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.
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.
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)

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;
    }
 
    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
?>

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 program to test above function
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

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 (10):
    print (catalan(i),end=" ")
# This code is contributed by Aditi Sharma

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.
?>

Output

1 1 2 5 14 42 132 429 1430 4862

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 program to test above functions
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 program to test above function 
    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)
 
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 program to test above function 
    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
?>

Output

1 1 2 5 14 42 132 429 1430 4862

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;
}

Output

1 1 2 5 14

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

we hahttps://youtu.be/2NZF2UKyh0g
https://youtu.be/MHbHYCJxKmA
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

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