Consider a n x n grid with indexes of top left corner as (0, 0). Dyck path is a staircase walk from bottom left, i.e., (n-1, 0) to top right, i.e., (0, n-1) that lies above the diagonal cells (or cells on line from bottom left to top right).
The task is to count the number of Dyck Paths from (n-1, 0) to (0, n-1).
Examples :Â
Input : n = 1 Output : 1 Input : n = 2 Output : 2 Input : n = 3 Output : 5 Input : n = 4 Output : 14
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The number of Dyck paths from (n-1, 0) to (0, n-1) can be given by the Catalan numberC(n).
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Below are the implementations to find count of Dyck Paths (or n’th Catalan number).
C++
// C++ program to count // number of Dyck Paths#include<iostream>using namespace std;Â
// Returns count Dyck // paths in n x n gridint countDyckPaths(unsigned int n){    // Compute value of 2nCn    int res = 1;    for (int i = 0; i < n; ++i)    {        res *= (2 * n - i);        res /= (i + 1);    }Â
    // return 2nCn/(n+1)    return res / (n+1);}Â
// Driver Codeint main(){Â Â Â Â int n = 4;Â Â Â Â cout << "Number of Dyck Paths is "Â Â Â Â Â Â Â Â Â << countDyckPaths(n);Â Â Â Â return 0;} |
Java
// Java program to count// number of Dyck Pathsclass GFG{    // Returns count Dyck     // paths in n x n grid    public static int countDyckPaths(int n)    {        // Compute value of 2nCn        int res = 1;        for (int i = 0; i < n; ++i)        {            res *= (2 * n - i);            res /= (i + 1);        }Â
        // return 2nCn/(n+1)        return res / (n + 1);    }Â
    // Driver code    public static void main(String args[])    {        int n = 4;        System.out.println("Number of Dyck Paths is " +                                    countDyckPaths(n));    }} |
Python3
# Python3 program to count # number of Dyck PathsÂ
# Returns count Dyck # paths in n x n griddef countDyckPaths(n):         # Compute value of 2nCn    res = 1    for i in range(0, n):        res *= (2 * n - i)        res /= (i + 1)Â
    # return 2nCn/(n+1)    return res / (n+1)Â
# Driver Coden = 4print("Number of Dyck Paths is ",    str(int(countDyckPaths(n))))Â
# This code is contributed by# Prasad Kshirsagar |
Javascript
<script>Â
// JavaScript program to count// number of Dyck PathsÂ
    // Returns count Dyck     // paths in n x n grid    function countDyckPaths(n)    {             // Compute value of 2nCn        let res = 1;        for (let i = 0; i < n; ++i)        {            res *= (2 * n - i);            res /= (i + 1);        }           // return 2nCn/(n+1)        return res / (n + 1);    }Â
// Driver CodeÂ
        let n = 4;        document.write("Number of Dyck Paths is " +                                    countDyckPaths(n));         // This code is contributed by target_2.</script> |
C#
// C# program to count// number of Dyck Pathsusing System;Â
class GFG {         // Returns count Dyck     // paths in n x n grid    static int countDyckPaths(int n)    {                 // Compute value of 2nCn        int res = 1;        for (int i = 0; i < n; ++i)        {            res *= (2 * n - i);            res /= (i + 1);        }Â
        // return 2nCn/(n+1)        return res / (n + 1);    }Â
    // Driver code    public static void Main()    {        int n = 4;        Console.WriteLine("Number of "                  + "Dyck Paths is " +                   countDyckPaths(n));    }}Â
// This code is contributed by anuj_67. |
PHP
<?php// PHP program to count // number of Dyck PathsÂ
// Returns count Dyck // paths in n x n gridfunction countDyckPaths( $n){    // Compute value of 2nCn    $res = 1;    for ( $i = 0; $i < $n; ++$i)    {        $res *= (2 * $n - $i);        $res /= ($i + 1);    }Â
    // return 2nCn/(n+1)    return $res / ($n + 1);}Â
// Driver Code$n = 4;echo "Number of Dyck Paths is " , Â Â Â Â Â Â Â Â Â Â Â Â Â Â countDyckPaths($n);Â
// This code is contributed by anuj_67.?> |
Output
Number of Dyck Paths is 14
Time complexity: O(n).
Auxiliary space: O(1).
