i), end=" ")
# This code is contributed by
# Nikhil Kumar Singh (nickzuck_007)

C#

// A recursive C# program to find
// nth catalan number
using System;
 
class GFG {
 
    // A recursive function to find
    // nth catalan number
    static 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;
    }
 
    // Driver Code
    public static void Main()
    {
        for (int i = 0; i < 10; i++)
            Console.Write(catalan(i) + " ");
    }
}
 
// This code is contributed by
// nitin mittal.

PHP

<?php
// PHP Program for nth 
// Catalan Number
 
// A recursive function to
// find nth catalan number
function catalan($n)
{
     
    // Base case
    if ($n <= 1)
        return 1;
 
    // catalan(n) is sum of 
    // catalan(i)*catalan(n-i-1)
    $res = 0;
    for($i = 0; $i < $n; $i++)
        $res += catalan($i) * 
                catalan($n - $i - 1);
 
    return $res;
}
 
// Driver Code
for ($i = 0; $i < 10; $i++)
    echo catalan($i), " ";
 
// This code is contributed aj_36
?>

Javascript

<script>
 
// Javascript Program for nth 
// Catalan Number
 
// A recursive function to
// find nth catalan number
function catalan(n)
{
     
    // Base case
    if (n <= 1)
        return 1;
 
    // catalan(n) is sum of 
    // catalan(i)*catalan(n-i-1)
    let res = 0;
    for(let i = 0; i < n; i++)
        res += catalan(i) * 
                catalan(n - i - 1);
 
    return res;
}
 
// Driver Code
for (let i = 0; i < 10; i++)
    document.write(catalan(i) + " ");
 
// This code is contributed _saurabh_jaiswal
 
</script>

Output

1 1 2 5 14 42 132 429 1430 4862

Time Complexity: The above implementation is equivalent to nth Catalan number.

$T(n)=\sum_{i=0}^{n-1}T(i)*T(n-i-1) \ for \ n\geq 1;$

The value of nth Catalan number is exponential which makes the time complexity exponential.
Auxiliary Space: O(n)

Dynamic Programming Solution for Catalan number:

We can observe that the above recursive implementation does a lot of repeated work. Since there are overlapping subproblems, we can use dynamic programming for this.

Below is the implementation of the above idea:

Create an array catalan[] for storing ith Catalan number.
Initialize, catalan[0] and catalan[1] = 1
Loop through i = 2 to the given Catalan number n.
- Loop through j = 0 to j < i and Keep adding value of catalan[j] * catalan[i – j – 1] into catalan[i].
Finally, return catalan[n]

Follow the steps below to implement the above approach:

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 code
int main()
{
    for (int i = 0; i < 10; i++)
        cout << catalanDP(i) << " ";
    return 0;
}

Java

import java.io.*;
class GFG {
 
    // A dynamic programming based function to find nth
    // Catalan number
    static int catalanDP(int n)
    {
        // Table to store results of subproblems
        int catalan[] = new int[n + 2];
 
        // Initialize first two values in table
        catalan[0] = 1;
        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 code
    public static void main(String[] args)
    {
        for (int i = 0; i < 10; i++) {
            System.out.print(catalanDP(i) + " ");
        }
    }
}
// This code contributed by Rajput-Ji

Python3

# 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]*(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):
        for j in range(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), end=" ")
# This code is contributed by Ediga_manisha

C#

using System;
 
class GFG {
 
    // A dynamic programming based
    // function to find nth
    // Catalan number
    static uint catalanDP(uint n)
    {
        // Table to store results of subproblems
        uint[] catalan = new uint[n + 2];
 
        // Initialize first two values in table
        catalan[0] = catalan[1] = 1;
 
        // Fill entries in catalan[]
        // using recursive formula
        for (uint i = 2; i <= n; i++) {
            catalan[i] = 0;
            for (uint j = 0; j < i; j++)
                catalan[i]
                    += catalan[j] * catalan[i - j - 1];
        }
 
        // Return last entry
        return catalan[n];
    }
 
    // Driver code
    static void Main()
    {
        for (uint i = 0; i < 10; i++)
            Console.Write(catalanDP(i) + " ");
    }
}
 
// This code is contributed by Chandan_jnu

PHP

<?php
// PHP program for nth Catalan Number
 
// A dynamic programming based function 
// to find nth Catalan number
function catalanDP( $n)
{
     
    // Table to store results 
    // of subproblems
    $catalan= array();
 
    // Initialize first two 
    // values in table
    $catalan[0] = $catalan[1] = 1;
 
    // Fill entries in catalan[] 
    // using recursive formula
    for ($i = 2; $i <= $n; $i++)
    {
        $catalan[$i] = 0;
        for ( $j = 0; $j < $i; $j++)
            $catalan[$i] += $catalan[$j] * 
                   $catalan[$i - $j - 1];
    }
 
