# Program for nth Catalan Number

Last Updated : 22 Nov, 2023

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 Cn, is found in the following formula:

The first few Catalan numbers for n = 0, 1, 2, 3, â€¦ 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

## Program for nth Catalan Number using Recursion:

Catalan numbers satisfy the following recursive formula:

Step-by-step approach:

• 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 using namespace std; // A recursive function to find nth catalan numberunsigned 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 codeint 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 Codefor 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 numberusing 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.

## Javascript

 

## PHP

 

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)

## Program for nth Catalan Number using Dynamic Programming:

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.

Step-by-step approach:

• 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 using namespace std; // A dynamic programming based function to find nth// Catalan numberunsigned 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 codeint 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 codefor 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

## Javascript

 

## PHP

 

Output
1 1 2 5 14 42 132 429 1430 4862


Time Complexity: O(n2)
Auxiliary Space: O(n)

## Program for nth Catalan Number using Binomial Coefficient:

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 nCr.

• 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)*â€¦*((n-r+1)/r), which is equal to nCr.

Below are steps to calculate Catalan numbers using the formula: 2nCn/(n+1)

• Calculate 2nCn using the similar steps that we use to calculate nCr
• Return the value 2nCn/ (n + 1)

Below is the implementation of the above approach:

## C++

 // C++ program for nth Catalan Number#include 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) timeunsigned 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 codeint 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 Codefor i in range(10):    print(catalan(i), end=" ") # This code is contributed by Aditi Sharma

## C#

 // C# program for nth Catalan Numberusing 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

## Javascript

 

## PHP

  $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: O(n).
Auxiliary Space: O(1)

We can also use the below formulas to find nth Catalan number in O(n) time.

## Program for nth Catalan Number using the (n-1)th Catalan Number:

We already know how to calculate the nth Catalan Number using the below formula,

This formula can be further simplified to express the nth Catalan Number in the terms of (n-1)th Catalan Number,

Below are steps to calculate Catalan numbers using the above formula:

• Initialize a variable res = 1
• Print 1 as the first Catalan Number
• Iterate from i = 1 to i < n
• Update res with res = (res * (4 * i – 2)) / (i + 1)
• print res

Below is the implementation for the above approach:

## C++

 #include using namespace std; // Function to print first n Catalan numbersvoid catalan(int n){    int res = 1;    cout << res << " ";    // Iterate till n    for (int i = 1; i < n; i++) {        // Calculate the ith Catalan number        res = (res * (4 * i - 2)) / (i + 1);           cout << res << " ";    }} // Driver codeint main(){    int n = 10;     // Function call    catalan(n);    return 0;}

## Java

 import java.util.*;class GFG {     // Function to print first n Catalan numbers    static void catalan(int n)    {        int res = 1;        System.out.print(1 + " ");        // Iterate till N        for (int i = 1; i < n; i++) {            // Calculate the ith Catalan Number            res = (res * (4 * i - 2)) / (i + 1);            System.out.print(res + " ");        }    }     // Driver code    public static void main(String args[])    {        int n = 10;         // Function call        catalan(n);    }} // This code is contributed by Debojyoti Mandal

## Python3

 # Function to print first n Catalan numbersdef catalan(n):     res = 1     print(1, end=" ")    # Iterate till N    for i in range(1, n):         # Calculate the ith Catalan number        res = (res * (4 * i - 2)) // (i + 1)        print(res, end=" ")  # Driver coden = 10 # Function callcatalan(n) # This code is contributed by rohan07

## C#

 using System; public class GFG {     // Function to print first n Catalan numbers    static void catalan(int n)    {        int res = 1;         Console.Write(1 + " ");        // Iterate till N        for (int i = 1; i < n; i++) {            // Calculate the ith Catalana Number            res = (res * (4 * i - 2)) / (i + 1);            Console.Write(res + " ");        }    }     // Driver code    public static void Main(String[] args)    {        int n = 10;        // Function call        catalan(n);    }} // This code is contributed by Rajput-Ji

## Javascript

 

Output
1 1 2 5 14 42 132 429 1430 4862


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

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