Exercise :Â Â
- Find number of sequences of 1 and -1 such that every sequence follows below constraints :Â
a) The length of a sequence is 2nÂ
b) There are equal number of 1’s and -1’s, i.e., n 1’s, n -1sÂ
c) Sum of prefix of every sequence is greater than or equal to 0. For example, 1, -1, 1, -1 and 1, 1, -1, -1 are valid, but -1, -1, 1, 1 is not valid. - Number of paths of length m + n from (m-1, 0) to (0, n-1) that are restricted to east and north steps.
Approach 2:-approach to count the number of Dyck paths –In this implementation, we generate all possible Dyck paths of length n by generating all binary numbers with n bits. We then traverse through each bit in the binary representation of the number and update the depth accordingly. If at any point the depth becomes negative, then the path is not a Dyck path, so we break out of the loop. If we reach the end of the path and the depth is zero, then the path is a Dyck path, so we increment the count. Finally, we return the count of Dyck paths.
C++
#include <iostream>Â
using namespace std;Â
// Function to calculate the factorial of a given numberint factorial(int n) {Â Â Â Â int fact = 1;Â Â Â Â for (int i = 1; i <= n; i++) {Â Â Â Â Â Â Â Â fact *= i;Â Â Â Â }Â Â Â Â return fact;}Â
// Function to calculate the number of Dyck paths of length n using the 2 approachint dyck_paths_2(int n) {Â Â Â Â int numerator = factorial(2 * n);Â Â Â Â int denominator = factorial(n + 1) * factorial(n);Â Â Â Â return numerator / denominator;}Â
int main() {Â Â Â Â int n = 4;Â
    cout << "Number of Dyck paths is " << n << ": " << dyck_paths_2(n) << endl;Â
    return 0;} |
Java
import java.util.*;Â
public class DyckPaths {Â
  // Function to calculate the factorial of a given number  public static int factorial(int n)  {    int fact = 1;    for (int i = 1; i <= n; i++) {      fact *= i;    }    return fact;  }Â
  // Function to calculate the number of Dyck paths of  // length n using the 2 approach  public static int dyck_paths_2(int n)  {    int numerator = factorial(2 * n);    int denominator = factorial(n + 1) * factorial(n);    return numerator / denominator;  }Â
  public static void main(String[] args)  {    int n = 4;Â
    System.out.println("Number of Dyck paths is " + n                       + ": " + dyck_paths_2(n));  }}Â
// This code is contributed by Prajwal Kandekar |
Python3
# Function to calculate the factorial of a given numberdef factorial(n):    fact = 1    for i in range(1, n + 1):        fact *= i    return factÂ
# Function to calculate the number of Dyck paths of length n using the 2 approachdef dyck_paths_2(n):Â Â Â Â numerator = factorial(2 * n)Â Â Â Â denominator = factorial(n + 1) * factorial(n)Â Â Â Â return numerator // denominatorÂ
if __name__ == '__main__':Â Â Â Â n = 4Â Â Â Â print("Number of Dyck paths is {}: {}".format(n, dyck_paths_2(n))) |
Javascript
function factorial(n) {Â Â let fact = 1;Â Â for (let i = 1; i <= n; i++) {Â Â Â Â fact *= i;Â Â }Â Â return fact;}Â
function dyckPaths2(n) {Â Â const numerator = factorial(2 * n);Â Â const denominator = factorial(n + 1) * factorial(n);Â Â return numerator / denominator;}Â
const n = 4;console.log(`Number of Dyck paths is ${n}: ${dyckPaths2(n)}`); |
C#
using System;Â
class Program {    // Function to calculate the factorial of a given number    static int Factorial(int n)    {        int fact = 1;        for (int i = 1; i <= n; i++) {            fact *= i;        }        return fact;    }Â
    // Function to calculate the number of Dyck paths of    // length n using the 2 approach    static int DyckPaths2(int n)    {        int numerator = Factorial(2 * n);        int denominator = Factorial(n + 1) * Factorial(n);        return numerator / denominator;    }Â
    static void Main(string[] args)    {        int n = 4;Â
        Console.WriteLine("Number of Dyck paths is " + n                          + ": " + DyckPaths2(n));    }} |
Output
Number of Dyck paths is 4: 14
Time complexity: O(n).
Auxiliary space: O(1).
This article is contributed by Aditya Chatterjee. If you like neveropen and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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