    // Return last entry
    return $catalan[$n];
}
 
    // Driver Code
    for ($i = 0; $i < 10; $i++)
        echo catalanDP($i) , " ";
 
// This code is contributed anuj_67.
?>

Javascript

<script>
// Javascript program for nth Catalan Number
 
// A dynamic programming based function 
// to find nth Catalan number
function catalanDP(n)
{
     
    // Table to store results 
    // of subproblems
    let catalan= [];
 
    // Initialize first two 
    // values in table
    catalan[0] = catalan[1] = 1;
 
    // Fill entries in catalan[] 
    // using recursive formula
    for (let i = 2; i <= n; i++)
    {
        catalan[i] = 0;
        for (let j = 0; j < i; j++)
            catalan[i] += catalan[j] * 
                   catalan[i - j - 1];
    }
 
    // Return last entry
    return catalan[n];
}
 
    // Driver Code
    for (let i = 0; i < 10; i++)
        document.write(catalanDP(i) + " ");
 
// This code is contributed _saurabh_jaiswal
</script>

Output

1 1 2 5 14 42 132 429 1430 4862

Time Complexity: O(n²)
Auxiliary Space: O(n)

Binomial Coefficient Solution for Catalan number:

We can also use the below formula to find nth Catalan number in O(n) time.

$C_n=\frac{1}{n+1}\binom{2n}{n}$

Here is the proof for the Expression:–

This is the expression for which we are going to see the proof

In the pascal triangle,

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

Pascal’s triangle as binomial coefficients

the formula for a cell of Pascal’s triangle

Pascal’s triangle

.

Below are the steps for calculating nC_r.

Create a variable to store the answer and change r to n – r if r is greater than n – r because we know that C(n, r) = C(n, n-r) if r > n – r
Run a loop from 0 to r-1
- In every iteration update ans as (ans*(n-i))/(i+1), where i is the loop counter.
So the answer will be equal to ((n/1)*((n-1)/2)*…*((n-r+1)/r), which is equal to nC_r.

Below are steps to calculate Catalan numbers using the formula: 2nC_n/(n+1)

Calculate 2nC_n using the similar steps that we use to calculate nC_r
Return the value 2nC_n/ (n + 1)\

Below is the implementation of the above approach:

C++

// C++ program for nth Catalan Number
#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 code
int main()
{
    for (int i = 0; i < 10; i++)
        cout << catalan(i) << " ";
    return 0;
}

Java

// Java program for nth Catalan Number
 
class GFG {
 
    // Returns value of Binomial Coefficient C(n, k)
    static long binomialCoeff(int n, int k)
    {
        long 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
    static long catalan(int n)
    {
        // Calculate value of 2nCn
        long c = binomialCoeff(2 * n, n);
 
        // return 2nCn/(n+1)
        return c / (n + 1);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        for (int i = 0; i < 10; i++) {
            System.out.print(catalan(i) + " ");
        }
    }
}

Python3

# Python program for nth Catalan Number
# 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)
 
 
# Driver Code
for i in range(10):
    print(catalan(i), end=" ")
 
# This code is contributed by Aditi Sharma

C#

// C# program for nth Catalan Number
using System;
class GFG {
 
    // Returns value of Binomial Coefficient C(n, k)
    static long binomialCoeff(int n, int k)
    {
        long 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
    static long catalan(int n)
    {
        // Calculate value of 2nCn
        long c = binomialCoeff(2 * n, n);
 
        // return 2nCn/(n+1)
        return c / (n + 1);
    }
 
    // Driver code
    public static void Main()
    {
        for (int i = 0; i < 10; i++) {
            Console.Write(catalan(i) + " ");
        }
    }
}
 
// This code is contributed
// by Akanksha Rai

PHP

<?php
// PHP program for nth Catalan Number 
 
// Returns value of Binomial 
// Coefficient C(n, k) 
function binomialCoeff($n, $k) 
{ 
    $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 ($i = 0; $i < $k; ++$i) 
    { 
        $res *= ($n - $i); 
        $res = floor($res / ($i + 1)); 
    } 
 
    return $res; 
} 
 
// A Binomial coefficient based function 
// to find nth catalan number in O(n) time 
function catalan($n) 
{ 
    // Calculate value of 2nCn 
    $c = binomialCoeff(2 * ($n), $n); 
 
    // return 2nCn/(n+1) 
    return floor($c / ($n + 1)); 
} 
 
// Driver code 
for ($i = 0; $i < 10; $i++) 
echo catalan($i), " " ; 
 
