Program for nth Catalan Number

Difficulty Level : Medium
Last Updated : 14 Apr, 2021

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

Count the number of expressions containing n pairs of parentheses which 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.
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 code
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;
    }
 
    // 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) + " ");
        }
    }
}

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 Code
for i in range(10):
    print catalan(i),
# 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 of 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 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 .

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

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>
// Javscript 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: 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++

// 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: Time complexity of above implementation is O(n).
We can also use below formula to find nth catalan number in O(n) time.

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

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

For example: N = 5
Initially set cat_=1 then, print cat_ ,
then, iterate from i = 1 to i < 5
for i = 1; cat_ = cat_ * (4*1-2)=1*2=2
cat_ = cat_ / (i+1)=2/2 = 1
For i = 2; cat_ = cat_ * (4*2-2)=1*6=6
cat_ = cat_ / (i+1)=6/3=2
For i = 3 :- cat_ = cat_ * (4*3-2)=2*10=20
cat_ = cat_ / (i+1)=20/4=5
For i = 4 :- cat_ = cat_ * (4*4-2)=5*14=70
cat_ = cat_ / (i+1)=70/5=14

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

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

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)

Another solution 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
Here we find solution using Binomial Coefficient method as discussed above

Java

import java.io.*;
import java.util.*;
import java.math.*;
 
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

Output

References:
http://en.wikipedia.org/wiki/Catalan_number
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

My Personal Notes

Related Articles

C++

Java

Python

C#

PHP

Javascript

C++

Java

Python3

C#

PHP

Javascript

C++

Java

Python3

C#

PHP

Javascript

C++

Java

Python3

Javascript

Java