Program for nth Catalan Number

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: 

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.

1. Count the number of expressions containing n pairs of parentheses that 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.
4. 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: 

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. 

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. 

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. 

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

42

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