// This code is contributed by Ryuga
?>

Javascript

<script>
// Javascript program for nth Catalan Number 
 
// Returns value of Binomial 
// Coefficient C(n, k) 
function binomialCoeff(n, k) 
{ 
    let 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 (let i = 0; i < k; ++i) 
    { 
        res *= (n - i); 
        res = Math.floor(res / (i + 1)); 
    } 
 
    return res; 
} 
 
// A Binomial coefficient based function 
// to find nth catalan number in O(n) time 
function catalan(n) 
{ 
 
    // Calculate value of 2nCn 
    c = binomialCoeff(2 * (n), n); 
 
    // return 2nCn/(n+1) 
    return Math.floor(c / (n + 1)); 
} 
 
// Driver code 
for (let i = 0; i < 10; i++) 
document.write(catalan(i) + " " ); 
 
// This code is contributed by _saurabh_jaiswal
</script>

Output

1 1 2 5 14 42 132 429 1430 4862

Time Complexity: O(n).
Auxiliary Space: O(1)

We can also use the below formulas to find nth Catalan number in O(n) time.

$C_n=\frac{(2n)!}{(n+1)!n!}=\prod_{k=2}^{n}\frac{n+k}{k} \ for \ n\geq 0$

$C_n = \frac{2(2n-1)}{n+1}*C_{n-1} \ \ \ {|} \ \ {n>0}$

Catalan number using the multi-Precision library:

In this method, we have used a boost multi-precision library, and the motive behind its use is just only to have precision meanwhile finding the large Catalan number and a generalized technique using for loop to calculate Catalan numbers.

Pseudocode:

a) initially set cat_=1 and print it
b) run a for loop i=1 to i<=n
            cat_ *= (4*i-2)
            cat_ /= (i+1)
            print cat_
c) end loop and exit

Below is the implementation using the multi-precision library:

C++

#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using boost::multiprecision::cpp_int;
using namespace std;
 
// Function to print the number
void catalan(int n)
{
    cpp_int cat_ = 1;
 
    // For the first number
    cout << cat_ << " "; // C(0)
 
    // Iterate till N
    for (cpp_int i = 1; i <= n; i++) {
        // Calculate the number
        // and print it
        cat_ *= (4 * i - 2);
        cat_ /= (i + 1);
        cout << cat_ << " ";
    }
}
 
// Driver code
int main()
{
    int n = 5;
 
    // Function call
    catalan(n);
    return 0;
}

Java

import java.util.*;
class GFG {
 
    // Function to print the number
    static void catalan(int n)
    {
        int cat_ = 1;
 
        // For the first number
        System.out.print(cat_ + " "); // C(0)
 
        // Iterate till N
        for (int i = 1; i < n; i++) {
            // Calculate the number
            // and print it
            cat_ *= (4 * i - 2);
            cat_ /= (i + 1);
            System.out.print(cat_ + " ");
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 5;
 
        // Function call
        catalan(n);
    }
}
 
// This code is contributed by Debojyoti Mandal

Python3

# Function to print the number
def catalan(n):
 
    cat_ = 1
 
    # For the first number
    print(cat_, " ", end='')  # C(0)
 
    # Iterate till N
    for i in range(1, n):
 
        # Calculate the number
        # and print it
        cat_ *= (4 * i - 2)
        cat_ //= (i + 1)
        print(cat_, " ", end='')
 
 
# Driver code
n = 5
 
# Function call
catalan(n)
 
# This code is contributed by rohan07

C#

using System;
 
public class GFG {
 
    // Function to print the number
    static void catalan(int n)
    {
        int cat_ = 1;
 
        // For the first number
        Console.Write(cat_ + " "); // C(0)
 
        // Iterate till N
        for (int i = 1; i < n; i++) {
            // Calculate the number
            // and print it
            cat_ *= (4 * i - 2);
            cat_ /= (i + 1);
            Console.Write(cat_ + " ");
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int n = 5;
 
        // Function call
        catalan(n);
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
 
// Function to print the number
function catalan(n)
{
    let cat_ = 1;
 
    // For the first number
    document.write(cat_ + " "); // C(0)
 
    // Iterate till N
    for (let i = 1; i < n; i++)
    {
        // Calculate the number
        // and print it
        cat_ *= (4 * i - 2);
        cat_ /= (i + 1);
        document.write(cat_ + " ");
    }
}
 
// Driver code
    let n = 5;
 
    // Function call
    catalan(n);
 
//This code is contributed by Mayank Tyagi
</script>

Output

1 1 2 5 14

Time Complexity: O(n)
Auxiliary Space: O(1), since no extra space has been taken.

Catalan number using BigInteger in java:

Finding values of Catalan numbers for N>80 is not possible even by using long in java, so we use BigInteger.

