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 :
pre>1 1 2 5 14 42 132 429 1430 4862

Time complexity of above implementation is equivalent to nth catalan number.
$T(n)=\sum_{i=0}^{n-1}T(i)*T(n-i) \ for \ n\geq 0;$

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 in C++.

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; 
} 

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 

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) + " "); 
        } 
  
    } 
} 

Python

#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 

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

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

C++

Java

Python

C#

PHP

C++

Python

PHP

C++

Java

Python

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