Dyck path

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 :

Input : n = 1
Output : 1

Input : n = 2
Output : 2

Input : n = 3
Output : 5

Input : n = 4
Output : 14

dyckpaths

The number of Dyck paths from (n-1, 0) to (0, n-1) can be given by the Catalan numberC(n).



C_n=\frac{(2n)!}{(n+1)!n1}=\prod_{k=2}^{n}\frac{n+k}{k} \ for\ n\geq 0

We strongly recommend that you click here and practice it, before moving on to the solution.


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;
  
// Returns count Dyck 
// paths in n x n grid
int 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);
    }
  
    // return 2nCn/(n+1)
    return res / (n+1);
}
  
// Driver Code
int main()
{
    int n = 4;
    cout << "Number of Dyck Paths is " 
         << countDyckPaths(n);
    return 0;
}

Java

// Java program to count
// number of Dyck Paths
class 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);
        }
  
        // return 2nCn/(n+1)
        return res / (n + 1);
    }
  
    // 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 Paths
  
# Returns count Dyck 
# paths in n x n grid
def countDyckPaths(n):
      
    # Compute value of 2nCn
    res = 1
    for i in range(0, n):
        res *= (2 * n - i)
        res /= (i + 1)
  
    # return 2nCn/(n+1)
    return res / (n+1)
  
# Driver Code
n = 4
print("Number of Dyck Paths is ",
    str(int(countDyckPaths(n))))
  
# This code is contributed by
# Prasad Kshirsagar

C#

// C# program to count
// number of Dyck Paths
using System;
  
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);
        }
  
        // return 2nCn/(n+1)
        return res / (n + 1);
    }
  
    // Driver code
    public static void Main()
    {
        int n = 4;
        Console.WriteLine("Number of "
                  + "Dyck Paths is " +
                   countDyckPaths(n));
    }
}
  
// This code is contributed by anuj_67.

PHP

<?php
// PHP program to count 
// number of Dyck Paths
  
// Returns count Dyck 
// paths in n x n grid
function countDyckPaths( $n)
{
    // Compute value of 2nCn
    $res = 1;
    for ( $i = 0; $i < $n; ++$i)
    {
        $res *= (2 * $n - $i);
        $res /= ($i + 1);
    }
  
    // return 2nCn/(n+1)
    return $res / ($n + 1);
}
  
// Driver Code
$n = 4;
echo "Number of Dyck Paths is "
              countDyckPaths($n);
  
// This code is contributed by anuj_67.
?>


Output :

Number of Dyck Paths is 14

Exercise :

  1. Find number of sequences of 1 and -1 such that every sequence follows below constraints :
    a) The length of a sequence is 2n
    b) There are equal number of 1’s and -1’s, i.e., n 1’s, n -1s
    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.
  2. .

  3. Number of paths of length m + n from (m-1, 0) to (0, n-1) that are restricted to east and north steps.

This article is contributed by Aditya Chatterjee. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



My Personal Notes arrow_drop_up

Improved By : vt_m, Prasad_Kshirsagar



Article Tags :
Practice Tags :

Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.