Follow the steps below for the implementation:

Create a BigInteger variable b and initialize it to 1.
Calculate n! and store it into b.
Calculate n! * n! and store into b.
Create another BigInteger variable d and initialize it to 1.
Calculate 2n! and store into d.
Calculate (2n)! / (n! * n!) into ans
Calculate ans / (n + 1) and return ans.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;
 
#define bigint long long int
 
bigint findCatalan(int n)
{
    bigint b = 1;
 
    // calculating n!
    for (int i = 1; i <= n; i++) {
        b = b * i;
    }
 
    // calculating n! * n!
    b = b * b;
 
    bigint d = 1;
 
    // calculating (2n)!
    for (int i = 1; i <= 2 * n; i++) {
        d = d * i;
    }
 
    // calculating (2n)! / (n! * n!)
    bigint ans = d / b;
 
    // calculating (2n)! / ((n! * n!) * (n+1))
    ans = ans / (n + 1);
    return ans;
}
 
// Driver Code
int main() {
    int n = 5;
    cout << findCatalan(n);
}
 
// This code is contributed by ajaymakvana.

Java

import java.io.*;
import java.math.*;
import java.util.*;
 
class GFG {
    public static BigInteger findCatalan(int n)
    {
        // using BigInteger to calculate large factorials
        BigInteger b = new BigInteger("1");
 
        // calculating n!
        for (int i = 1; i <= n; i++) {
            b = b.multiply(BigInteger.valueOf(i));
        }
 
        // calculating n! * n!
        b = b.multiply(b);
 
        BigInteger d = new BigInteger("1");
 
        // calculating (2n)!
        for (int i = 1; i <= 2 * n; i++) {
            d = d.multiply(BigInteger.valueOf(i));
        }
 
        // calculating (2n)! / (n! * n!)
        BigInteger ans = d.divide(b);
 
        // calculating (2n)! / ((n! * n!) * (n+1))
        ans = ans.divide(BigInteger.valueOf(n + 1));
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 5;
        System.out.println(findCatalan(n));
    }
}
// Contributed by Rohit Oberoi

Python3

def findCatalan(n):
    b = 1
 
    # calculating n!
    for i in range(1, n + 1, 1):
        b = b * i
 
    # calculating n! * n!
    b = b * b
    d = 1
 
    # calculating (2n)!
    for i in range(1, 2 * n + 1, 1):
        d = d * i
         
    # calculating (2n)! / (n! * n!)
    ans = d / b
     
    # calculating (2n)! / ((n! * n!) * (n+1))
    ans = ans / (n + 1)
 
    return ans
 
# Driver Code
n = 50
print(int(findCatalan(n)))
 
# This code is contributed by ajaymakavana.

C#

// C# code to implement the approach
using System;
using System.Numerics;
 
class GFG {
    public static BigInteger findCatalan(int n)
    {
        // using BigInteger to calculate large factorials
        BigInteger b = new BigInteger(1);
 
        // calculating n!
        for (int i = 1; i <= n; i++) {
            b = BigInteger.Multiply(b, new BigInteger(i));
        }
 
        // calculating n! * n!
        b = BigInteger.Multiply(b, b);
 
        BigInteger d = new BigInteger(1);
 
        // calculating (2n)!
        for (int i = 1; i <= 2 * n; i++) {
            d = BigInteger.Multiply(d, new BigInteger(i));
        }
 
        // calculating (2n)! / (n! * n!)
        BigInteger ans = BigInteger.Divide(d, b);
 
        // calculating (2n)! / ((n! * n!) * (n+1))
        ans = BigInteger.Divide(ans, new BigInteger(n + 1));
        return ans;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int n = 5;
        Console.WriteLine(findCatalan(n));
    }
}
 
// This code is contributed by phasing17.

Javascript

// JavaScript code to implement the approach
 
function findCatalan(n){
    let b = 1
     
    // calculating n!
    for(let i = 1; i <= n; i++){
        b = b * i;
    }
     
    // calculating n! * n!
    b = b * b;
    let d = 1;
     
    // calculating (2n)!
    for(let i = 1; i <= 2 * n; i++){
        d = d * i
    }
     
    // calculating (2n)! / (n! * n!)
    let ans = d/b;
     
    // calculating (2n)! / ((n! * n!) * (n+1))
    ans = ans / (n + 1);
     
    return ans;
}
 
let n = 5;
console.log(findCatalan(n));
 
// This code is contributed by lokeshmvs21.

Output

Time Complexity: O(n)
Auxiliary Space: O(1), since no extra space has been taken.

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Program for nth Catalan Number

Recursive Solution for Catalan number:

C++

Java

Python3

C#

PHP

Javascript

Dynamic Programming Solution for Catalan number:

C++

Java

Python3

C#

PHP

Javascript

Binomial Coefficient Solution for Catalan number:

C++

Java

Python3

C#

PHP

Javascript

Catalan number using the multi-Precision library:

C++

Java

Python3

C#

Javascript

Catalan number using BigInteger in java:

C++

Java

Python3

C#

Javascript

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