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.

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, …

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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

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

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

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

My Personal Notes

Nov	DEC	Jan
	06
2018	2019	2020

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python

C#

PHP

C++

Java

Python

C#

PHP

C++

Java

Python3

C#

PHP

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