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Find the number of valid parentheses expressions of given length
  • Difficulty Level : Medium
  • Last Updated : 01 Apr, 2021

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
#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
 
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

Python3




# Python3 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/
 
# Returns value of Binomial
# Coefficient C(n, k)
def 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 in range(k):
        res *= (n - i);
        res /= (i + 1);
 
    return int(res);
 
# A Binomial coefficient based
# function to find nth catalan 
# number in O(n) time
def catalan(n):
     
    # Calculate value of 2nCn
    c = binomialCoeff(2 * n, n);
 
    # return 2nCn/(n+1)
    return int(c / (n + 1));
 
# Function to find possible
# ways to put balanced parenthesis
# in an expression of length n
def findWays(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(int(n / 2));
 
# Driver Code
n = 6;
print("Total possible expressions of length",
                       n, "is", findWays(6));
     
# This code is contributed by mits

C#




// C# program to find valid paranthesisations
// of length n The majority of code is taken
// from method 3 of
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.

PHP




<?php
// PHP program to find valid
// paranthesisations of length n
// The majority of code is taken
// from method 3 of
 
// 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);
        $res /= ($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 $c / ($n + 1);
}
 
// Function to find possible
// ways to put balanced
// parenthesis in an expression
// of length n
function findWays($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 Code
    $n = 6;
    echo "Total possible expressions of length "
                    , $n , " is " , findWays(6);
     
// This code is contributed by nitin mittal
?>

Javascript




<script>
// Javascript program to find valid
// paranthesisations of length n
// The majority of code is taken
// from method 3 of
 
// Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeff(n, k)
{
    let 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 (let 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
function catalan(n)
{
     
    // Calculate value of 2nCn
    let 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
function findWays(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 Code
    let n = 6;
    document.write("Total possible expressions of length " +
                   n + " is " + findWays(6));
     
// This code is contributed by _saurabh_jaiswal
</script>

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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