Find the number of valid parentheses expressions of given length

Given a number n find the number of valid parentheses expressions of that length.
Examples :

Input: 2
Output: 1 
There is only possible valid expression of length 2, "()"

Input: 4
Output: 2 
Possible valid expression of length 4 are "(())" and "()()" 

Input: 6
Output: 5
Possible valid expressions are ((())), ()(()), ()()(), (())() and (()())

This is mainly an application of Catalan Numbers. Total possible valid expressions for input n is n/2’th Catalan Number if n is even and 0 if n is odd.



Below given is the implementation :

C++

// C++ program to find valid paranthesisations of length n
// The majority of code is taken from method 3 of
// https:// www.geeksforgeeks.org/program-nth-catalan-number/
#include <bits/stdc++.h>
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);
}

// Function to find possible ways to put balanced
// parenthesis in an expression of length n
unsigned long int findWays(unsigned n)
{
    // If n is odd, not possible to 
    // create any valid parentheses
    if (n & 1)
        return 0;

    // Otherwise return n/2'th Catalan Numer
    return catalan(n / 2);
}

// Driver program to test above functions
int main()
{
    int n = 6;
    cout << "Total possible expressions of length "
         << n << " is " << findWays(6);
    return 0;
}

Java

// Java program to find valid paranthesisations of length n
// The majority of code is taken from method 3 of
// https:// www.geeksforgeeks.org/program-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);
    }

    // Function to find possible ways to put balanced
    // parenthesis in an expression of length n
    static long findWays(int n)
    {
        // If n is odd, not possible to
        // create any valid parentheses
        if ((n & 1) != 0)
            return 0;

        // Otherwise return n/2'th Catalan Numer
        return catalan(n / 2);
    }

    // Driver program to test above functions
    public static void main(String[] args)
    {
        int n = 6;
        System.out.println("Total possible expressions of length " + 
                                          n + " is " + findWays(6));
    }
}

// This code is contributed by Smitha Dinesh Semwal

C#

// C# program to find valid paranthesisations
// of length n The majority of code is taken
// from method 3 of https:// www.geeksforgeeks
// .org/program-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);
    }

    // Function to find possible ways to put 
    // balanced parenthesis in an expression
    // of length n
    static long findWays(int n)
    {
        // If n is odd, not possible to
        // create any valid parentheses
        if ((n & 1) != 0)
            return 0;

        // Otherwise return n/2'th
        // Catalan Numer
        return catalan(n / 2);
    }

    // Driver program to test 
    // above functions
    public static void Main()
    {
        int n = 6;
        Console.Write("Total possible expressions"
                       + "of length " + n + " is " 
                                   + findWays(6));
    }
}

// This code is contributed by nitin mittal.


Output:
Total possible expressions of length 6 is 5

Time Complexity: O(n)

This article is contributed by Sachin. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above




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